Interpret the given information

The curve $y=f(x)$ has positive intercepts:

• on the $x$-axis at $x=2$, so $f(2)=0$
• on the $y$-axis at $y=0$? This wording means the intercept on the $y$-axis is $0$ units, so at $x=0$, we have $f(0)=0$.

Thus the curve meets both axes at $(0,0)$ and $(2,0)$.

Also, the area enclosed by the curve and the axes is given as $\int_0^2 f(x)\,dx=\frac34.$

Evaluate the required integral

We need to find $I=\int_0^2 x f'(x)\,dx.$

Use integration by parts: $\int x f'(x)\,dx = x f(x)-\int f(x)\,dx.$

Therefore, $I=\left[x f(x)\right]_0^2-\int_0^2 f(x)\,dx.$

Use the boundary values

Since $f(2)=0$ and $f(0)=0$, $\left[x f(x)\right]_0^2 = 2f(2)-0\cdot f(0)=0.$

Hence, $I=0-\int_0^2 f(x)\,dx = -\frac34.$

Final answer

$\int_0^2 x f'(x)\,dx=-\frac34.$

So the correct option is D.