Question

Evaluate the integral $\int\limits_1^2 {\left( {\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}} \right){e^{2x}}dx}$ using substitution.

Answer 1

Here, in the question, the given integral is a definite integral. The definite integral is denoted by $\int\limits_a^b {f\left( x \right)dx}$, where $a$ is the lower limit of the integral and $b$ is the upper limit of the integral. The definite integral is evaluated in the two ways mentioned: (i) The definite integral as the limit of the sum, (ii)$\int\limits_a^b {f\left( x \right)dx = F\left( b \right) - F\left( a \right)}$, if $F$ is an anti-derivative of $f\left( x \right)$. We will use the second method discussed to evaluate the given integral.
Formula Used:
$\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + C$

Complete step-by-step solution:
Let $I = \int\limits_1^2 {\left( {\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}} \right){e^{2x}}dx}$
Substituting the value of $x$ in form of $t$ as,
Put$2x = t$
$\Rightarrow 2dx = dt$
When $x = 1,t = 2$ and when $x = 2,t = 4$
Therefore, $I = \int\limits_2^4 {\dfrac{1}{2}\left( {\dfrac{2}{t} - \dfrac{2}{{{t^2}}}} \right){e^t}dt}$
Taking $2$ common from both the terms in bracket, we get,
$I = \int\limits_2^4 {\left( {\dfrac{1}{t} - \dfrac{1}{{{t^2}}}} \right){e^t}dt}$
Now, Let $\dfrac{1}{t} = f\left( t \right)$
Then, $f'\left( t \right) = - \dfrac{1}{{{t^2}}}$
Therefore,$I = \int\limits_2^4 {{e^t}\left[ {f\left( t \right) + f'\left( t \right)} \right]} dt$
Using the identity, $\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + C$, we get,
$I = \left[ {{e^t}f\left( t \right)} \right]_2^4$
Now, we have$f\left( t \right) = \dfrac{1}{t}$,

$$I = \left[ {{e^t} \times \dfrac{1}{t}} \right]_2^4 \\\ \Rightarrow I = \left[ {\dfrac{{{e^t}}}{t}} \right]_2^4 \\\$$

Putting the limits, we get,
$I = \left( {\dfrac{{{e^4}}}{4} - \dfrac{{{e^2}}}{2}} \right)$
Taking$\dfrac{{{e^2}}}{4}$common, we get,
$I = \dfrac{{{e^2}}}{4}\left( {{e^2} - 2} \right)$
Hence, $\int\limits_1^2 {\left( {\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}} \right){e^{2x}}dx} = \dfrac{{{e^2}}}{4}\left( {{e^2} - 2} \right)$
Additional information: In mathematics, integration by substitution, also known as variable change, is a method for evaluating integrals and antiderivatives. As chain rule is the most common method for differentiation, similarly, substitution is the most common method for evaluating integrals.

Note: The process of differentiation and integration are inverse of each other, i.e., $\dfrac{d}{{dx}}\int {f\left( x \right)dx = f\left( x \right)}$ and $\int {f'\left( x \right)dx = f\left( x \right) + C}$, where $C$ is any arbitrary constant. These integrals are known as indefinite integrals or general integrals, $C$ is called a constant of integration. All these integrals differ by a constant. We use indefinite integrals when there are no limits given to a particular function. We use definite integrals when both the upper and lower limits of that function are given.