Question

Calculus integration with substitution problem

Answer 1

General integration t^2 (t^3 - 3)^10 dt

is the u sub = t^3-3?
du/dt = 3t^2

du = 3t^2dx

Answer 2

du=3t^2dt, yah looks good so far.

From here, you want to solve for `t^2dt`, so you can replace that part in your integral. So get the 3 off of it! :O

Answer 3

do you mean when you divide by 1/3?

Answer 4

mutiply by 1/3, or divide by 3, yes.

Answer 5

i did 1/3 integral 3t^2 * t^2 (t^3-3)^10 dt

and 3t^2 dt is du

so 1/3 integral t^2 u^10 du?

Answer 6

\[\large du=3t^2 dt \qquad \rightarrow \qquad \color{orangered}{\frac{1}{3}du=t^2dt}\]I would have changed the 1/3 in this step, but the way you did it works fine also! :) It looks like you replaced dt with du, you didn't replace the t^2 hmm

Answer 7

Oh you have t^2 twice in there for some reason..? :O

Answer 8

\[\large \int \frac{1}{3}(t^3-3)^{10}\color{orangered}{3t^2\; dt}\]\[\large \color{orangered}{du=3t^2dt}\]

Answer 9

because the original equation has t^2 * (t^3-3)^10 dt

Answer 10

and u sub = 3t^2

Answer 11

So how did that change to, 1/3 integral 3t^2 * `t^2` (t^3-3)^10 dt?
See how you have an extra t^2?

Answer 12

\[\large \int\limits t^2(t^3-3)^{10}dt \qquad \rightarrow \qquad \frac{1}{3}\int\limits 3t^2(t^3-3)^{10}dt\]I understand what you were doing here, you put a 1/3 and 3 in to fix things up. But you also threw in a t^2! woops! :)

Answer 13

OH yea! i only put the 3! instead i put du there -_-

Answer 14

oh i see XD heh

Answer 15

oh ok so it's 1/3 integral 3t^2 (t^3-3)^10 dt

1/3 integral u^10 du?

Answer 16

yes looks good!