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JES2052106
Group Title
I need help with understanding the substitution part! thanks
Definite Integral w/ substitution
2t^2(14t^3) dt with the limits x=0 and x=2
 one year ago
 one year ago
JES2052106 Group Title
I need help with understanding the substitution part! thanks Definite Integral w/ substitution 2t^2(14t^3) dt with the limits x=0 and x=2
 one year ago
 one year ago

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myininaya Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{2}2t^2(14t^3) dt \] 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) \[\int\limits_{f(a)}^{f(b)}c u du =c \frac{u^2}{2}_{f(a)}^{f(b)}=\frac{c}{2}((f(b))^2f(a))^2)\]
 one year ago

JES2052106 Group TitleBest ResponseYou've already chosen the best response.0
thank you... the part I'm having a hard time with is the substitution... I know that i would state that u= 14t^3 and du = 12t^2 ??? so du/dt equals???
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
du=12t^2 dt
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
\[1 \int\limits_{2}^{0} 1 \cdot 2t^2(14t^3) dt \] By the by 1(1)=1 So we have dw:1352390289185:dw
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.0
u=f(t)=14t^3 So if t=0, then u=f(0)=14(0)^3=10=1 So if t=2, then u=f(2)=14(2)^3=14(8)=33
 one year ago

JES2052106 Group TitleBest ResponseYou've already chosen the best response.0
geez... i wish you went to my school.. hahaha... cuz the teacher i have for this class sucks!
 one year ago

JES2052106 Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{2}2t^2(14t^3) dt\] \[u=14t^3,u(0)=14\times 0^3=1,u(2)=14\times(2)^3=33\] \[du=12t^2dt\] \[\frac{1}{6}du=2t^2dt\] \[\frac{1}{6}\int_1^{33}udu\] etc
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
top line should be \[\int\limits_{0}^{2}2t^2(14t^3) dt\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
of course you can just multiply this one out and get \[\int_1^{2}(2t^28t^4)dt\] if you likke
 one year ago

JES2052106 Group TitleBest ResponseYou've already chosen the best response.0
where does the −1/6du=2t2dt come from.. how is that determined that is the part that i am getting confused on...
 one year ago
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