## swin2013 Group Title Calculus integration with substitution problem one year ago one year ago

1. swin2013

General integration t^2 (t^3 - 3)^10 dt is the u sub = t^3-3? du/dt = 3t^2 du = 3t^2dx

2. zepdrix

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

3. swin2013

do you mean when you divide by 1/3?

4. zepdrix

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

5. swin2013

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?

6. zepdrix

$\large du=3t^2 dt \qquad \rightarrow \qquad \color{orangered}{\frac{1}{3}du=t^2dt}$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

7. zepdrix

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

8. zepdrix

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

9. swin2013

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

10. swin2013

and u sub = 3t^2

11. zepdrix

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?

12. zepdrix

$\large \int\limits t^2(t^3-3)^{10}dt \qquad \rightarrow \qquad \frac{1}{3}\int\limits 3t^2(t^3-3)^{10}dt$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! :)

13. swin2013

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

14. zepdrix

oh i see XD heh

15. swin2013

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

16. zepdrix

yes looks good!