Finding $(a,b)\,,a>b$ such that $\int^b_a(6-x-x^2)\,dx$ is maximum

Question

Find 2 numbers $(a,b)$ $a>b$ such that we get the maximum value of the integral $\int^b_a(6-x-x^2)dx$.

attempt:

I tried this:

$f'(x) = 0 $

$ 6-x-x^2 = 0$

$ (x_1,x_2)=(2,-3)$

$(a,b) = (2,-3)$

$\int^b_a(6-x-x^2)dx = \int^2_{-3}(6-x-x^2)dx = 38.83$

Is it the correct answer? $(a,b) = (2,-3)$?

Thanks!!

Answer 1

Yes, you are correct. Note that $f(x)=(6-x-x^2)=-(x-2)(x+3)$ so in order the get the maximum value of $\int_a^bf(x)dx$ with $a<b$, it suffices to integrate over the interval $[a,b]$ where the integrand is non-negative that is $[-3,2]$.