g(x)=∫−1xt2(3−t)dt
By the Fundamental Theorem of Calculus part I
g′(x)=x2(3−x)
Find the critical number(s)
g′(x)=0,x>−1x2(3−x)=0x1=0,x2=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 ∫−1xt2(3−t)dt has the absolute maximum at x=3.
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