Suppose $a<b$. The maximum value of the integral $$\int_{a}^{b}(\frac{3}{4}-x-x^{2})dx$$ over all possible values of $a$ and $b$.
note- We can see that the function inside the integral is a decreasing function so what interval do we have to look at to get the answer. I thought the interval to be should be of such length such that $a-b$ becomes minimum so that the integral can be maximised but no definite answer comes to mind. Thank you!
$$I=\int_{a}^{b}(\frac{3}{4}-x-x^{2})dx$$
$$f(x)=\frac{3}{4}-x-x^{2}=\frac{1}{4}(-4x^2-4x+3)=\frac{-1}{4}\left(2x-1\right)\left(2x+3\right)$$
The maximum area under this integral would be $$I=\int_{\frac{-3}{2}}^{\frac{1}{2}}(\frac{3}{4}-x-x^{2})dx=\frac{3x}{4}-\frac{x^2}{2}-\frac{x^3}{3}|_{\frac{-3}{2}}^{\frac{1}{2}}\approx0.667$$