Evaluation of Definite Integrals by Substitution

Evaluation of Definite Integrals by Substitution

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity y concerning another quantity x is called the derivative or differential coefficient of y about x. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

Substitution is one of the basic methods for calculating indefinite integrals. This technique transforms a complex integral into a simpler one by changing the variable of integration. It is especially useful for integrals involving composite functions where a direct integration approach is difficult.

We have already learned to find Indefinite Integration by using the substitution method. But in the case of definite integration, we also need to change the limits of integration 'a' and 'b'. If we substitute x = g(t), then g(t) must be continuous in the interval [a, b].

Definite integration calculates the area under a curve between two specific points on the x-axis.

Let f be a function of x defined on the closed interval [a, b] and F be another function such that $\frac{d}{d x}(F(x))=f(x)$ for all x in the domain of f, then $\int_a^b f(x) d x=[F(x)+c]_a^b=F(b)-F(a)$is called the definite integral of the function f(x) over the interval [a, b], where a is called the lower limit of the integral and b is called the upper limit of the integral.

Let's see some examples to see how such questions are solved.

Example 1

Compute the integral $\int_0^{\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$

Let $\quad I=\int_{x=0}^{x=\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$
Divide numerator and denominator by $\cos ^2 x$

$
=\int_{x=0}^{x=\pi / 2} \frac{\sec ^2 x d x}{a^2+b^2 \tan ^2 x}
$

Put $\quad \tan x=t \Rightarrow \sec ^2 x d x=d t$

$
\therefore \quad I=\int_{t=0}^{t=\infty} \frac{d t}{a^2+b^2 t^2}
$
We find the new limits of integration $t=\tan x \Rightarrow t=0$ when $x=0$ and $t=\infty$ when $x=\pi / 2$
\begin{aligned}
\Rightarrow \quad I & =\frac{1}{b^2} \int_0^{\infty} \frac{d t}{\left(\frac{a}{b}\right)^2+t^2}=\frac{1}{b^2} \cdot \frac{1}{a / b}\left[\tan ^{-1} \frac{b t}{a}\right]_0^{\infty} \\
& =\frac{1}{a b}\left[\frac{\pi}{2}-0\right]=\frac{\pi}{2 a b}
\end{aligned}

Solved Examples Based on Evaluation Of Definite Integrals By Substitution

Example 1: The integral $\int_1^2 e^x \cdot x^x\left(2+\log _e x\right) d x$equals:

1) $e(4 e+1)$
2) $4 e^2-1$
3) ${ }^{e(4 e-1)}$
4) $e(2 e-1)$

Solution

$\begin{aligned} & \int_1^2 e^x \cdot x^x\left(2+\log _e x\right) d x \\ & \int_1^2 e^x\left(2 x^x+x^x \log _e x\right) d x \\ & \int_1^2 e^x(\underbrace{x^x}_{f(x)}+\underbrace{x^x\left(1+\log _e x\right)}_{f^{\prime}(x)}) d x \\ & \left(e^x \cdot x^x\right)_1^2=4 e^2-e\end{aligned}$

Hence, the answer is the option 3.

Example 2: Let f be a twice differentiable function defined on R such that $f(0)=1, f^{\prime}(0)=2$ and $f^{\prime}(x) \neq 0$ for all$x \in R$. If $\left|\begin{array}{cc}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$ for all $x \in R$ then the value of $f(1)$ lies in the interval:

1) $(0,3)$
2) $(9,12)$
3) $(6,9)$
4) $(3,6)$

Solution

$\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}=0$

$\begin{aligned} & f(x) f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2=0 \\ & \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)} \\ & \ln \left(f^{\prime}(x)\right)=\ln f(x)+\ln c\end{aligned}$

$\begin{aligned} & f^{\prime}(x)=c f(x) \\ & \frac{f^{\prime}(x)}{f(x)}=c \\ & \ln f(x)=c x+k\end{aligned}$

$\begin{aligned} & f(x)=k e^{c x} \\ & f(0)=1=k \\ & f^{\prime}(0)=c=2 \\ & f(x)=e^{2 x} \\ & f(1)=e^2 \in(6,9)\end{aligned}$

Hence, the answer is the option 3.

Example 3: For x>0 , if $f(x)=\int_1^x \frac{\log _e t}{1+t} d t$, then $f(e)+f\left(\frac{1}{e}\right)$ is equal to

1) 1

2) -1

3) $\frac{1}{2}$

4) 0

Solution

$\begin{aligned} & f(x)=\int_1^x \frac{\log _e t}{(1+t)} d t \\ & f\left(\frac{1}{x}\right)=\int_1^{1 / x} \frac{\log _e t}{1+t} d t \\ & \operatorname{let} t=\frac{1}{y}\end{aligned}$

$\begin{aligned} & =\int_1^x \frac{\log _e y}{1+y} \cdot \frac{y}{y^2} d y \\ & =\int_1^x \frac{\log _e y}{y(1+y)} d y\end{aligned}$

Hence,

$f(x)+f\left(\frac{1}{x}\right)=\int_1^x \frac{(1+t) \log _c t}{t(1+t)} d t=\int_1^x \frac{\log _e t}{t} d t$

put $u=\log _{\mathrm{c}} t \Rightarrow \frac{d u}{d t}=\frac{1}{t}$

$\begin{gathered}=\frac{1}{2} \log _{\mathrm{e}}^2(\mathrm{x}) \\ \text { so } f(\mathrm{e})+f\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{2}\end{gathered}$

Hence, the answer is the option 3.

Example 4: Let $f(x)=\int_0^x e^t f(t) d t+e^x$ be a differentiable function for all $x \in R$. Then f(x) equals:

1) $
2 e^{\left(\mathrm{e}^x-1\right)}-1
$
2) $e^{\left(e^x-1\right)}$
3) $e^{e^x}-1$
4) $2 e^{e^x}-1$

Solution

$f(\mathrm{x})=\int_0^{\mathrm{x}} \mathrm{e}^{\mathrm{t}} f(\mathrm{t}) \mathrm{dt}+\mathrm{e}^{\mathrm{x}} \Rightarrow f(0)=1$

differentiating with respect to x

$\begin{aligned} & f(x)=e^x f(x)+e^x \\ & f^{\prime}(x)=e^x(f(x)+1) \\ & \int_0^x \frac{f^{\prime}(x)}{f(x)+1} d x=\int_0^x e^x d x \\ & \left.\ln (f(x)+1)\right|_0 ^x=\left.e^x\right|_0 ^x\end{aligned}$

$\begin{aligned} & n(f(x)+1)-\ln (f(0)+1)=e^2-1 \\ & \ln \left(\frac{f(x)+1}{2}\right)-e^x-1 \quad(\text { as } f(0)=1) \\ & f(x)=2 e^{\left(c^2-1\right)}-1\end{aligned}$

Hence, the answer is the option 1.

Example 5: If $I=\int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x$ . Then find the value of $\mathrm{I}$

1) $2 \ln 2$
2) $\ln 2$
3) $2 \ln \frac{\pi}{2}$
4) $\ln \frac{\pi}{2}$

Solution

Let $u=1+\sin x$

$d u=\cos x d x$

When $x=0, u=1+\sin (0)=1$
When $x=\pi / 2, \quad u=1+\sin \left(\frac{\pi}{2}\right)=2$

Then

$\int_0^{\pi / 2} \frac{\cos x}{1+\sin x}=\int_1^2 u^{-1} d u=\ln |u|_1^2=[\ln 2-\ln 1]-\ln 2$

Hence, the answer is the option 2.

List of Topics Related to Definite Integrals by Substitution

Explore important topics related to definite integrals by substitution, including applications of integrals, integrals of particular functions, indefinite integrals, integration by parts, and inequalities in definite integration. Mastery of these interconnected concepts will enhance your integral calculus skills and problem-solving ability.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by parts

Application of Inequality in Definite Integration

NCERT Resources

Access comprehensive NCERT resources for Class 12 Maths Chapter 7 on Integrals, including detailed notes, complete solutions, and exemplar problem sets. These resources are designed to help students strengthen their understanding and excel in their board and competitive exams.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Definite Integral by Substitution

Sharpen your calculus skills with practice questions focused on definite integrals solved using substitution. These exercises will help reinforce key concepts and improve problem-solving speed for exams like JEE Main and board tests.

Evaluation Of Definite Integrals By Substitution- Practice Question MCQ

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