An integral with upper and lower limits is known as definite integral. We can represent a definite integral as shown below-
∫ Here p, q, and x are complex Numbers. We calculate the definite integrals in terms of indefinite integrals, let F be the indefinite integral for a continuous function f ( x ). Therefore it is calculated as-
∫ So when we evaluate the definite integral then first thing we do is calculate the indefinite integral and then apply limits. Our function should be continuous in the interval of Integration; this is most important property of a definite integral. We need to keep above things in mind to evaluate the definite integral- Let us take an example to understand the concept of definite integrals. The example is shown below-
Example 1) We have a integral ∫ Solution) So after integrating, we get-
F ( x ) = [ x
= [ ( 2 ) = 8 / 3 + 2, = 14 / 3,
Example 2) Find the definite integral of ∫ Solution) After integrating the above function we get-
=> F ( x ) = [ x4 + x3 ] Now applying limits, we get- => F ( 3 ) - F ( 2 ),
=> F ( 3 ) - F ( 2 ) = [ ( 3 ) => F ( 3 ) - F ( 2 ) = [ 108 – 27 ], = 84. So this is how we evaluate definite integrals. |

The Fundamental Theorem of Integral Calculus is one of the most interesting thing that one after having done with lot of differentiation and Integration concepts would love to read. This Theorem relates the concepts of differentiation and integration too closely, which helps the users to understand its concepts more predominantly.

If f is integrable on [ p , q ] and F is the actual anti derivative of f on [ p , q ] which are continuous on [ p , q ] then we can say :

∫_{p}^{q} f(x) dx = F( q ) – F( p ).

The result from the above equation proves the logic that we have been hearing for long time that integration is inverse of differentiation. The Definite Integral of a function f may be evaluated easily, under the condition that a function F whose derivative is f.

We can write this as,

F( x )∣_{p}^{q}

for the form F ( q ) – F( p )

∫_{p}^{q} f(x) dx = F( x )∣_{p}^{q}

Now let’s move to some real examples to understand the nature of these questions:

Example 1

F (z) = 1/3 z^{3}

is an anti-derivative of the function f(z) = z^{2},

∫_{0}^{1} z^{2} dx = 1/3 z^{3} ∣_{0}^{1 P} =1/3 – 0 = 1/3

Thus the area under the Parabola which is y=x^{2} and the above mentioned interval [0,1] on the base axis (x-axis) is exactly 1/3.

Example 2

Since, F(z) = 1/3 z^{3} +z is an anti derivative of g(z) = z^{2P} +1

∫_{-1}^{2} ( z^{2} + 1 )dz = ( 1 / 3 z^{3} +z) ∣_{-1}^{2} = ( 8 / 3 +2 ) - ( -1 / 3 -1 ) = 6

Example 3

Since, F(z) = 4 / 3 z^{3} - ½ z^{2} +2z.

This equation is an anti-derivative of the equation 4z^{2} – 4z +2,

Now, moving to the integration:

∫_{-2}^{3} ( 4z^{2} – z +2 ) dz = ( 4 / 3z^{3} -1/2z^{2} +2z ) |_{-2}^{3}

(108 / 3 – 9 / 2 +6) – (32 / 3 - 2- 4),

= 325 / 6.

Example 4

F(p) = 2 / 3 p^{3/2}

Is an anti derivative of f(p) = √p, we have

∫_{0}^{4} √p dp = 2 / 3 p^{3/2} |_{0}^{4} = 16 / 3 – 0 =16 / 3.

Example 5

If A is the Area under one arch of the curve y = cos(x), than we can say:

A = ∫_{0}^{∏} cos (x) dx.

Since F(x) = sin(x) is an anti derivative of f(x)= sin(x), we have

A= ∫_{0}^{∏} cos( x ) dx = sin( x )∣_{0}^{∏} = sin( ∏ ) - sin ( 0 ) = 0 – 0,

= 0.

As we can see from the above examples that the fundamentals of integral calculus , supplements us with the power to evaluate definite integrals exactly. Thus to evaluate this we must evaluate the anti Derivatives first which we are integrating.

The Riemann integral was developed by Bernhard Riemann. So the name of formula is Riemann integral on the bases of developer name. Bernhard Riemann gives the definition of the integral in Calculus where the function has interval. This formula is mostly applicable in the practical purpose and it is a rare case that we can apply it on theoretical problems. We can also evaluate the Riemann integral in calculus by using fundamental theorem. In the field of engineering and physics we use the Riemann Definite Integral formula. There is other type of integral that is Lebesgue integral but this type of integral is not applied on advanced mathematics. The Jordan measure is used to drive the Riemann integral in the calculus by taking the limit of Riemann sum. Mathematically we can express the interval as, if x and y are the interval of the function then we can write this as:

∫_{x}^{y} f(s) ds.

This is basic idea of the Riemann integral at any given interval of given function. In Riemann integral the limit of sum we find the area of under the any curve. The function f is positive or negatives for the integral. The area which is above the x-axis is positive and negative for the below of x-axis. The Riemann integral equations are given below,

∫_{x}^{y}f(a)da = lim_{max∆ak→0 }^{n}∑_{k=1} f(a_{k})∆a_{k},

∫∫f (a,b) dX = lim_{max∆Xk→0 }^{n}∑_{k=1} f(a_{k+}b_{k})∆X_{k},

∫∫∫f (a, b, c)dZ= lim_{max∆Zk→0 }^{n}∑_{k=1} f(a_{k+}b_{k+}c_{k})∆Z_{k},

Where x and y are the interval of the given function and a_{k }, b_{k }, c_{k } are the arbitrary points. The max ∆a_{k} is known as mesh size. ∆a, ∆X, ∆Z are the subinterval. Now the Riemann sum formula is,

^{k-1 }∑_{i=0}f(t_{i})(z_{i+1}-z_{i})

This is the formula for real valued function which has interval (x, y) with partition Z_{0}, .. ,Z_{n.}

=> ∑_{k=1}^{n} [(7 * (k^{2}/n^{2})]* 1/n + ∑_{k=1}^{n} [1]* 1/n

Now we discuss applications of Riemann sum to integral: Riemann sum integral is used for calculating area of calculating the approximation area under certain curve. We use following steps to calculate approximation area under certain curve -

Step 1: For calculating the approximation area, first of all we analyze the equation. In this process we analyze that equation is first order or second order or n order.

Step 2: After analysis process, we break the equation into certain intervals like we break equation in lower bound limit and upper bound limit.

Step 3: after these two steps, we perform Riemann sum integral.

Now we take some examples to understand the steps of _{Riemann Sum Integral method -}

Example 1: Find the approximation area of given ∫_{0}^{1}_{ (7x}^{2} _{+ 1) dx = ?}

Solution:

Step 1: Given equation is 7x^{2} _{+ 1 , which is a second order equation.}

Step 2: Now we break the equation into certain intervals:

Because here given intervals are 0 to 1 So, we divide interval between 0 to 1 into n parts -

_{0 – 1/n = -1/n}

_{Step 3: Now we put x = x/n in above equation}

7 * (x/n)^{2} + 1 => 7*x^{2}/n^{2} + 1 ….......equation(1)

_{Now we perform Riemann sum on this equation -}

S = ∑_{k=1}^{n} f(x) * 1/n

=> ∑_{k=1}^{n} [(7 * (k^{2}/n^{2}) + 1] * 1/n

Now we divide sum part into two parts

=> ∑_{k=1}^{n} [(7 * (k^{2}/n^{3})] + ∑_{k=1}^{n} 1/n

=> (7/n^{3})*∑_{k=1}^{n} (k^{2}) + (1/n) * ∑_{k=1}^{n} (1) …......equation(2)

As we all know that ∑_{k=1}^{n} (k^{2}) = n(n+1)(2n+1)/6 and ∑_{k=1}^{n} (1) = n

So, we put these values in equation(2)

=> (7/n^{3}) * n(n+1)(2n+1)/6 + (1/n) * n

=> (7/n^{3}) * n^{3}(1+1/n)(2+1/n)/6 + (1)

=> 7 * (1 + 1/n)(2 + 1/n) /6 + 1

=> 14/6 + 1

=> 2.33 + 1

=> 3.33

So, approximation area of above equation ∫_{0}^{1}_{ (7x}^{2} _{+ 1) dx = 3.33 ANSWER}

_{Example 2: Find the approximation area of given} ∫_{0}^{1}_{ (4x}^{2}_{ – 2x + 3) dx = ?}

_{Solution:}

_{Step 1: given equation is} _{4x}^{2}_{ – 2x + 3} _{, which is a second order equation.}

_{Step 2: now we break the equation into certain intervals:}

_{Because here given intervals are 0 to 1 So, we divide interval between 0 to 1 into n parts -}

_{0 – 1/n = -1/n}

_{Step 3: Now we put x = x/n in above equation}

4 * (x/n)^{2} – 2 * (x/n) + 1 => 4*x^{2}/n^{2} – 2*(x/n) + 1 ….......equation(1)

_{Now we perform Riemann sum on this equation -}

S = ∑_{k=1}^{n} f(x) * 1/n

=> ∑_{k=1}^{n} [(4 * (k^{2}/n^{2}) – 2*(k/n) + 1] * 1/n

Now we divide sum part into two parts

=> ∑_{k=1}^{n} [(4 * (k^{2}/n^{2})]* 1/n - ∑_{k=1}^{n} [(2 * (k/n)]* 1/n + ∑_{k=1}^{n} [1]* 1/n

=> ∑_{k=1}^{n} [(4 * (k^{2}/n^{3})] - ∑_{k=1}^{n} [(2 * (k/n^{2}) + ∑_{k=1}^{n} 1/n

=> (4/n^{3})*∑_{k=1}^{n} (k^{2}) - (2/n^{2}) * ∑_{k=1}^{n} (k) + 1/n * ∑_{k=1}^{n} (1) …......equation(2)

as we all know that ∑_{k=1}^{n} (k^{2}) = n(n+1)(2n+1)/6, ∑_{k=1}^{n} (k) = n(n+1)/2 and ∑_{k=1}^{n} (1) = n

So, we put these values in equation(2)

=> (4/n^{3}) * n(n+1)(2n+1)/6 - (2/n^{2})* n(n+1)/2 + (1/n) * n

=> (7/n^{3}) * n^{3}(1+1/n)(2+1/n)/6 - (2/n^{2})* n^{2}(1+1/n)/2 + (1)

=> 7 * (1 + 1/n)(2 + 1/n) /6 – (1 + 1/n) + 1

=> 14/6 – 1 + 1

=> 2.33

=> 2.33

So, approximation area of above equation ∫_{0}^{1}_{ (4x}^{2}_{ – 2x + 3) dx = 2.33.}

These are examples which shows procedure of calculating approximate area of given equation under certain intervals and by using of Riemann sum integral method we can easily calculate Integration of any curve under certain interval.

Find the area of the region bounded by interval in Integration is one of the important topics of integration Calculus. In calculus when two Functions are intersect each other and the area lie between the intervals then it means the area is bounded by interval. Suppose we have two function if we take the interval between to two function is [a, b] then formula of area of the region will be given as:

Area of the region bounded A = _{a}∫^{b} (f(x) – g(x)) dx (where we that assume f(x) function ≥ g(x) function)

The area of region create two case first case: First upper function and second is lower function

In this situation formula will create like that A = _{a}∫^{b} ((upper function)-(lower function)) dx

Where interval a≤ x ≤b

Now second case: first right function and second is left function

So in this case formula looks like that A = _{a}∫^{b} ((right function)-(left function)) dx

Where interval a≤ x ≤b

So that is the situation of area of region bounded region.

Area can be bounded can be represent in two cases which are given below.

Case 1: If area bounded by x –axis then limits are use for x function and function of x.

Area A = _{a}∫^{b}(f(x)-g(x))dx where interval a≤ x ≤b

Case 2: If area bounded by y-axis then limits are use for y fun and the function of y.

Area A = _{c}∫^{d}(f(y)-g(y)) dy where interval c≤ y ≤d .

Let’s see the Area region bounded function based example.

Example 1: Find the area region bounded between the function where first function y = x and second function is y = x^{2}?

Solution: For finding area of the region bounded, we need to follow the following steps:

Step 1: In this step first we would solve the both function and find the interval between these function.

So write the both function, first function y = x ,

Second function y = x^{2},

So we simplify both function variable x =x^{2}

x^{2} = x , x =0 , x=1

After solving we got the points first Point is x =0, and another is 1 and these points are worked as interval [0,1] for area region bounded.

Step 2: in this step we are integrated the both function with interval

We know Area of the region bounded between function A = _{a}∫^{b} (f(x) – g(x)) dx

: f(x) = x

: g(x) = x^{2}

Area of curve A = _{0}∫^{1} (x- x^{2})dx

A = [x^{2}/2 – x^{3}/(3)]_{0}^{1}

A = [(1/2 – 1/3) – (0-0)]

A= [(3-2)/6] = 1/6

Now we got the Area of the region bounded A = 1/6

Example 2: Find the area of the bounded between the function and also find the interval between the function, where first function y = x and second function is y = x^{2} +3x?

Solution:

Step 1: In this step first we would solve the both function and find the interval between these function.

So write the both function, first function y = x,

Second function y = x^{2} +3x

So we simplify both function x =x^{2} +3x

x= x(x +3)

1 = (x +3)

x +3 -1 = 0

x=0, x= 2

After solving we got the points first point is x =0, and another is 2 and these points are worked as interval [0,2] for area of the region bounded.

Step 2: In this step we are integrated the both function with interval points

We know Area of the region bounded between function A = _{a}∫^{b} (f(x) – g(x)) dx

: f(x) = x

: g(x) = x^{2} +3x

Area of the region bounded function A = _{0}∫^{2} (x- (x^{2}+3x))dx

A = [x^{2}/2 – (x^{3}/3 + 3x^{2}/2) ]_{1}^{2}

A = [(4/2-(8/3 +12/2))] –[ (1/2) – (1/3 + 3/2)]

A= [2 –(16 +36)/6 ] – [1/2 – (2+9)/2]

A = [2- 52/6]- [1/2 – 11/2]

A = (12-52)/6 –(-10/2)

A = 40/6 +10/2

A = 70/6 = 35/3

Now we got the Area of the region bounded A = 35/3

Example 3: Find the area region bounded between the function where first function x = -y and second function is

x = √y and interval is [0 ,1]?

Solution: Step 1: in this step first we would solve the both curve and find the interval between these function.

So write the both curve first function x = -y,

Second function 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 function with interval

We know Area of the region bounded between function A = _{a}∫^{b} (f(y) – g(y)) dy

: f(y) = -y

: g(y) = √y

Area region bounded A = _{0}∫^{1} (-y)- (√y)dy

A = [(-y)^{2}/2 – y^{3/2}/(3/2)]_{0}^{1}

A = [(1)^{2}/2 – 1^{3/2}/(3/2)] - [(0)^{2}/2 – (0)^{3/2}/(3/2)]

A = [1/2 -2/3 ] = 3-4 /3 = -1/3

A =-1/3

Area is always positive so we do not take negative value it is considered as positive

Now we got the Area of the region bounded A = -1/3

Example 4: Find the area region bounded between the function where first function y = 4x -2 and second function is y = x^{2}?

Solution: for finding area region bounds we need to follow below steps.

Step 1: In this step first we would solve the both function and find the interval between these function.

So write the both function, first function y = 4x -2,

Second curve y = x^{2}

So we simplify both curve 4x-2 =x^{2} +1

x^{2} - 4x +2 +1 =0

x^{2} –4x +3 = 0

x^{2} -3x - x +3 = 0

x(x-3) +1(x-3)=0

(x-3)(x+1)=0

x =3 , x=-1

After solving we got the points first point is x =-1, and another is 3 and these points are worked as interval [-1,3] for area of the region bounded.

Step 2 :in this step we are integrated the both function with interval

We know Area of the region bounded between function A = _{a}∫^{b} (f(x) – g(x)) dx

: f(x) = 4x -2

: g(x) = x^{2}

Area of the region bounded curve A = _{-1}∫^{3} (4x-2)- (x^{2})dx

A = [(4x^{2}/2 -2x) –(x^{3}/3)]_{-1}^{3}

A = [(4x3^{2}/2 -2x3) –(2^{3}/3)] - [(4/2 +2) – (-1)^{3}/3]

A = [36/2 -6 - 8/3 ] –[4 +1/3 ]

A = [18 -6 - 8/3] – [(12+1)/3]

A = [12 – 8/3]-[13/3]

A = 36-8/3 -13/3

A = (28-13)/3

A = 15/3 = 5

Now we got the Area of region bounded function A = 5

Example 5: Find the area region bounds between the function where first function y = x^{2} + 3x and second function is y = x^{2} +2x .Where function is bounded by interval [0,1]?

Solution: for finding area region bounds we need to go through following steps:.

So write the both function, first function y = x^{2} + 3x,

Second function y = x^{2} +2x

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 function with interval

We know Area of the region bounded between curves A = _{a}∫^{b} (f(x) – g(x)) dx

: f(x) = x^{2} +3x

: g(x) = x^{2} +2x

Area of curve A = _{0}∫^{1} (x^{2} +3x)-( x^{2} +2x)dx

A = [(x^{3}/3 + 3x^{2}/2)– (x^{3}/3 + 2x^{2}/2]_{0}^{1}

A = [(1/3 – 3/2) – (1/3 + 1)] – [0 + 0 – 0+ 0]

A= [2-9/6 - (4/3)]

A = -7/6 -4/3 = (-7 -8)/6 = -15/6

Now we got the Area region bound A = -15/6

Example 6: Find the area region bounds between the function where first function y = x^{3} + 3x^{2} +x and second function is y = x^{3} +2x^{2} + x .Where function is bounded by interval [0,1]?

Solution:

So write the both function, first function y = x^{3} + 3x^{2} +x,

Second function y = x^{3} +2x^{2} + x,

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 function with interval

We know Area of the region bounded between curves A = _{a}∫^{b} (f(x) – g(x)) dx

: f(x) = x^{3} +3x^{2} + x

: g(x) = x^{3} +2x^{2} + x

Area of the region bounded curve A = _{0}∫^{1} (x^{3} +3x^{2}+x)-( x^{3} +2x^{2} +x)dx

A = [(x^{4}/4 + 3x^{3}/3 + x^{2}/2)– (x^{4}/4 + 2x^{3}/3 + x^{2}/2)]_{0}^{1}

A = [(1/4 + 3/3 +1/2) – (1/4 +2/3 +1/2 )] – [0 + 0 +0 +0 -0+0]

A= [ (3+4 +6 )/12 - (3 +8 +6)/12]

A = [13/12 – 17/12]

A = [-3/12]

A = -1/4

Area is always positive so we do not consider negative value. Now we get the area region bound A = -1/4.

To understand definite Integration in the Calculus, we have to first go through the definition of the integration, it is the opposite process of the differentiation and always used to find the sum of areas of small parts of any irregular and big body for example an unknown field. We have generally two types of the integrals in calculus. First type of integration is the Definite Integral and the other one is the indefinite integral. The definite integration is the integration type in which we perform the integration for a specific and particular definite range. For example let's we have a function f(x) and an interval of variable x as [a, b], means the variable x is defined over the interval [a, b] and thus the definite integral of f(x) can be given as:

∫_{a}^{b} f(x) dx

This integral is the area of a region bounded by the graph of the function f(x), the x-axis, and the vertical lines x = a, and x = b. Here a is the upper limit of the function and b is the lower limit of the function. We integrate the integral and apply the certain limits for the given interval in the definite integration and subtract the lower limit value with the upper limit value. The definite integral in the term of explanation can be given as:

∫_{a}^{b} f(x) = f(b) - f(a)

Talking about the other type of the integration, indefinite integral is, any function whose derivative is the given function itself. The indefinite integral can be given by

F = ∫ f(x) dx

Now talking about the properties of definite integrals we have certain definite Integral Properties. The properties of definite integrals make the calculation easy and makes problem simpler. The several properties of definite integral can be given as:

The first property shows that integral of any function having same upper and lower limits for the integral, always results in the zero. So:

1. ∫_{a}^{a} f(x) dx = 0

The proving of this property can be given as:

∫_{a}^{a}f(x) dx = f(a) – f(a) = 0

The second property is for a constant number. This states that the integration of the constant number is constant itself multiplied with difference of the limits.

2. ∫_{a}^{b} k dx = k ( b – a ), where k is any constant.

The property third and fourth is for the addition or subtraction of two different Functions. This property is very useful in the conditions where we have a need to add or subtract two different Functions in the integration. We have to integrate both of the terms individually and then after we can add or subtract the both of the results to get the result of Combination. Both of these properties can be used to get the area of the combination body. So these properties can be given as:

3. ∫_{a}^{b} (f(x) + g(x)) dx = ∫_{a}^{b} f(x) dx + ∫_{a}^{b} g(x) dx

4. ∫_{a}^{b} (f(x) - g(x) ) dx = ∫_{a}^{b} f(x) dx - ∫_{a}^{b} g(x) d

The property five is for interchange of the limits. We can inter change the upper and lower limits and the resulting integration will have a negative sign due to the change in the limits.

5. ∫_{a}^{b} f(x) dx = - ∫_{b}^{a} f(x) dx

Proof for this property can be given as:

∫_{a}^{b} f(x) dx = f(b) - f(a)

= - (f(a) – f(b))

So we can write it in the form of:

∫_{a}^{b} f(x) dx = - ∫_{b}^{a} f(x) dx

Next property tells that if we have any constant common in the integral function then we can take it out of the integral and there is no need to integrate the constant one.

6. ∫_{a}^{b} k f(x) = k ∫_{a}^{b} f(x) dx, such that k is an arbitrary constant.

The other and always used properties of the definite integral are for some of the inequalities. The first property tells that we can break any of the big intervals in to two or more sub intervals lies in the bigger limit range. We can break a bigger limit [a, b] in to limits [a, c] and [c, b] such that the third number lies in between of range [a, b]. We can break it in number of intervals as per our simplification required.

1. If the value of limits a < c < b, then for these limits we can say that:

∫_{a}^{b} f(x) dx = ∫_{a}^{c} f(x) dx + ∫_{c}^{b} f(x) dx

In the terms of the inequalities we can also say that if the function f(x) < g(x) then we can also take it for an interval [a, b] as:

2. If f(x) >= zero in the interval [a, b] then ∫_{a}^{b} f(x) dx >= 0

3. If f(x) =< zero in the interval [a, b] then ∫_{a}^{b} f(x) dx <= 0

Now talking about the next property, if one function f(x) is larger than any other function g(x) then it is also fact that the definite integral of the bigger function is larger than that of the smaller function. So we can give this property as:

if 0 <= f(x) <= g(x) then the definite integral:

4. 0 <= ∫_{a}^{b} f(x) dx <= ∫_{a}^{b} g(x) dx

We also have a property through which we can change the dependent variable of the integral function. The property of change of the dependent variable in the function can be given as:

1. ∫_{a}^{b} f(x) dx = ∫_{a}^{b} f(t) dt

Property means that the value of the variable is dependent on the variable of the integration.

The other limit related properties of the definite integration can be given as:

2. ∫_{0}^{a} f(x) dx = ∫_{0}^{a} f(a-x) dx

3. ∫_{a}^{b} f(x) dx = ∫_{a}^{b} f(a + b - x) dx

The other properties are for some of the specific type of the intervals. If the interval is like [+a, -a] then the properties for such intervals can be given as:

4. ∫_{-a}^{+a} f(x) dx = 2∫_{0}^{a} f(x) dx, this is true if the function f is an even function and

5. ∫_{-a}^{+a} f(x) dx = 0 is true in the case of if the function f is an odd function.

We can also break any limit of the definite integral into some of the inner lying intervals as:

6. ∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(2a - x) dx

The Mean value theorem for the definite integral can be given as:

If function f(x) is continuous in the interval of [a, b] then at least one value of the the variable x=y in the interval [a, b] such that

∫_{a}^{b}f(x) = (b - a) f(c)

so we can again say that the value of f(c) can be given as:

f(c) = 1/(b-a) ∫_{a}^{b} f(x) dx

Here the function f(c) is the mean value of the function f(x) in the interval of [a,b].

here we can take some of the examples for these properties:

Example: Evaluate ∫_{16}^{9} √x dx if ∫_{9}^{16} √x dx = 74/3

solution: We can write it as in the form of ∫_{16}^{9} √x dx = - ∫_{9}^{16} √x dx

so the answer for this example after putting the value of the function and the limits will be:

∫_{16}^{9}√x dx = -74/3

Example: find the integration of the function ∫_{4}^{4} √(4x^{2} - 7x + 5) dx for given intervals.

Solution: Now here we can use one of the properties of definite integral as int aa f(x) so by using this property we have:

∫_{a}^{a}f(x) dx = 0

Applying the property we have:

∫_{4}^{4}√(4x^{2} - 7x + 5) dx = 0

Answer = 0.

Example: What will be the result of the definite integral ∫_{2}^{5} (x^{2} - 3x) dx

Solution: This function is the combination of the subtraction of the two different functions, and for this function we can apply the property of addition and subtraction. It can be written as:

= ∫_{2}^{5} (x^{2} - 3x) dx

= ∫_{2}^{5} x^{2} dx - ∫_{2}^{5} 3x dx

and after this we can solve it properly as we solves other integrals:

= ∫_{2}^{5} (x2 - 3x) dx = ∫_{2}^{5} (x2) dx - ∫_{2}^{5} 3x dx

= [x^{3} /3]_{2}^{5} - [3x^{2} /2]_{2}^{5}

Example: It is given that ∫_{1}^{2} f(x) dx = 2, and ∫_{1}^{4} f(x) dx = -1, so find the value of the integral ∫_{2}^{4} f(x) dx.

Solution: We have a property of the breaking any definite interval into the several sub intervals of its limits. So we can write this function ∫_{1}^{4} f(x) dx in to its sub intervals as:

∫_{1}^{4} f(x) dx = ∫_{1}^{2} f(x) dx + ∫_{2}^{4} f(x) dx,

So we can give the solution as:

-1 = 2 + ∫_{2}^{4} f(x) dx,

So the value of the integral ∫_{2}^{4} f(x) dx can be given as:

∫_{2}^{4} f(x) dx = 2 + 1 = 3,

Example: If the value of ∫_{2}^{7} f(x) dx = 20, and ∫_{4}^{7} f(x) dx = 15, so find the value of the integral ∫_{2}^{4} f(x) dx.

Solution: We have the property of sub intervals as:

∫_{2}^{7} f(x) dx = ∫_{2}^{4} f(x) dx + ∫_{4}^{7} f(x) dx,

Putting all the given values in the above expressions:

20 = ∫_{2}^{4} f(x) dx + 15,

∫_{2}^{4} f(x) dx = 20 - 15,

∫_{2}^{4} f(x) dx = 5.