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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral\[\int\limits_0^1(xe^x+\sin\frac{\pi\;x}{4})dx\]

$\begin{array}{1 1} 1+\large \frac{4}{\pi}-\frac{2 \sqrt 2}{\pi} \\1-\large \frac{4}{\pi}+\frac{2 \sqrt 2}{\pi} \\ 1-\large \frac{2}{\pi}+\frac{4 \sqrt 2}{\pi} \\ 1-\large \frac{2}{\pi}-\frac{2 \sqrt 4}{\pi} \end{array} $

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1 Answer

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-f(a)$
  • (ii) Methods of integration by parts: $\int udv=uv-\int vdu$
  • (iii)$ \int \sin x dx=-\cos x+c$
Given $\int\limits_0^1(xe^x+\sin\frac{\pi\;x}{4})dx$
 
On seperating the terms
 
$I=\int \limits_0^1 xe^xdx+\int\limits_0^1\sin \frac{\pi x}{4}dx$
 
Consider $I_1=\int \limits_0^1 x e^xdx$
 
This is of the form $\int udv$
 
Where $ \int udv=uv-\int vdu$
 
Let u=x
 
On differentiating with respect to x,
 
du=dx
 
Let $dv=e^xdx$
 
On integrating we get,
 
$v=e^x$
 
$\int \limits_0^1 xe^x=(xe^x)_0^1-\int \limits_0^1e^x dx$
 
On integrating we get
 
$ I_1=(xe^x)_0^1-(e^x)_0^1$
 
Consider $ I_2=\int \limits_0^1 \sin \frac{\pi}{4}x$
 
$=\frac{1}{\pi/4}(\cos \pi/4 x)_0^1$
 
$I=I_1+I_2$
 
$=(xe^x)_0^1-(e^x)_0^1+(-\frac{4}{\pi} \cos \frac{\pi}{4}x)_0^1$
 
On applying limits
 
$[1.e^1-0]-[e^1-e^0]+[(-\frac{4}{\pi}.\cos \frac{\pi}{4}-(-\frac{4}{\pi} \cos 0)]$
 
$e^0=1;\cos \frac{\pi}{4}=\frac{1}{\sqrt 2}; \cos 0=1$
 
Therefoer $ I=1-\frac{4}{\pi}.\frac{1}{\sqrt 2}+\frac{4}{\pi}$
 
$\int \limits _0^1 (xe^x+\sin \large\frac{\pi x}{4})dx=1+\frac{4}{\pi}-\frac{2 \sqrt 2}{\pi}$

 

 

answered Feb 12, 2013 by meena.p
 
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