Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JES2052106

  • 2 years ago

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

  • This Question is Closed
  1. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{0}^{2}2t^2(1-4t^3) dt \] 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) \[\int\limits_{f(a)}^{f(b)}c u du =c \frac{u^2}{2}|_{f(a)}^{f(b)}=\frac{c}{2}((f(b))^2-f(a))^2)\]

  2. JES2052106
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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???

  3. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    du=-12t^2 dt

  4. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[-1 \int\limits_{-2}^{0} -1 \cdot 2t^2(1-4t^3) dt \] By the by -1(-1)=1 So we have |dw:1352390289185:dw|

  5. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  6. JES2052106
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    geez... i wish you went to my school.. hahaha... cuz the teacher i have for this class sucks!

  7. JES2052106
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  8. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{0}^{2}2t^2(1-4t^3) dt\] \[u=1-4t^3,u(0)=1-4\times 0^3=1,u(-2)=1-4\times(-2)^3=33\] \[du=-12t^2dt\] \[-\frac{1}{6}du=2t^2dt\] \[-\frac{1}{6}\int_1^{33}udu\] etc

  9. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    top line should be \[\int\limits_{0}^{-2}2t^2(1-4t^3) dt\]

  10. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    of course you can just multiply this one out and get \[\int_1^{-2}(2t^2-8t^4)dt\] if you likke

  11. JES2052106
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where does the −1/6du=2t2dt come from.. how is that determined that is the part that i am getting confused on...

  12. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.