Mathematics Evaluation of Definite Integrals by Substitution For CBSE-NCERT

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Class 12 Chapter 7 - Integrals

Evaluation of Definite Integrals by Substitution For CBSE-NCERT

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`color{blue}star` Evaluation of Definite Integrals by Substitution

`color{blue}star` Evaluation of Definite Integrals by Substitution

Evaluation of Definite Integrals by Substitution

To evaluate `color {red} {int_a^b f(x) dx}` , by substitution, the steps could be as follows:

1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form.

2. Integrate the new integrand with respect to the new variable without mentioning the constant of integration.

3. Resubstitute for the new variable and write the answer in terms of the original variable.

4. Find the values of answers obtained in (3) at the given limits of integral and find the difference of the values at the upper and lower limits.

In order to quicken this method, we can proceed as follows: After performing steps 1, and 2, there is no need of step 3. Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step.

To evaluate `color {red} {int_a^b f(x) dx}` , by substitution, the steps could be as follows:

1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form.

2. Integrate the new integrand with respect to the new variable without mentioning the constant of integration.

3. Resubstitute for the new variable and write the answer in terms of the original variable.

4. Find the values of answers obtained in (3) at the given limits of integral and find the difference of the values at the upper and lower limits.

In order to quicken this method, we can proceed as follows: After performing steps 1, and 2, there is no need of step 3. Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step.

Q 3115580460

Evaluate ` ∫_(0)^1 (tan^(-1)x )/( 1+x^2) dx`
Class 12 Chapter 7 Example 29

Solution:

Let `t = tan^( – 1) x`, then `dt = 1/(1+x^2) dx` . The new limits are, when x = 0, t = 0 and

when `x =1 , t = pi/4 `. Thus, as x varies from 0 to 1, t varies from 0 to `pi/4`

Therefore ` ∫_(0)^1 (tan^(-1) x)/(1+x^2) dx = ∫_(0)^(pi/4) t dt [t^2/2]_(0)^(pi/4) =1/2 [(pi^2)/16 -0 ] = (pi^2 )/(32)`

Q 3115480369

Evaluate ` int_(-1)^1 5x^4 sqrt (x^5 +1 ) dx`
Class 12 Chapter 7 Example 28

Solution:

Put `t = x^5 + 1`, then `dt = 5x^4 dx`.

Therefore, ` ∫ 5x^4 sqrt (x^5 +1) dx = ∫ sqrt t dt = 2/3 t^3/2 = 2/3 (x^5 +1)^3/2`

Hence, ` ∫_(-1)^1 5x^4 sqrt (x^5 +1)dx = 2/3 [ (x^5 +1 )^3/2]_(-1)^1`

`= 2/3 [ (1^5 +1 )^3/2 - ( (-1)^5 + 1 )^3/2]`

` = 2/3 [2^3/2 - 0^3/2] = 2/3 (2 sqrt 2) = (4 sqrt 2 )/3`

Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits.

Let `t = x^5 + 1`. Then `dt = 5 x^4 dx`.

Note that, when x = – 1, t = 0 and when x = 1, t = 2

Thus, as x varies from – 1 to 1, t varies from 0 to 2

Therefore ` ∫_(-1)^1 5x^4 sqrt (x^5 +1) dx = ∫_(0)^2 sqrt t dt`

` =2/3 [t^3/2]_(0)^2 = 2/3 [2^3/2 - 0^3/2] = 2/3 (2 sqrt 2) = (4 sqrt 2)/3`

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