Question

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

Answer 1

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

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)

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

Answer 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???

Answer 3

du=-12t^2 dt

Answer 4

\[-1 \int\limits_{-2}^{0} -1 \cdot 2t^2(1-4t^3) dt \]

By the by -1(-1)=1
So we have