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

Hint: 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.