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swin2013
Calculus integration with substitution problem
General integration t^2 (t^3 - 3)^10 dt is the u sub = t^3-3? du/dt = 3t^2 du = 3t^2dx
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
do you mean when you divide by 1/3?
mutiply by 1/3, or divide by 3, yes.
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?
\[\large du=3t^2 dt \qquad \rightarrow \qquad \color{orangered}{\frac{1}{3}du=t^2dt}\]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
Oh you have t^2 twice in there for some reason..? :O
\[\large \int \frac{1}{3}(t^3-3)^{10}\color{orangered}{3t^2\; dt}\]\[\large \color{orangered}{du=3t^2dt}\]
because the original equation has t^2 * (t^3-3)^10 dt
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?
\[\large \int\limits t^2(t^3-3)^{10}dt \qquad \rightarrow \qquad \frac{1}{3}\int\limits 3t^2(t^3-3)^{10}dt\]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! :)
OH yea! i only put the 3! instead i put du there -_-
oh ok so it's 1/3 integral 3t^2 (t^3-3)^10 dt 1/3 integral u^10 du?