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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int_1^2 \bigg( \large\frac{1}{x} - \frac{1}{x^2} \bigg) $$e^x \: dx$

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1 Answer

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Toolbox:
  • $\int e^x[f(x)+f'(x)]dx=e^x(f(x))+c$
Step 1:
$\int_1^2(\large\frac{1}{x}-\frac{1}{x^2})$$e^xdx$
Clearly I is in the form of
$\int e^x[f(x)+f'(x)]dx=e^x(f(x))+c$
Here $f(x)=\large\frac{1}{x}$
$f'(x)=-\large\frac{1}{x^2}$
$\therefore I=\big[e^x(\large\frac{1}{x})\big]_1^2$
Step 2:
On applying the limits we get,
$I=\bigg[e^2(\large\frac{1}{2})$$-e^1(1)\bigg]$
$\;\;=\large\frac{e^2}{2}$$-e$
$\;\;=e(\large\frac{e}{2}$$-1)$
answered Nov 7, 2013 by sreemathi.v
 
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