If y=f(x) makes positive intercepts of 2 and 0 units on the x and y axes respectively, and…

Question

If $y=f(x)$ makes positive intercepts of $2$ and $0$ units on the $x$ and $y$ axes respectively, and encloses an area of $3/4$ square unit with the axes, then $\displaystyle\int_0^2 xf'(x)\,dx$ is:

Accepted Answer

Given $\displaystyle\int_0^2 f(x)\,dx = \dfrac34$ and $f(2)=0$. Integrating by parts:

$$\int_0^2 xf'(x)\,dx = \left[xf(x)\right]_0^2 - \int_0^2 f(x)\,dx = 2f(2) - \frac34 = 0-\frac34 = -\frac34$$

Solutions · 1