Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

I need help with understanding the substitution part! thanks Definite Integral w/ substitution 2t^2(1-4t^3) dt with the limits x=0 and x=-2

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\int\limits_{0}^{2}2t^2(1-4t^3) dt \] So we have something like this: \[\int\limits_{a}^{b}c f'(t) f(t) dt \] u=f(t) du/dt=f'(t) du=f'(t) dt If t=a, u=f(a) If t=b, u=f(b) \[\int\limits_{f(a)}^{f(b)}c u du =c \frac{u^2}{2}|_{f(a)}^{f(b)}=\frac{c}{2}((f(b))^2-f(a))^2)\]
thank you... the part I'm having a hard time with is the substitution... I know that i would state that u= 1-4t^3 and du = 12t^2 ??? so du/dt equals???
du=-12t^2 dt

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[-1 \int\limits_{-2}^{0} -1 \cdot 2t^2(1-4t^3) dt \] By the by -1(-1)=1 So we have |dw:1352390289185:dw|
u=f(t)=1-4t^3 So if t=0, then u=f(0)=1-4(0)^3=1-0=1 So if t=-2, then u=f(-2)=1-4(-2)^3=1-4(-8)=33
geez... i wish you went to my school.. hahaha... cuz the teacher i have for this class sucks!
can you explain this part.. but with the numbers plugged in and how you got them?? u=f(t) du/dt=f'(t) du=f'(t) dt
\[\int\limits_{0}^{2}2t^2(1-4t^3) dt\] \[u=1-4t^3,u(0)=1-4\times 0^3=1,u(-2)=1-4\times(-2)^3=33\] \[du=-12t^2dt\] \[-\frac{1}{6}du=2t^2dt\] \[-\frac{1}{6}\int_1^{33}udu\] etc
top line should be \[\int\limits_{0}^{-2}2t^2(1-4t^3) dt\]
of course you can just multiply this one out and get \[\int_1^{-2}(2t^2-8t^4)dt\] if you likke
where does the −1/6du=2t2dt come from.. how is that determined that is the part that i am getting confused on...

Not the answer you are looking for?

Search for more explanations.

Ask your own question