Evaluate the integral _1^2 ( 1 x - 1 2 x^2 ) e^ 2x dx using substitution.

Let I = _1^2 ( 1 x - 1 2 x^2 ) e^ 2x dx …(i) Putting 2x = t 2 = dt dx 2dx = dt dx = dt 2 Limits of integration when x = 1,t = 2 1pt 1 = 2 and when x = 2,t = 2 2 = 4 From eq. (i), I = _2^4 ( 1 t 2 - 1 2 ( t 2 ) ^2 ) e^t dt 2 = _2^4 ( 2 t - 2 t^2 ) e^t dt 2 = _2^4 1 2 .2 ( 1 t - 1 t^2 ) e^t dt = _2^4 ( 1 t - 1 t^2 ) e^t dt = _2^4 f ( t ) + f' ( t ) e^t dt = e^t f ( t ) _2^4 = ( e^t t )_2^4 = e^4 4 - e^2 2 = e^4 - 2 e^2 4

Evaluate the integral _1^2 ( 1 x - 1 2 x^2 ) e^ 2x dx using subst…

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Question

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

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Answer

Let $I = \int\limits_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right){e^{2x}}dx} $…(i)
Putting 2x = t
$ \Rightarrow 2 = \frac{{dt}}{{dx}}$
$ \Rightarrow 2dx = dt$
$ \Rightarrow dx = \frac{{dt}}{2}$
Limits of integration when $x = 1,t = 2{\kern 1pt} \times 1 = 2$ and when $x = 2,t = 2 \times 2 = 4$
$\therefore$ From eq. (i),
$I = \int\limits_2^4 {\left( {\frac{1}{{\frac{t}{2}}} - \frac{1}{{2{{\left( {\frac{t}{2}} \right)}^2}}}} \right){e^t}\frac{{dt}}{2}} $
$= \int\limits_2^4 {\left( {\frac{2}{t} - \frac{2}{{{t^2}}}} \right){e^t}\frac{{dt}}{2}} $
$= \int\limits_2^4 {\frac{1}{2}.2\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right){e^t}{{dt}}{}} $
$= \int\limits_2^4 {\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right){e^t}{{dt}}{}} $
$= \int\limits_2^4 {\left\{ {f\left( t \right) + f'\left( t \right)} \right\}{e^t}{{dt}}{}} $
$= \left\{ {{e^t}f\left( t \right)} \right\}_2^4$
$= \left( {\frac{{{e^t}}}{t}} \right)_2^4$
$= \frac{{{e^4}}}{4} - \frac{{{e^2}}}{2}$
$= \frac{{{e^4} - 2{e^2}}}{4}$

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