Question

Integration Help Calculus AP

Answer 1

If \[\int\limits_{2}^{4} 3x ^{2}\sqrt{x ^{3}-1} dx\]=\[\int\limits_{a}^{b}\sqrt{u}du\] then what are the values of a and b

Answer 2

Please dont give the answer i need help

Answer 3

\[\large\rm \int\limits_2^4 3x^2\sqrt{x^3-1}~dx\quad=\quad \int\limits_2^4\sqrt{x^3-1}\left(3x^2dx\right)\]The correct substitution for this problem is\[\large\rm u=x^3-1\]right?
It looks like you figured that out by now,
unless that was given.

Answer 4

\[\large\rm du=3x^2dx\]So that works out nicely.

Answer 5

yah so that right there is u-subsitution

Answer 6

To find your `new limits of integration`,
Plug the old limits into your substitution.\[\large\rm x=2:\qquad u=(2)^3-1\]\[\large\rm x=4:\qquad u=(4)^3-1\]

Answer 7

those are the equations i use to find them and then that is it?

Answer 8

So your new lower limit is 2^3-1
while your new upper limit is 4^3-1
simplify those values :)

Answer 9

a = 7 and b=63

Answer 10

Yay good job!

Answer 11

Thank you sir!!! have a great day!!

Answer 12

You too