Sub Topics

Anti derivative of function f is the function F whose derivative is function f. We can understand it by an equation as F'=f. This process is also known as anti differentiation. This term is related to the definite integrals by using the Functions of Calculus. It can be understand by an example as the function F(x)=x^{3}/3 is an anti derivative of the function f=x^{2} means x2 have indefinite number of anti Derivatives because the derivative of a constant is zero so x^{3}/3+3,x^{3/3}+78,x^{3}/3+453 and so on. By it we can understand that when the value of the constant is changed then the new anti derivative is obtained as F(x)=x^{3}/3+c here c is a arbitrary constant. The Applications of Antiderivatives is understand with an example of acceleration and velocity in physics as v=u+at ,where v and u are velocity and a is acceleration as anti derivative of this equation means Integration of the acceleration yields the velocity and a constant as: a=dv / dt + c and ∫_{t1}^{t2 }a(t) dt = v (t2) v (t1) .This same pattern is applied to all other parts of equation as Position ,velocity, acceleration and so on .These are some essential application of anti derivative. These anti derivative are important to compute the definite integrals and using the formulas of calculus. If a function F is anti derivative of function f then it is shown as ∫ f(x) dx = F(b)F(a). We also define the anti derivative F of a function of f that have interval as then each anti derivative is different from other because of the value of the constant such as G (x) = F (x) + c here constant c has the different values on that basis the value of anti derivatives so c is known as arbitrary constant of integration. We can find the anti derivative of a function F(a) if F' (a) = 4 – 3 ( 1 + a^{2}) ^{1} and f (1) = 0. The general anti derivative of function is F ( a) = 4a  3 arc tan ( a ) + c For an arbitrary constant c the derivative of arc tan ( x) is ( 1 + x^{2 }) . For find the specific Antiderivative, we evaluate: F(1)=4  3 arc tan ( 1 ) + c = 0 so c = 3 arc tan(1) 4 = π / 4  4 Thus we have F (a ) =4a  3 arc tan( a ) + π / 4  4 if we find the G ( a) then G ' (a) = a ( 6 + 5a) ^{1/2} and G ( 1 ) = 10. So this is the way of generate the anti derivative of the Functions. We can take another example for the finding the anti derivative as a ball dropped from a roof and hits the ground at 120 fts / s .What is the height of the roof ? (Note that the acceleration due to gravity is a(T) = 32 ft / s^{2}) Solution of it as follows: we have a (t) =  32 ft / s^{2} v (t) = 32 t + c here c is an arbitrary constant ,to find c we evaluate getting v (0) = c = 0 But the initial velocity is 0.Thus we have v(t)=32t. Next we calculate height d (t) = 16 t^{2 }+ c Note that when t = 0 we have d ( 0 ) = c ,and so c is the height of the roof . In particular for getting the answer, we need to decide c . As we know that when the ball hits the ground, we have v (t) = 32t = 120 And so it follows that the ball hits the ground when t 120 / 32 = 15 / 4 . Since the ball gets the height 0 when it hits the ground, we have d ( 15 / 4 ) = 16.225 / 16 + c = 0 and so c = 225. So the height of the roof is 250 ft. For defining the application of anti derivative we take some more example of it: The most general form of anti derivative as follows f(a)=a^{2} The most general anti derivative is F(a) =a^{3}+c for an arbitrary constant c. g(a) = 5 – 4a^{3 }+ 2a^{6 }/ a When initially we solve it we get g(a) = 5 / a^{6 } 4 / a^{3 }+ 2 = 5a^{6 –} 4a^{3 }+ 2 Anti derivative of the most general type is g(a) = a^{5 }+ 2a^{2 }+ 2a + c For an arbitrary constant c. So these are some examples that show the application of anti derivative. There is also some other application of anti derivatives as Anti derivatives and differential equation: anti derivative can be used in finding the general solution of the differential equation dp / dq = f ( q ). Another example of application of anti derivative is as: A brick is dropped from an apartment whose window is 20 meters higher from the ground. Suppose that the acceleration of the brick is −16 m / sec^{2 }then we have to calculate some of the answers for the following questions: (1) The velocity of the brick based on the function of time t; (2) The position d ( t) ; (3) Time of the brick when hits the ground and the corresponding velocity. Solution of the above questions as follows :(1) The motion is described by the equation dv / dt = −16 m / sec^{2} . The general solution of this equation is given by v (t) = −16 t + C. Since the initial velocity is v (0) = 0, we find C = 0 and therefore v ( t ) = −16 t. (2) The solution to the equation dh / dt= −16 the anti derivative h ( t) = −8 t^{2 }+ C . But h (0) is the initial distance of the brick from the ground, that is h (0) = 20 . Thus, 20 = C and therefore h (t) = −8 t^{2} + 20. (3) The brick hits the ground when h (t) = 0 that is, −8 t^{2} + 20 = 0. Solving for t > 0 we find t ≈ 1.6 sec. At that time, v = −16 (1.6) = −25.6 m / sec. (The velocity is negative because we are decide up as positive and down as negative.) We can take another example of anti derivative as we have a function ∫ cos (3a) da=? So as the solution of this function as the derivative of the sine is cosine so anti derivative of cosine is sine. For solution of it we can take the bottom up approach that the anti derivative of the function most probably is sin (3x).So if we take the derivative sin (3x) then what do: d / da sin (3a)=cos (3a) d/ da (3a) = 3 cos (3a) So the anti derivative of it is almost sin (3x) but there is some other factor of 3 which comes at differentiation. It is because of Chain Rule of differentiation. Then we calculate little bit of changing function of sine as: d / da sin (3a) / 3 = 1 /3 cos (3a) d / da (3a)=1 / 3 .3 cos (3a) = cos (3a). So the derivative of sin (3a) / 3 equals to cos (3a), that is as follows: ∫cos (3a) da= sin(3a)/3 + c, The problem of initial value that is defined by an example as da / db =4x32x+1 and b=3 when a=0 then find b as a function of a. b=∫ [4a^{3} 2a + 1] da =4 ∫a^{3} da 2 ∫a da + ∫ da, c = 4 . a^{4}/ 4  2.a^{2}/2 + a + c, y = a^{4 } a^{2 }+ a + c, Initial condition that is b=3 when a=0 we use for solving the constant of integration: b = a^{4 } a^{2 }+ a + c. 3=0^{4 }0^{2 }+0 +c 3=c so, b=a^{4 } a^{2 }+ a +3. 
Area A = _{p}∫^{q} g (x) dx,
For the area problems some properties of definite integrals is here:
If a function g (x) is defined at x = p then,
_{p}∫^{p} g (x) dx = 0,
If the function g (x) is defined on the interval [p, q], then
_{p}∫^{q} g (x) dx =  _{p}∫^{q} g (x) dx,
If the function g (x) ≥ 0 on the interval [p, q] then the area,
Area A = + _{p}∫^{q} g (x) dx,
If the function g (x) ≤ 0 on the interval [p, q] then the area is:
Area A =  _{p}∫^{q} g (x) dx,
If there are two Functions f (x) and g (x) are continuous on the interval [p,q] and given that
g (x) ≤ f (x) for all x in the interval [p, q], then the area of the region bounded by the graphs of function f and g and the vertical lines x = and x = q is:
Area A = _{p}∫^{q} [f (x)  g (x)] dx (1),
Now if we consider the two Functions x = f (y) and g (y) on the interval [p,q] and its given that g (y) ≤ f (y), then the enclosed area in this case is
Area A = _{p}∫^{q} [f (y)  g (y)] dy (2)
Now notice both the equations. Almost both the curves of both the equations are same. So it can be said that the area is the difference between the larger function to the smaller function.
For the first equation it can be written:
A = _{p}∫^{q} (upper function) – (lower function) dx; p ≤ x ≤ q
For the second equation we can use the formula
A = _{a}∫^{b} (right function) – (left function) dy; a ≤ y ≤ b
To find the area of the region bounded by the curves these formulas are helpful. When these formulas are used make sure that the consideration of the larger and smaller function should be correct because it will affect the whole question.
If we consider a function g (x) = 1 over the interval [a,b] then,
_{p}∫^{q} g (x) dx = _{p}∫^{q} 1 dx = x_{a}^{b} = b – a.
Here the integral of the function g (x) = 1 is nothing but just the length of the interval [a, b]. This can be obtained to the area of a Rectangle having height 1 and length (b – a) but here it's interpreted as the length of the interval [a,b].
The same trick is also applicable for the double Integration. The integral of a function g (x,y) over a region D can be interpreted as the volume covered under the surface z= g (x,y) over the region D. If the function is considered g (x,y) = 1 over the region D then the above trick can be applied and this is called as the area of the region D. If A is the area of the region D then it can also be written as:
Area A = ∬_{D} dA.
In Calculus when two curves intersects each other then area is also intersected by these two curves. Suppose we have two curve where first curve is f(x) and second curve is g(x) ,so area between two curve is A = f(x) – g(x). That is area between two curves if we take the interval between to two curve is [a,b] then formula would below:
A = _{a}∫^{b} (f(x) – g(x)) dx (where we that assume f(x) ≥ g(x))
This situation can be created two case first cases: first upper curve and second is lower curve. In this situation formula will create like that A = _{a}∫^{b} ((upper curve)(lower curve)) dx
Where interval a≤ x ≤b
Now second case: first right curve and second is left curve
So in this case formula looks like that A = _{c}∫^{d} ((right curve)(left curve)) dx
Where interval a≤ x ≤b
So that was the situation of curve
Now we see that how to area can be bound: area can be bound in two cases which are given below.
Case 1: If area bounded by x –axis then limits are use for x and curves are function of x.
Area A = _{a}∫^{b}(f(x)g(x))dx where interval a≤ x ≤b
Case 2: if area bounded by yaxis then limits are use for y and the curves are function of y.
Area A = _{c}∫^{d}(f(y)g(y)) dy where interval c≤ y ≤d .
Let’s see the Area between curves based examples.
Example 1: Find the area bounds between the curve where first curve x = y and second curve is
x = √y and interval is [1 ,1]?
Solution: Use the given steps to solve such type of problems:
Step 1: In this step first we would solve the both curve and find the interval between these curve.
So write the both curve first curve x = y,
Second curve x = √y
In this problem we have given interval so we do not find the interval we can direct used in Integration
Step 2: In this step we are integrated the both curve with interval
We know Area between curves A = _{a}∫^{b} (f(y) – g(y)) dy ,
f(y) = y,
g(y) = √y ,
Area of curve A = _{1}∫^{1} (y) (√y)dy,
A = [(y)^{2}/2 – y^{3/2}/(3/2)]_{1}^{1}
A = [(1)^{2}/2 – 1^{3/2}/(3/2)]  [(1)^{2}/2 – (1)^{3/2}/(3/2)],
A = [1/2 2/3 ] – [1/2 + 2/3] = [1/2 2/3  1/2 2/3] = 4/3,
A =4/3,
Now we got the Area of curve A = 4/3.
Example 4: Find the area bounds between the curve where first curve y = x +2 and second curve is y = x^{2}?
Solution: Step 1: In this step first we would solve the both curve and find the interval between these curve.
So write the both curve first curve y = x +2,
Second curve y = x^{2},
So we simplify both curve x+2 =x^{2},
x^{2}  x 2 =0,
x^{2} 2x +x 2 = 0,
x(x2) +1(x2)=0,
(x2)(x+1)=0,
x =2 , x=1.
After solving we got the points first Point is x =1, and another is 2 and these points are worked as interval [1,2]
Step 2: In this step we are integrated the both curve with interval
We know Area between curves A = _{a}∫^{b} (f(x) – g(x)) dx ,
f(x) = x +2,
g(x) = x^{2} ,
Area of curve A = _{1}∫^{2} (x+2) (x^{2})dx,
A = [(x^{2}/2 +2x) –(x^{3}/3)]_{1}^{2}
A = [(2^{2}/2 +2x2) –(2^{3}/3)]  [(1/2 2) – (1)^{3}/3],
A = [4/2 +4 – 8/3][1/2 2 + 1/3],
A = [6 8/3] – [5/2 1/3],
A = [10/3] – [(151)/6] = 10/3 +16/6 = (20 16) /16 = 4/16,
A = ¼,
A = ¼,
Now we got the Area of curve A = 1/4.
If a plane area is revolved about a fixed line in its own plane then the body or Solid that is formed by the revolution of the plane area is called as solid of revolutions and the fixed line about which the area revolves is known as the axis of revolution, it can be used to find the area of the region enclosed by the curves.
In other words we can say that, if a region is revolved about a line then solid formed is known as a solid of revolution. The solid is generated by the region and the axis of revolution is the line about which the revolution takes place.
The volume of this solid is given by
V = _{p}∫^{q} A (x) dx,
V = _{p}∫^{q} A (y) dy,
Here A (x) and A (y) is the cross sectional area of the solid.
The formula for the area of the ring can be written as
Area A = [(outer radius)^{2} – (inner radius)^{2}].
These radii are dependent upon the Functions given and the Axis of Rotation.
In mathematics the volume of a solid of revolution can be obtained by rotating a Plane Curve around some Straight Line that lies on the same plane. Suppose that the curve doesn’t cross the axis then the volume of the solid is equal to the length of the Circle that is described by the centroid multiplied by the area of the figure. A disk can be considered as a three dimensional volume element of a solid of revolution and this element can be created by rotating a Line Segment of the length ‘w’ around the given axis located ‘r’ units away so that a cylindrical volume of π / r^{2}w units is enclosed by the disks.
The general methods to find the volume of a solid of revolution are the disk method and the shell method of Integration. Besides that cross section method and washer methods are also used. The first step to solve the area or volume related problems with the help of disk method is to draw the curve first then calculate the enclosed area that revolved about the axis of revolution then calculate the volume of disk shaped slice of the solid having thickness ‘δx’ or a cylindrical shell of width ‘δx’ and then the final step is to find the limiting sum of these volumes as ‘δx’ approaches to the value zero and this value can be found by evaluating a suitable integral. Besides that washer method and cross section method is also used to calculate the volume of the solid.
Now suppose the region between the graph of a continuous function y = f (x) and the x axis from x = p to x = q then the volume of the solid can be given as
Volume V = _{p}∫^{q} π (radius)^{2} dx = _{p}∫^{q} π (f (x))^{2} dx.
If the equation of the developing curve is given in the polar coordinates assume
r = f (Θ) and if the curve revolves about the x axis then the generated volume is
Volume V = π _{a}∫^{b} (y)^{2} dx,
Volume V = π _{Θ1}∫ ^{Θ2} (y)^{2} (dx / dΘ) . dΘ,
or
Volume V = π _{Θ1}∫ ^{Θ2} (x)^{2} (dy / dΘ) . dΘ.
Where Θ_{1} and Θ_{2} are the values of Θ at the points x = a and x = b.
Now let's assume that x = r cos Θ and y = r sin Θ. Then the enclosed volume is rewritten as
Volume V = π _{Θ1}∫ ^{Θ2} (r)^{2} sin^{2}Θ (d / dΘ) . r cos Θ dΘ.
Where the value of ‘r’ in terms of ‘Θ’ must be substituted from the equation of the curve given in the problem.
The volume of the solid generated by the revolution of the area enclosed by the curve r = g (Θ) and the radii vectors given by Θ = Θ_{1} and Θ = Θ_{2}.
About the initial line Θ = 0 which always refers the x axis is
Volume V = _{Θ1}∫ ^{Θ2} ( 2/3 ) (π r^{2}) sinΘ dΘ.
About the line π = π/2 which refers the y axis is
Volume V = _{Θ1}∫ ^{Θ2} ( 2/3 ) (π r^{3}) cosΘ dΘ.
About the line Θ = γ, the volume of revolution is
Volume V = _{Θ1}∫ ^{Θ2} ( 2/3 ) (π r^{3}) sin (Θ  γ) dΘ.
As illustrated above in each of the above three formulae the value of ‘r’ must be in terms of ‘Θ’ that is substituted from the equation of the given curve.
If two Functions f (x) and g (x) are such that f (x) ≥ g (x) for all ‘x’ in the interval [p,q], then the volume of the solid generated by revolving two functions around the x axis the region bounded by the graphs of function f (x) and h (x) between the values of x = p and x = q is given by the integral,
Volume V = _{p}∫^{q} h [f (x)^{2} – g (x)^{2}] dx.
if ‘h’ is a function such that x = h (y) ≥ 0 for all y in the interval [p, q], then the volume of the solid generated by revolution, around the y axis, the region enclosed by the graph of ‘h’, the y axis (x = 0) and the horizontal lines y = p and y = q can be written by the following integral
Volume V = _{p}∫^{q} [h (y)]^{2} dy.
If two functions ‘z’ and ‘w’ are such that f (y) ≥ g (y) for all ‘y’ in the interval [p, q] then the volume of the solid generated by the revolution of the region enclosed by the graphs of ‘f’ and ‘g’ between y = p and y = q around the y axis is given by the following integral
Volume V = _{p}∫^{q} h [f (y)^{2} – g (y)^{2}] dy.
If the cross section is not a disk but is a washer then the area of the washer must be written first by subtracting the area of the inner cross section from the area of the outer cross section and make sure that the rotation is either about the x axis or y axis i.e.
A = π (outer radius)^{2} – π (inner radius)^{2}
The washer is formed by revolving a Rectangle about an axis. This is used because disc method cannot be used for the solids having a hole. In this situation Easher method is preferred instead of disc method.
The volume of the solids formed by rotating the area between the curves of function f (x) and g (x) and the lines x = p and x = q about the y axis is given by
Volume V = 2 π _{p}∫^{q} [f (x) – g (x)]dx
As explained above if r and R are the inner and outer radii of the washer and w is the width iof the washer then the volume of the washer can be written as
Volume V = π (R^{2} – r^{2}) w
The volume of the solid of revolution generated by washer can be written as
Volume V = π _{p}∫^{q} (R(x)^{2} – r(x)^{2}) dx
The integral involving the inner radius refers the volume of the hole and is subtracted from the integral having the outer radius.
The cross section method is used for the solid of any shape because disc method is used for a circular cross section of area πr^{2} . The volume by cross section method is
Volume V = _{p}∫^{q} f (x) dx,
or
Volume V = _{p}∫^{q} f (y) dy.
The cylinder method to find the volume of revolution of solids is used when the slice that was drawn is parallel to the axis of revolution means integrating perpendicular to the axis of revolution. It is also known as shell method. The volume of the shell can be written as
Volume V = π (d + w/2)^{2} h  π (d + w/2)^{2} h,
Volume V = 2π dhw,
Volume V = 2π (average radius) (height) (Thickness).
Where ‘w’ is the width of the rectangle, ‘h’ is the height of the rectangle and ‘d’ is the distance between the axis of revolution and the center of the rectangle. When this rectangle is revolved about its axis of revolution this generates a cylinder shell of width ‘w’.
For horizontal axis of revolution the Volume is
Volume V = 2π _{p}∫^{q} f (y) h )(y) dy.
Another theorem about the volume of solid of revolution states that
“If a closed plane curve revolves about a straight line in its plane which does not intersect it, then the volume of the ring that is obtained by the revolution would equal to the area of the region enclosed by the curve multiplied by the length of the path described by the centroid of the region.
This is how we find volumes of solid of revolutions.
Area of region bounded by polar curve, in this curve are lies between an interval [a ,b] and function is continuous at this interval, and f(x) ≥ 0 then the area of the region between the function and the xaxis is given by Area = _{a}∫^{b} f(x) dx.
If yaxis curve f(y) is continuous and another yaxis curve g(y) is also continuous, where g(y) ≤ f(y), ‘y’ is the subset of interval [a, b] then area of the region bounded by curve is A =_{a}∫^{b} (f(y) –g(y)) dx.
If a curve f(x) lies on xaxis and another curve g(y) lies on yaxis and both are subset of interval [a, b], then area of region with interval [a,b] is given by
A = _{a}∫^{b} (f(x) –g(y)) dx ,
If we use the common origin and take initial positive xaxis , then rectangular polar coordinate (x, y) can be represented as x = r cos ѳ and y = r sin ѳ, here x ,y can be converted to the Polar Coordinates r and ѳ by this equation.
x^{2} + y^{2} = r^{2},
r= √(x^{2} + y^{2}) (by pythagoras theorem),
r= polar equation f(ѳ),so f(r, ѳ) =0,
r^{2}cos^{2}ѳ + r^{2}sin^{2}ѳ = r^{2},
r =√(r^{2} cos^{2} ѳ + r^{2} sin^{2} ѳ),
y/x = tan ѳ (tan ѳ is angle between xaxis and yaxis).
In this equation if ‘r’ is negative then cos (ѳ +180) = cos ѳ, sin (ѳ +180) = sin ѳ the above equations are used to find a Cartesian equation, using above equation we can find Area A =_{a}∫^{b} ½ r^{2} dѳ.
Suppose we a have curve and this curve lies on closed interval [a, b] so area of region bounded by curve A = 1/2 _{a}∫^{b} f(ѳ)^{2} dѳ, where angle ‘ѳ’ of the curve lies between a<ѳ< b.
Above equation is also called polar equation.
Now let’s see area of region bounded by polar curves based example:
Example: Find the area of the inner loop of r= 1+2cos ѳ where interval is [π/2, π/2]?
Solution: We need to follow the steps shown below to understand the concept
Step 1: In first step we will write the given Cartesian equation.
r = 1+2cos ѳ with interval [π/2, π/2 ]
Step 2: In this step we will write the formula of area bounded with given interval
A = _{a}∫^{b} ½ r^{2} dѳ,
A = _{π/2}∫^{π/2} ½ r^{2} dѳ,
A = _{π/2}∫^{π/2} ½ (1+2 cos ѳ)^{2}dѳ,
A = _{π/2}∫^{π/2 }½ (1 + 4 cos ѳ + 4cos^{2} ѳ)dѳ,
A = _{π/2}∫^{π/2} 1/2 dѳ + _{π/2}∫^{π/2} 2 cos ѳ dѳ + _{π/2}∫^{π/2} 2cos^{2}ѳ dѳ,
A = [ 1/2 ѳ] _{ π/2}^{π/2} +[2 sin ѳ] _{π/2}^{π/2} + 2_{π/2}∫^{π/2} ( ѳ/2 + ¼ cos 2ѳ ),
A = [ 1/2 ѳ] _{π/2}^{π/2} +[2 sin ѳ] _{π/2}^{π/2} + _{π/2}∫^{π/2} ( ѳ + 1/2 cos 2ѳ ),
A = [π/4 + π/4 ] +2[sin π/2 – sin (π/2 )] + [π/2  π/2] + 1/2[cos π – cos (π)],
A = [π/4 + π/4 ] +2[sin π/2 – sin (π/2 )] + [π/2 + π/2] + 1/2[cos π – cos (π)],
A = π/2 + 2[1 +1] + [π] + ½ [1 +1],
A = π/2 +4 + π +1 = 3π/2 + 5 = (3π +10)/2,
So, the area of polar curve is A =(3π +10)/2.
Let’s see another example of area region bounded polar curve:
Example 2: Find the area of Polar curve where curves are represented by f(ѳ) = 1+sin ѳ ,g(ѳ) = 3+ cos ѳ with interval [π/2 , π/2] ?
Solution: In this problem we have two curves with interval, so we will use the above formula.
Step 1: First we will write the given polar curve with interval then we check the continuity of the curve
First curve f(ѳ) = 1+sin ѳ (this curve is a continuous curve with interval [π/2 ,π/2]),
Second curve g(ѳ) = 3+ cos ѳ (continuous curve with interval [π/2 , π/2]),
Step 2: In this step we will integrate the given curve and solve with given interval,
First we integrate the first curve: _{π/2}∫ ^{π/2}f(ѳ) =_{π/2}∫ ^{π/2} (1+sin ѳ) dѳ,
_{π/2}∫ ^{π/2}f(ѳ)= [ѳ] _{π/2} ^{π/2 }  cos ѳ _{π/2}^{π/2},
: [ π/2 + π/2]  [cos(π/2) – cos (π/2)],
_{π/2}∫ ^{π/2} f(ѳ)= π+ 0 = π.
Now we will integrate the second curve: _{π/2}∫ ^{π/2}g(ѳ) = _{π/2}∫ ^{π/2} (3+ cos ѳ) dѳ,
_{π/2}∫ ^{π/2} g(ѳ) = [3ѳ]_{π/2} ^{π/2 } + [sin ѳ] _{π/2} ^{π/2},
[3π/2 + 3π/2] + [sin (π/2) – sin (π/2)],
_{π/2}∫ ^{π/2} g(ѳ) = [ 3 π ] + [1+1] = 3 π +2.
Step 3: In this step we will solve the first curve and second curve with interval .
Now the area of polar curve: A = _{π/2}∫ ^{π/2} f(ѳ) – g(ѳ) dѳ,
A = _{π/2}∫ ^{π/2} g(ѳ)  _{π/2}∫ ^{π/2} f(ѳ),
A = 3 π +2  π ,
A= 2 π + 2.
Therefore the area of polar curve is A = 2π + 2.
Example 3: Find the area of Polar curve where curves are represented as f(ѳ) = 2sin ѳ ,g(ѳ) = 3cos ѳ with interval [π/2< ѳ< π/2]?
Solution: In this problem we have two curves with interval, steps to solve this problem are shown below
Step 1: First we write the given polar curve with interval, then we will check the continuity of the curve
First curve f(ѳ) = 2sin ѳ (this curve is continuous curve with interval [π/2 , π/2])
Second curve g(ѳ) = 3cos ѳ (continuous curve with interval [π/2 , π/2])
Step 2: In this step we will integrate the given curve and solve with interval,
First we will integrate the first curve: _{π/2}∫ ^{π/2} f(ѳ) =_{π/2}∫ ^{π/2} (2sin ѳ) dѳ
_{π/2}∫ ^{π/2}f(ѳ)= [ 2cos ѳ ]_{π/2}^{π/2},
= [(2cos(π/2)) – (2cos (π/2))],
_{π/2}∫ ^{π/2} f(ѳ) = 0+ 0 = 0.
Now we will integrate the second curve: _{π/2}∫ ^{π/2} g(ѳ) = _{π/2}∫ ^{π/2} (3cos ѳ) dѳ
_{π/2}∫ ^{π/2} g(ѳ) = [3sin ѳ] _{π/2} ^{π/2},
= [3sin (π/2) – 3sin (π/2)],
_{π/2}∫ ^{π/2} g(ѳ) = [3 + 3] = 6.
Step 3: In this step we will solve the first curve and second curve with interval.
Now the area of polar curve: A = _{π/2}∫ ^{π/2} g(ѳ) – f(ѳ) dѳ,
A = _{π/2}∫ ^{π/2} g(ѳ)  _{π/2}∫ ^{π/2} f(ѳ),
A = 60 = 6.
Therefore area of polar curve is A = 4.
Example 4: Find the area of Polar curve where curves are represented by f(ѳ) = 2sin ѳ , g(ѳ) = 3 + sin ѳ, with interval [π/2, π/2]?
Solution: In this problem we have two curves with interval.
Step 1: First we will write the given polar curve with interval then we will check the continuity of the curve
First curve f(ѳ) = 2sin ѳ (this curve is a continuous curve with interval [π/2 , π/2])
Second curve g(ѳ) = 3 + sin ѳ (continuous curve with interval [π/2 , π/2])
Step 2: In this step we will integrate the given curve and solve with interval,
First we will integrate the first curve: _{π/2}∫ ^{π/2}f(ѳ) =_{π/2}∫ ^{π/2} (2sin ѳ) dѳ,
_{π/2}∫ ^{π/2}f(ѳ)= [ 2cos ѳ ]_{π/2}^{π/2},
= [(2cos(π/2)) – (2cos (π/2))],
_{π/2}∫ ^{π/2} f(ѳ)= 0 + 0 = 0.
Now we will solve the second curve: _{π/2}∫ ^{π/2} g(ѳ) = _{π/2}∫ ^{π/2} (3 +sin ѳ) dѳ
_{π/2}∫ ^{π/2} g(ѳ) = [ 3 ѳ  cos ѳ] _{π/2} ^{π/2} ,
= [3(π/2) – cos (π/2)] –[3(π/2) – cos (π/2)],
= [3π/2 +3 π/2] = 3π,
_{π/2}∫ ^{π/2} g(ѳ) = 3π.
Step 3: In this step we will solve the first curve and second curve with interval.
Now the area of polar curve: A = _{π/2}∫ ^{π/2} g(ѳ) – f(ѳ) dѳ,
A = _{π/2}∫ ^{π/2} g(ѳ)  _{π/2}∫ ^{π/2} f(ѳ),
A = 3π  0 = 3π.
Therefore area of polar curve is A = 3π.
This is how we can find Area of region bounded by parametrically defined or Polar Curves.
The topic "Arc length" is generally related to the geometries but sometimes we use it in the Calculus Geometry also. As the name suggests “Arc Length”, means a part of a curve in any region. To understand the arc length we first have to understand that what is arc and then arc length, so Arc is curved line made by two distinct points on the curve, and the arc length is the calculation of the distance of the arc or to find the length of the curved line which forms the arc on the curve. The length of the arc in the curve is greater than the Chord of the curve because the chord is the distance between two end points of the curve and arc is the length of the curve from one Point to the other point on the curve. For example, we have a Circle and the circle has its arc as ‘AB’ on the circle. Assume the length of the AB is 5 then it is read as “The length of the arc is 5”. We calculate the arc length with the help of some formulas in geometry and for other we use arc length Integration to get arc length of the body.
For instance, if the central angle of the circle or curve is A and radius is R then the arc length of the curve will be:
Arc length = 2 π R (A/360)
Where A is the central angle, R is the radius of the arc in circle, and π is the constant term and it is equal to 22/7, or 3.1414. In some other form of the formula if the central angle of the curve is in the radians units then we can write the formula for arc length of the circle as:
Arc length = R x A
Here again R is the radius of arc and A is the central angle of arc in radian unit. The arc length in degree unit and in radians unit are both same but the single difference is that the conversion of the degree quantity in to radian quantity.
Finding the arc length of a segment is the process of rectification of the curve. The arc length is always given by L and it may be either finite or infinite.
While finding the arc length integration we assume a part of the curve in the plane. Here in this curve a region [a, b] is taken and the curve is varying in the defined region.
Then the formula for the arc length on the interval [a, b] will be given by:
If the function y = f(x) (f(x) should be a continuous function), then arc length (L) of y = f(x) on interval [a, b] will be (L) = ∫a^{b} √(1 + f’(x)) dx,
If the curve is in the form of x(t), and y(t) then the arc length = ∫a^{b} √((dx/dt)^{2} + (dy/dt)^{2} ) dx, similarly if the curve is given in the Polar Coordinates then the arc length (L) will be = ∫a^{b} √((r)^{2} + (dr/dθ)) dθ,
Where θ is the angle and r = f(θ).
Velocity is defined as speed with direction. Speed is equals to the distance covered per unit time. Speed is defined in Scalar terms without direction and if we consider the direction, it is said to be the velocity of the object. Velocity and acceleration vectors of planar curves are formed in the 2 dimensional planes. Velocity always has direction associated with it, it is a vector quantity as direction is also associated with it. In mathematical terms, we can represent the velocity as,r = x I + y J,
It is called the radius vector or Position vector as it describes the position of any Point on the plane.
The position vector or radius vector is from the origin to the point (x, y).
Here, ‘I’ and ‘j’ are the unit vectors.
‘I’ and ‘j’ are unit vectors defined along ‘x’ and ‘y’ axis in the coordinate system.
Unit vectors represent the direction and have unit magnitude. It shows only direction with unit magnitude.
Radius vector is defined in the 2  d planes.
A 2 – d plane can be parametrically defined as,
x = x ( u ),
y= y ( u ),
Here, the Functions x (u) and y (u) are the Functions which are continuous in The Range ‘u’. They can be finite or infinite range.
The Plane Curve can be plotted on a 2 – d plane using the following equation,
r ( u ) = x (u ) i + y ( u ) j,
As we differentiate a normal Scalar Function, we can also differentiate a vector function.
For example,
The derivative of the above equation is,
d r /d u = ( d x/ d u ) .i + ( d y / d t ) . j,
Here, d r / d u ( u ) is a Tangent which is a vector with a direction.
It is the tangent to the above curve.
Here, ‘u’ is taken as a time variable, then d r (u) / d t is known as the velocity.
As we have discussed the radius vector, we can now discuss the velocity and acceleration of a moving point.
Let us take a point V (x, y) in 2 – d motion moving along the curve formed by the following equation,
x = x ( u ),
y = y ( u ),
Here, ‘u’ represents the time.
As we have discussed about the position vector already,
The position vector of a point from origin to a point is given by ,
r = x ( u ) i + y ( u ) j                        ( equation 1 )
position vector is also called as radius vector.
This is the vector equation of a point with unit vectors ‘I’ and ‘j’.
As we know, speed is determined by differentiating the distance with respect to time.
In terms of vectors, velocity can be determined by differentiating the same distance equation.
Differentiating once equation 1, we get the velocity of the moving particle.
V = d r / d t
‘r’ is the radius vector of the particle.
d r / d t = ( d x / d t) . i + ( d y / d t) . j                    ( equation 2 )
‘I’ and ‘j’ are unit vectors.
We know that a vector has two quantities,
1 ) Magnitude part known as scalar part.
2 ) Direction part known as vector part.
Velocity is a vector quantity. Hence, it must have the two parts, the magnitude part and the direction part.
The magnitude part of the velocity is represented by the mod symbol.
Hence,  v  = √ ( v ) . ( v ),
= √ ( v _{x} ^{2} + v _{y} ^{2} ).
Here, ‘x’ is the velocity calculated in the ‘x’ direction and ‘y’ is the velocity component in ‘y’ direction.
We can obtain ‘x’ and ‘y’ components of velocity individually by partially differentiating the equation of the particle.
As,
v ( x component ) = d x / d u
v ( y component ) = d y / d u
Here, we have differentiated the equation with respect to time variable ‘u’.
As ‘v’ is a vector, so we also have to find its second component.
The direction of velocity at the point is the direction of the tangent at that point at time ‘u’.
The acceleration of the point at time ‘u‘ is determined by differentiating the equation of distance twice.
In other words, acceleration is obtained by differentiating the velocity equation once.
Note that acceleration cannot be defined without direction.
But velocity has its magnitude part which can be defined without its direction, known a speed.
As we determined the velocity of the particle, we can also define the acceleration of that particle.
Acceleration= a = d v / d u = d ^{2} r / d t ^{2},
= d ^{2} x / d u^{2} . i + d ^{2} y / d u ^{2} . j.
Here, ‘I’ and ‘j’ have their usual meanings. They are the unit vectors which represent the direction at the time ‘u’.
We have differentiated the displacement equation twice to get the acceleration.
We are taking displacement in place of distance because we are talking about the vector quantities.
Displacement is the vector of distance as it has a direction associated with it.
We can now obtain the magnitude part of acceleration.
 a  = √ a . a,
In other words,
 a  = √ a _{x} ^{2} + a _{y} ^{2},
Where ‘a _{x}’ is the x – component of acceleration in the x – axis.
‘a _{y}’ is the y – component of acceleration in the y – axis.
We can partially differentiate the equation to obtain the ‘x’ and ‘y’ parts individually.
a _{x} = d ^{2} x / d t ^{2},
a _{y} = d ^{2} y / d t ^{2},
The direction is same as the direction of the velocity.
Let us take a trigonometric equation of the position of any particle.
r ( u ) = cos u i + sin u j + u k,
Here, we have taken the position of the particle in a 3 d plane.
Here, note that we have taken another unit vector ‘j’. It shows the direction towards z axis.
i , j , k are unit vectors with unit magnitude along x , y and z axis respectively.
As we have discussed earlier, the first order differentiation of the given particle equation will produce velocity and second order differentiation will give the acceleration of the particle.
Differentiating the equation, we get
v ( u ) =  sin u i + cos u j + k,
Again differentiating the equation, we get the acceleration
a ( u ) =  cos u i – sin u j + 0k,
Here, we have zero magnitude on the z axis.
Hence, the direction vectors for the acceleration are in x – y plane.
Using the Calculus, we now prove that the derivative of constant speed is zero.
Let us take example of a moving particle which is moving with a constant speed ‘v’ but its velocity is different from speed.
The direction vector is changing.
For now, don't think about the direction vector.
The magnitude of the particle is ‘v’.
As we studied earlier,
The magnitude of velocity is equals to
 v   v  =   v ^{2}  ,
= k,
Here, k is a constant as ‘v’ is a constant .
v .v = k                     ( equation 1 )
Differentiating with respect to time variable ‘u’, we get
d ( v . v ) / d u = d k / d u
Using multiplicative derivation rule, we can solve this equation as
= d v / d u . v + v . dv / d u,
= 0.
As ‘v’ is a constant and differentiation of a constant is zero.
It implies ,
2 ( v . d v / d t ) = 0,
Further solving,
v . d v / d u = 0                       ( equation 2 )
‘u’ is time variable.
We know that, differentiating the velocity component, we get acceleration.
Hence d v / d u = a (acceleration)
Using this in equation 2 , we get
v. a = 0,
Here ‘v’ and ‘a’ are velocity and acceleration vectors.
The dot product of ‘v’ and ‘a’ is equals to zero, therefore we can say that, if the magnitude of velocity is ‘v’ and it is constant, then according to the properties of vectors, the velocity vector and acceleration vectors are at 90 degrees to each other.
Let us understand the acceleration in terms of force.
According to the Newton’s second law of motion ,
F = m a
F = force applied
m = mass of the object
a = acceleration of the object
We can also calculate the instantaneous acceleration as it is equal to the rate of change of velocity at that particular instant.
Let us discuss a practical example which will clear the concept of vectors.
Example: A car is tracing a circular pattern on a ground of radius 100 m with a constant speed.
If the car takes 120 s to complete one revolution, evaluate the following.
1 ) The speed of the car?
2 ) The instantaneous velocity at the point M?
3 ) The instantaneous velocity at the point N ?
Solution 1 ) We have to calculate the speed of the car. As speed is a scalar quantity, we don't have to consider the direction vectors.
We will simply find the magnitude.
As we know,
Total distance covered in 1 revolution =circumference of the Circle
d = 2 ∏ r,
r = radius of circle = 100 m,
hence, d = 2 * 3.14 * 100,
d = 6 2 8 . 3 meters,
We have distance covered by car and time taken by the car. We can calculate the speed of the car.
Speed = distance travelled / time taken,
Speed = 628.3 / 120,
As time t = 120 s given,
Speed = 5.23 meters per second.
2) As we know that the car is moving at a constant speed.
So, at point ‘M’, the magnitude of velocity will be 5.23 meters per second.
 Velocity  = speed,
Now, we have to find the direction also, as velocity is a vector and a direction vector is associated with it.
The direction of the car at point ‘M’ will be the direction of the tangent at M towards west as shown in the diagram.
3) As we find the velocity of car at point ‘M’, we can find the velocity at point ‘N’.
Magnitude of velocity = speed = 5.23 meters per second.
Velocity is a vector, hence we have to find its direction also.
Now, we have to find the direction of car at point ‘N’.
The direction of a vector is decided by the tangent passing through that point.
The direction of the car will be towards the direction of the tangent at point ‘N’ in south direction.
This is how we find Velocity and acceleration vectors of Planar curves.
Other applications involving the use of integral of rates can be understood with the help of various examples from our day to day life. Some of the examples for better understanding of the implementation of the above property are:
We will start with a small example from ancient civilization: The art of modern pottery which is still around us from the time of Ancient civilization is an area where we can see the practical application of Integration of rates.
Example 1: The desired shape of the side of the pottery vase can be described by:
Velocity = 5.0 + 2 sin (hei. / 4) (0 < = hei. < = 8 ∏) ( Using the Riemann sums Property),
Where “ hei.“ is the height.
A base for the vase can be formed by placing it on the potter's wheel. How much clay can be added to the base to form their vase if the inside radius is always 1 inch less than the external radius. Here with this example it can be related with two things: First is the rate of change of radius and the second is the rate of change of height.
After last example we have a better idea about the importance of Riemann sums and its applications. It is not just any intellectual exercise, it is thus natural way to calculate mathematical and physical quantities that primarily seem to be irregular as a whole, but can be fragmented into regular pieces.
Now we calculate the values for the regular pieces using the known formulas and latter sum them to calculate the value of irregular as a whole. The observation from these problems are very tedious.
With the concepts of Calculus it becomes more predominant to get the exact answers for these problems with almost no efforts from our side.
Some more applications of integrals of rates are given below
A. Linear Motion Revisited: In many applications, the integral is defined to be the net change over time. The best example to understand this is distance traveling.
Example 2 Interpret Velocity function.
The velocity value is dx / dt = v(t) = t^{2 –} 8 / ( t + 1 )^{2} cm / sec, for a particle moving along a horizontal x – axis for 0 < = t < = 5. Describe the motion?
Solution: Solve graphically: The graph of ‘v’, from the figure, starts with v( 0 ) =  8, which we will assume by saying that the particle has an initial velocity of 8 cm / sec to the left. It slows at the rate of 1.25 sec, after which it moves to the right (v > 0) with the increasing value of speed. And latter we will come down to the Point where, the value of the velocity v( 5 ) is approximately equals to 24.8 cm / sec at the end.
Example 3: Calculating the Total distance Traveled
We have, the following Set of information:
(  t^{2} – 8/ (t+1)^{2}  , t, 0, 5 ) = 42.59.
Here we can constitute that by Integrating, the absolute value of velocity we get the total distance covered (Total Area with the time axis and the velocity curve).
B. General Strategy: To calculate the fragmenting net effect from the finite sums of estimated changes in the given subject than we can do, Cardiac output, air pollution and the volume, what has added into our knowledge, many of these sums as Riemann sums and latter express them to limits. There are two main advantages of this:
1. We can do the evaluation in one of the integrals to get the desired result in less time, than to crank out the estimate from its finite sum.
2. The integral by itself becomes a formula. That empowers us to solve the similar problems, without repeating the modeling step.
Example 5: Modeling different implications of Acceleration.
Let us assume a couple is traveling in a car with an initial velocity of 10 mph accelerates at the rate of a(t) = 4t mph for time Period of 16 seconds.
Find:
1. What is the speed of the car moving when 16 seconds are up ?
2. Distance covered by the car when 16 seconds are passed ?
Solution: Let us look at first the effect of acceleration on the car's velocity.
STEP 1: When the acceleration is constant,
Velocity change = acceleration * time applied
The net change in velocity for the time interval 0 < = t < = 16.
STEP 2: Write a Definite Integral. The limit of these sums is:
∫_{0}^{16} a(t) dt.
STEP 3: Now, Evaluate the integral.
Net velocity change = ∫_{0}^{16} 4 t dt = 2t^{2} _{0}^{16} = 512 mph.
We also know that the initial velocity is 10 mph, so the total will be: 512 + 10 = 522.
(2) Now, applying acceleration for any length at time ‘t’ adds to,
∫_{0}^{t} 4 u du mph , where ‘u’ is just a dummy variable here, therefore we get
v(t) = 5 + ∫_{0}^{t} 4u du = 5 + 1.2 u^{2} mph.
The total distance traveled from t = 0 to t = 16 sec is:
∫_{0}^{t}  v(t)  dt = ∫_{0}^{t} ( 5+ 4 t^{2} ) dt,
= [ 5t + 1.2t^{3}]_{0}^{8},
= 654.4.
C. Consumption over Time The integral works as natural tool to calculate the total accumulation and net change of more quantities in general. Integration is used to find the cumulative effect of varying rate of change and latter to integrate the term.
In daily life, motion is common thing to us, we walk, run and ride is a called as motion. General meaning of motion is change with respect to time like earth rotates once in every 24 hours.
In real life, when we change our Position in straight line, then this type of motion is called as a Straight Line motion and difference between one position to another position is called as a distance 
Distance = (position)_{2 } (position)_{1}
We can define motion with the help of frame of reference, which include all the axis means xaxis, yaxis and zaxis. So, when a particle moves along a xaxis with respect to time, than this kind of motion is called as a straight motion or we can say that a motion along a straight line.
Now we discuss what distance is:
Distance is path length of a particle like we assume a motion of car along a straight line.
We choose the xaxis such that it coincides with the path of the car’s motion and origin of axis as the Point from where the car started moving, point O(x = 0, t = 0) to point P(x,t). So, path length from O to P is called as a distance of car. In mathematical point of view distance is 
Distance = OP = P(x, t)  O(0, 0),
Distance is different from displacement because distance is calculated with certain directions and on other hand displacement is a Scalar quantity means directionless quantity.
Now we discuss what velocity is and what difference exists between velocity and speed:
We know motion is change with respect to time, but how fast motion can change with respect to time, we don't know. So, Scientist Newton introduced velocity concept, which tells us that distance per unit time, is called as velocity 
Velocity = total distance/total time,
Let us assume a car moves from a origin position O(0, 0) to particular position P(2, 3) in time t = 3 sec. So, for calculating how fast position of car is changing with respect to origin in 1 sec, we use velocity formula here 
Velocity = change in position / time interval = (√(3  0)^{2} + (2 – 0)^{2}) / (3 – 0) = (√ 9 + 4) / 3 = √ 13 / 3.
So, velocity of car is √ 13 / 3 km/sec.
Difference between velocity and speed: Speed is a dimensionless quantity means speed is equals to “total motion/total time” and other hand velocity is a quantity which has certain dimensions, so for calculating velocity, it is necessary to calculate distance first because distance is also a quantity which is calculated in certain directions. Therefore velocity is equals to total distance/total time. For calculating distance and velocity, we use Newton's laws of motion:
Newton's first law of motion says that a particle remains at rest until an external force acts on that particle. In mathematical point of view –
v = u + at …......................equation(1).
Where ‘u’ is initial velocity, ‘a’ is acceleration and ‘t’ is time.
Newton's second law of motion says that if there is an external force acting on particle than that force is directly proportional to its acceleration. In mathematical point of view 
S = ut + ½ at^{2 }….........................equation(2).
Here ‘S’ is total displacement, ‘u’ is initial speed, ‘a’ is acceleration and ‘t’ is total time.
Now we discuss Newton’s third law of motion, which says that there is an equal and opposite reaction to every action. In mathematical point of view 
v^{2 } u^{2 }= 2as ….........................equation(3).
Where ‘v’ is final speed, ‘u’ is initial speed, ‘a’ is acceleration and ‘s’ is total displacement
Using above laws we can solve velocity and distance problems involving motion along a line.