# integration homework help

• Dec 11th 2012, 10:15 AM
asilvester635
integration homework help
I rewrote this as

= t(t - 4)^1/3.... do I combine them and integrate after?
• Dec 11th 2012, 10:35 AM
Zouzz
Re: integration homework help
Integration by parts, like the other one i commented on u(x)=(t-4)^(1/3), v(x)=t :D
• Dec 11th 2012, 10:37 AM
skeeter
Re: integration homework help
$\displaystyle \int t \sqrt[3]{t-4} \, dt$

$\displaystyle u = t-4$

$\displaystyle t = u+4$

$\displaystyle du = dt$

substitute ...

$\displaystyle \int (u+4) \cdot \sqrt[3]{u} \, du$

can you finish?
• Dec 11th 2012, 10:41 AM
Zouzz
Re: integration homework help
...or that XD
• Dec 11th 2012, 10:56 AM
asilvester635
Re: integration homework help
this is different on how i learned it.... so i integrate the right side??? sorry idk x/
• Dec 11th 2012, 12:01 PM
asilvester635
Re: integration homework help
wait i got it... im working on it at the moment....
okay heres what i've got so far

u = t - 4
du = dt
t = u + 4

= ^3√u ( u + 4) + c
= u^1/3 (u + 4) + c

now do i multiply u^1/3 (u + 4) together and integrate?
• Dec 11th 2012, 02:44 PM
Prove It
Re: integration homework help
Yes.
• Dec 19th 2012, 05:20 AM
x3bnm
Re: integration homework help
$\displaystyle \text{We know that } \int u \, dv=uv-\int v \, du.\!$

So for $\displaystyle \int t\sqrt[3]{t-4} \;\;dt$ let $\displaystyle u = t \therefore du = dt$
And let $\displaystyle dv = \sqrt[3]{t-4} \;\;dt \therefore v = \frac{3}{4}(t - 4)^{\frac{4}{3}}$

\displaystyle \begin{align*}\int t\sqrt[3]{t-4} \;\;dt =& \frac{3}{4}t(t-4)^{\frac{4}{3}} - \int \frac{3}{4}(t - 4)^{\frac{4}{3}} \\ =& \frac{3}{4}t(t-4)^{\frac{4}{3}} - \frac{9}{28}(t-4)^{\frac{7}{3}} + C \\ =& \frac{3}{7}(t-4)^{\frac{4}{3}}[\frac{7t}{4} - \frac{3t}{4} + \frac{12}{4}] + C \\ =& \frac{3}{7}(t-4)^{\frac{4}{3}}(t+3) + C\end{align*}

$\displaystyle \therefore \int t\sqrt[3]{t-4} \;\;dt = \frac{3}{7}(t-4)^{\frac{4}{3}}(t+3) + C$

Check:

int x(x-4)^(1/3) dx - Wolfram|Alpha
• Dec 19th 2012, 11:10 AM
hollywood
Re: integration homework help
Or starting from skeeter's post,

$\displaystyle \int (u+4) \cdot \sqrt[3]{u} \, du =\int u^\frac{4}{3} + 4u^\frac{1}{3} \, du =$

$\displaystyle \frac{3}{7}u^\frac{7}{3}+ 3 u^\frac{4}{3} + C = \frac{3}{7}u^\frac{4}{3} (u + 7) + C=$

$\displaystyle \frac{3}{7} (t-4)^\frac{4}{3} (t+3) + C$

- Hollywood
• Mar 29th 2013, 04:53 AM
vicwilson123
Re: integration homework help
• Mar 29th 2013, 06:56 AM
HallsofIvy
Re: integration homework help
Quote:

Originally Posted by asilvester635
wait i got it... im working on it at the moment....
okay heres what i've got so far

u = t - 4
du = dt
t = u + 4

= ^3√u ( u + 4) + c
= u^1/3 (u + 4) + c

now do i multiply u^1/3 (u + 4) together and integrate?

You have not yet integrated so there should be no "+ c".