Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.7

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.7

Question 1. Evaluate the following
(i) \( \int_{0}^{∞} x^5 e^{-3x} dx \)
(ii) \( \int_{0}^{\pi/2} \frac{e^{-\tan x}}{\cos^6 x} dx \)
Answer:
(i) To evaluate \( \int_{0}^{∞} x^5 e^{-3x} dx \), we use a standard Gamma function integral formula.
The formula is: \( \int_{0}^{\infty} e^{-ax} x^n dx = \frac{n!}{a^{n+1}} \)
Comparing our integral \( \int_{0}^{∞} x^5 e^{-3x} dx \) with the formula, we have \( n=5 \) and \( a=3 \).
Substitute these values into the formula:
\( \int_{0}^{∞} x^5 e^{-3x} dx = \frac{5!}{3^{5+1}} \)
\( \implies \int_{0}^{∞} x^5 e^{-3x} dx = \frac{5!}{3^6} \)
This formula is very useful for solving certain types of definite integrals quickly.

(ii) To evaluate \( I = \int_{0}^{\pi/2} \frac{e^{-\tan x}}{\cos^6 x} dx \):
First, rewrite the integral in terms of \( \sec x \):
\( I = \int_{0}^{\pi/2} e^{-\tan x} \cdot \frac{1}{\cos^6 x} dx \)
\( \implies I = \int_{0}^{\pi/2} e^{-\tan x} \sec^6 x dx \)
Next, split \( \sec^6 x \) into \( \sec^4 x \cdot \sec^2 x \). Also, use the identity \( \sec^2 x = 1 + \tan^2 x \):
\( I = \int_{0}^{\pi/2} e^{-\tan x} \sec^4 x \sec^2 x dx \)
\( \implies I = \int_{0}^{\pi/2} e^{-\tan x} (1 + \tan^2 x)^2 \sec^2 x dx \)
Now, we use a substitution to simplify the integral. Let \( t = \tan x \).
Then, differentiate \( t \) with respect to \( x \): \( dt = \sec^2 x dx \).
We also need to change the limits of integration for \( t \):
When \( x = 0 \), \( t = \tan(0) = 0 \).
When \( x = \pi/2 \), \( t = \tan(\pi/2) = \infty \).
Substitute \( t \) and \( dt \) into the integral with the new limits:
\( I = \int_{0}^{\infty} e^{-t} (1 + t^2)^2 dt \)
Expand the term \( (1 + t^2)^2 \):
\( (1 + t^2)^2 = 1^2 + 2(1)(t^2) + (t^2)^2 = 1 + 2t^2 + t^4 \)
Substitute this back into the integral:
\( I = \int_{0}^{\infty} e^{-t} (1 + 2t^2 + t^4) dt \)
Split this into three separate integrals:
\( I = \int_{0}^{\infty} e^{-t} dt + 2 \int_{0}^{\infty} e^{-t} t^2 dt + \int_{0}^{\infty} e^{-t} t^4 dt \)
We will use the Gamma function integral formula again: \( \int_{0}^{\infty} e^{-at} t^n dt = \frac{n!}{a^{n+1}} \). In these integrals, \( a=1 \).
For the first integral \( \int_{0}^{\infty} e^{-t} dt \), \( n=0 \) and \( a=1 \): \( \frac{0!}{1^{0+1}} = \frac{1}{1} = 1 \)
For the second integral \( 2 \int_{0}^{\infty} e^{-t} t^2 dt \), \( n=2 \) and \( a=1 \): \( 2 \times \frac{2!}{1^{2+1}} = 2 \times \frac{2}{1} = 4 \)
For the third integral \( \int_{0}^{\infty} e^{-t} t^4 dt \), \( n=4 \) and \( a=1 \): \( \frac{4!}{1^{4+1}} = \frac{24}{1} = 24 \)
Add the results of the three integrals together:
\( I = 1 + 4 + 24 \)
\( \implies I = 29 \)
This method of substitution and formula application makes complex integrals manageable.
In simple words: For the first part, we used a special formula for integrals with \( e^{-ax} \) and \( x^n \) to find the answer quickly. For the second part, we changed \( \sec^6 x \) to \( (1+\tan^2 x)^2 \sec^2 x \). Then, we made a substitution \( t = \tan x \) to change the integral and its limits. After expanding and splitting it, we used the same special formula again for each part and added them up to get 29.

🎯 Exam Tip: Remember the Gamma function integral formula \( \int_{0}^{\infty} e^{-ax} x^n dx = \frac{n!}{a^{n+1}} \) as it is often useful. Also, when using substitution in definite integrals, always remember to change the limits of integration according to the new variable.

Question 2. \( \int_{0}^{∞} e^{-ax^2} x^3 dx = 32 \), \( a > 0 \), find \( a \)
Answer:
We are given the integral \( \int_{0}^{∞} e^{-ax^2} x^3 dx = 32 \) and that \( a > 0 \). We need to find the value of \( a \).
Let \( I = \int_{0}^{∞} e^{-ax^2} x^3 dx \).
To solve this, we will use a substitution. Let \( t = x^2 \).
Now, find the differential \( dt \) in terms of \( dx \):
\( dt = 2x dx \)
This means \( x dx = \frac{dt}{2} \).
We also need to express \( x^3 dx \) in terms of \( t \) and \( dt \):
\( x^3 dx = x^2 \cdot x dx \)
Substitute \( t = x^2 \) and \( x dx = \frac{dt}{2} \):
\( x^3 dx = t \cdot \frac{dt}{2} \)
Next, change the limits of integration for \( t \):
When \( x = 0 \), \( t = (0)^2 = 0 \).
When \( x = ∞ \), \( t = (∞)^2 = ∞ \).
Substitute these into the integral \( I \):
\( I = \int_{0}^{∞} e^{-a(x^2)} (x^2) (\frac{1}{2} dt) \)
\( \implies I = \int_{0}^{∞} e^{-at} t \frac{1}{2} dt \)
Pull the constant \( \frac{1}{2} \) outside the integral:
\( I = \frac{1}{2} \int_{0}^{∞} e^{-at} t^1 dt \)
This integral now matches the Gamma function integral formula \( \int_{0}^{\infty} e^{-at} t^n dt = \frac{n!}{a^{n+1}} \), where \( n=1 \).
Apply the formula:
\( I = \frac{1}{2} \left( \frac{1!}{a^{1+1}} \right) \)
\( \implies I = \frac{1}{2} \left( \frac{1}{a^2} \right) \)
\( \implies I = \frac{1}{2a^2} \)
We are given that \( I = 32 \). So, set the expression equal to 32:
\( \frac{1}{2a^2} = 32 \)
Multiply both sides by \( 2a^2 \):
\( 1 = 32 \times 2a^2 \)
\( \implies 1 = 64a^2 \)
Divide by 64 to isolate \( a^2 \):
\( a^2 = \frac{1}{64} \)
Take the square root of both sides:
\( a = \pm \sqrt{\frac{1}{64}} \)
\( \implies a = \pm \frac{1}{8} \)
The problem states that \( a > 0 \). So, we choose the positive value for \( a \).
\( a = \frac{1}{8} \)
The condition \( a > 0 \) ensures the integral converges to a finite value.
In simple words: We had an integral with an unknown number 'a'. We changed the variable from \( x \) to \( t = x^2 \). This made the integral easier to solve using a special formula. After solving, we found an equation with 'a'. Since we were told the answer to the integral is 32 and 'a' must be positive, we calculated 'a' to be \( \frac{1}{8} \).

🎯 Exam Tip: When making a substitution in a definite integral, always remember to convert not just the variable \( dx \) and the function, but also the limits of integration. This is a common point where errors can occur.