anonymous
  • anonymous
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
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katieb
  • katieb
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myininaya
  • myininaya
\[\int\limits_{0}^{2}2t^2(1-4t^3) dt \] 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) \[\int\limits_{f(a)}^{f(b)}c u du =c \frac{u^2}{2}|_{f(a)}^{f(b)}=\frac{c}{2}((f(b))^2-f(a))^2)\]
anonymous
  • anonymous
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???
myininaya
  • myininaya
du=-12t^2 dt

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myininaya
  • myininaya
\[-1 \int\limits_{-2}^{0} -1 \cdot 2t^2(1-4t^3) dt \] By the by -1(-1)=1 So we have |dw:1352390289185:dw|
myininaya
  • myininaya
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
anonymous
  • anonymous
geez... i wish you went to my school.. hahaha... cuz the teacher i have for this class sucks!
anonymous
  • anonymous
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
anonymous
  • anonymous
\[\int\limits_{0}^{2}2t^2(1-4t^3) dt\] \[u=1-4t^3,u(0)=1-4\times 0^3=1,u(-2)=1-4\times(-2)^3=33\] \[du=-12t^2dt\] \[-\frac{1}{6}du=2t^2dt\] \[-\frac{1}{6}\int_1^{33}udu\] etc
anonymous
  • anonymous
top line should be \[\int\limits_{0}^{-2}2t^2(1-4t^3) dt\]
anonymous
  • anonymous
of course you can just multiply this one out and get \[\int_1^{-2}(2t^2-8t^4)dt\] if you likke
anonymous
  • anonymous
where does the −1/6du=2t2dt come from.. how is that determined that is the part that i am getting confused on...

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