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Choose the correct answer If f(x)=$\int\limits_0^x t\sin t\;dt,$ then f'(x) is

$\begin{array}{1 1} (A) \;(cos x + x\; sin\; x) \\ (B)\; x \;\sin\; x \\ (C)\; x\; \cos x \\ (D)\; \sin x + x \cos x\end{array} $

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  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$ \int udv=uv-\int vdu$
  • (iii)$product \;rule: u\frac{dv}{dx}+v\frac{du}{dx}=\frac{d}{dx}(uv)$
Given $f(x)=\int\limits_0^x t\sin t\;dt,$
This is in the form of $\int udv$. Hence
$ \int udv=uv-\int vdu$
Here let u=t=>du=dt
let $dv=\sin t dt $
On integrating ,
$v=-\cos t$
Substituting for u,v,dv and dt
$ \int \limits_0^x t \sin tdt=t(-\cos t)_0^x+\cos t.dt$
On integrating we get
$-(t \cos t)_0^x+(\sin t)_0^x$
On applying limits
$(x \cos x-0)+(\sin x-\sin 0)$
$\sin x-x cos x$
Hence $ I=\sin x-x \cos x$
$=>f(x)=\sin x-x \cos x$
To find f'(x),let us differentiate
w.r.t by apply product rule:$f'(x)=\cos x-x(-\sin x)-\cos x$
$=x\sin x$
Hence the correct answer is B


answered Feb 13, 2013 by meena.p
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