INTEGRATE:
1) (x3+3)/x6(x2+1).
Upper limit=∞;Lower limit=0
2) xe-x/(√1-e-2x).
Upper limit=0;Lower limit=∞.
3) If f(x) be a function satisfying f'(x)=f(x) with f((0)=1 and g be the function satisfying f(x) + g(x)=x2,then prove that the value of the integral ∫f(x)g(x)dx , upper and lower limits being 1 & 0 is -(e2-2e+3)/2.
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3 Answers
1357third one is easy
f(x)=ex, g(x)=x2-ex
Sushovan Halder i did the 3rd one.i made a simple mistake in calculation.Upvote·0· Reply ·2014-07-24 04:51:07
1708\hspace{-16}\bf{(2)\;::\; }$ Given $\bf{\int_{\infty}^{0}\frac{x\cdot e^{-x}}{\sqrt{1-e^{-2x}}}dx = -\int_{0}^{\infty}\frac{x\cdot e^{-x}}{\sqrt{1-e^{-2x}}}dx}$\\\\\\ Let $\bf{e^{-x}=\sin \phi\;,}$ Then $\bf{-e^{-x}dx = \cos \phi d\phi\Rightarrow dx = -\frac{\cos \phi}{\sin \phi}d\phi}$\\\\\\ and Changing Limits......\\\\\\ So Limit is $\bf{=-\int_{0}^{\frac{\pi}{2}}\ln(\sin \phi)d\phi = \frac{\pi}{2}\ln(2)}$\\\\\\ Here $\bf{\int_{0}^{\frac{\pi}{2}}\ln (\sin \phi)d\phi}$ is a Standard Definite Integral.......\\\\\\ Which is $\bf{= -\frac{\pi}{2}\ln(2)}$ and You can calculate easily.....\\\\\\
- Sushovan Halder thanks a lot.
- Manish Shankar good one :)