JEE MAIN - Mathematics (2002 - No. 13)
If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx} $$ is
$$3/2$$
$$1$$
$$5/4$$
$$-3/4$$
Explanation
We have $$\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$$
$$\int\limits_0^2 {xf'\left( x \right)} dx$$
$$ = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$$
$$ = \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}$$
$$ = 2f\left( 2 \right) - {3 \over 4}$$
$$ = 0 - {3 \over 4}$$
$$\left( {} \right.$$ As $$f\left( 2 \right) = 0$$ $$\left. {} \right)$$ $$ = - {3 \over 4}.$$
$$\int\limits_0^2 {xf'\left( x \right)} dx$$
$$ = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$$
$$ = \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}$$
$$ = 2f\left( 2 \right) - {3 \over 4}$$
$$ = 0 - {3 \over 4}$$
$$\left( {} \right.$$ As $$f\left( 2 \right) = 0$$ $$\left. {} \right)$$ $$ = - {3 \over 4}.$$
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