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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the integral using substitution\[\int\limits_1^2\bigg(\frac{1}{x}-\frac{1}{2x^2}\bigg)e^{2x}dx\]

$\begin{array}{1 1}\large\frac{e^2(e^2-2)}{4} \\ \large\frac{e^2(e^2-2)}{2} \\\large\frac{e^2(e-2)}{4} \\\large\frac{e^2(e-2)}{2} \end{array} $

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int e^x[f(x)+f'(x)]dx=e^xf(x)+c$
Given$ \int\limits_1^2\bigg(\frac{1}{x}-\frac{1}{2x^2}\bigg)e^{2x}dx$
 
Let 2x=t
 
On differentiating with respect to x
 
$ 2dx=dt \qquad=>dx=dt/2$
 
As $x=2,t=4\;and\; As\; x=1,t=2$
 
Therefore $\int \limits_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx=\frac{1}{2}\int \limits_2^4(\frac{2}{t}-\frac{2}{t^2}).e^t.dt$
 
$=\int \limits_2^4(\frac{1}{t}-\frac{1}{t^2})e^t.dt$
 
Let $f(t)=\frac{1}{t},then\;f'(t)=-\frac{1}{t^2}$
 
Therefore $\int \limits_2^4 (\frac{1}{t}-\frac{1}{t^2}e^t.dt=[e^t.(\frac{1}{t})]_2^4$
 
On applying limits,
 
$=e^4(\frac{1}{4})-e^2(\frac{1}{2})^2$
 
$=\large\frac{e^4}{4}-\frac{e^2}{2}$
 
$=\large\frac{e^2(e^2-2)}{4}$

 

answered Feb 13, 2013 by meena.p
 
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