Question

What value of x > -1 maximizes the integral ∫_(-1)^x▒〖t^2 (3-t) 〗 dt?

Answer 1

$$g(x)=\displaystyle\int_{-1}^xt^2(3-t)dt$$

By the Fundamental Theorem of Calculus part I

$$g'(x)=x^2 (3-x)$$

Find the critical number(s)

$$g'(x)=0, x>-1$$ $$x^2 (3-x)=0$$ $$x_1=0, x_2=3$$

Critical numbers: $0, 3.$

If $-1<x<0, g'(x)>0 , g(x)$  increases.

If $0<x<3, g'(x)>0 , g(x)$  increases.

If $x>3, g'(x)<0 , g(x)$ decreases.

The function $g(x)$ has a local maximum at $x=3.$

Since the function $g(x)$ has the only extremum for $x>-1,$ then the function $g(x)$ has the absolute maximum at $x=3.$

The integral $\displaystyle\int_{-1}^xt^2(3-t)dt$ has the absolute maximum at $x=3.$