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Integral Calculus

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12 Feb 2009 09:58:43 IST

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Two functions

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Suppose a>0, and f(x) is continuous in , define for . Show that

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Anant Kumar

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Joined: 10 Jul 2008

Posts: 598

1 Mar 2009 02:07:05 IST

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Since $g(x)=\int_x^a\dfrac{f(t)}{t}\ \mathrm{d}t$ and $f$ is continuous, it follows that $g$ is a differentiable function. Therefore, differentiating with respect to $x$ , we obtain
$g^\prime(x)=-\dfrac{f(x)}{x}$
$\Rightarrow \ -f(x) = x g^\prime(x)$
Hence,
$-\int_0^a f(x)\ \mathrm{d}x = \int_0^a x g^\prime(x)\ \mathrm{d}x$
Integrate the right-side by parts taking $x$ as the first function and $g^\prime(x)$ as the second:
$\int_0^a x g^\prime(x)\ \mathrm{d}x = \big[xg(x)\big]_0^a - \int_0^a g(x)\ \mathrm{d}x=0-\int_0^a g(x)\ \mathrm{d}x$
since $g(a)=0$ . Hence
$\int_0^a x g^\prime(x)\ \mathrm{d}x = -\int_0^a g(x)\ \mathrm{d}x$
So, $-\int_0^a f(x)\ \mathrm{d}x = -\int_0^a g(x)\ \mathrm{d}x$
giving us the required equality.

Hari Shankar

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Joined: 28 Feb 2007

Posts: 2173

1 Mar 2009 06:45:52 IST

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Yeah, thats all it takes. It was such a simple problem i am disturbed that none of our jee hopefuls touched it for so long! Must be board exams pressure

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