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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

JEE · Math · previous-year question

  1. A.$$3/2$$
  2. B.$$1$$
  3. C.$$5/4$$
  4. D.$$-3/4$$correct

Answer

D. $$-3/4$$

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 $$f\left( 2 \right) = 0$$ $$\left. {} \right)$$ $$ = - {3 \over 4}.$$

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