If y=f(x) makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 …

Question

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

Accepted Answer

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

Solutions · 1