Answer to Question #177464 in Calculus for Moel Tariburu

Question #177464

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


1
Expert's answer
2021-04-15T06:56:40-0400
g(x)=1xt2(3t)dtg(x)=\displaystyle\int_{-1}^xt^2(3-t)dt


By the Fundamental Theorem of Calculus part I


g(x)=x2(3x)g'(x)=x^2 (3-x)


Find the critical number(s)


g(x)=0,x>1g'(x)=0, x>-1x2(3x)=0x^2 (3-x)=0x1=0,x2=3x_1=0, x_2=3


Critical numbers: 0,3.0, 3.

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

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


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

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

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


The integral 1xt2(3t)dt\displaystyle\int_{-1}^xt^2(3-t)dt has the absolute maximum at x=3.x=3.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment