If $y = f(x)$ makes positive intercepts of 2 and 1 unit on x and y coordinate axes and encloses an area of 3/4 square unit with 2 the axes, then $\int_0^2x f'(x) dx$, is

$-\frac{3}{4}$

Clearly, $y = f (x)$ passes through (2,0) and (0, 1).

$∴0=f(2)$ and $1=f(0)$

Also, $\int\limits_0^2f(x)dx=\frac{3}{4}$   [Given]

$\int\limits_0^2\underset{I}{x}\underset{II}{f'(x)}dx=\left[xf(x)\right]_0^2-\int\limits_0^2f(x)dx$

$⇒xf'(x)= [2 \,f (2) -0\, f (0)]-\frac{3}{4}=2×0-0×1-\frac{3}{3}=-\frac{3}{4}$