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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
 2 years ago
 2 years 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
 2 years ago
 2 years 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) \[\int\limits_{f(a)}^{f(b)}c u du =c \frac{u^2}{2}_{f(a)}^{f(b)}=\frac{c}{2}((f(b))^2f(a))^2)\]
 2 years 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???
 2 years ago

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

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