# Math Help - definite integral

1. ## definite integral

just want to ask what will i do first if i will be given a problem like this:

$\int_7^7 \sqrt[3](|t|+1)dt$

btw the lower limit must be -7 and the |t|+1 must be inside the cube root. I dont know how to put in in latex sorry

thanks

2. Originally Posted by FailCalculus
just want to ask what will i do first if i will be given a problem like this:

$\int_7^7 \sqrt[3](|t|+1)dt$

btw the lower limit must be -7 and the |t|+1 must be inside the cube root. I dont know how to put in in latex sorry

thanks
You will need to treat $|t|$ as a hybrid function.

Note that

$|t| = \begin{cases}\phantom{-}t\textrm{ if }t \geq 0\\ -t\textrm{ if }t < 0\end{cases}$.

Therefore

$\int_{-7}^7{\sqrt[3]{|t| + 1}\,dt} = \int_{-7}^0{\sqrt[3]{-t + 1}\,dt} + \int_0^7{\sqrt[3]{t + 1}\,dt}$.

Can you go from here? You will need to use $u$ substitutions...

3. ok thanks just to clarify things. so i need to use u=|t|+1 and use the general integration formula?

4. $\displaystyle \int_{-7}^7 \sqrt[3]{|t|+1}dt$

note that for $f(t) = \sqrt[3]{|t|+1}$ , $f(t) = f(-t)$ , an even function.

using symmetry about the y-axis ...

$\displaystyle \int_{-7}^7 \sqrt[3]{|t|+1}\,dt = 2 \int_0^7 \sqrt[3]{t+1}\,dt = \frac{3}{2}\left[(t+1)^{\frac{4}{3}\right]_0^7 = \frac{45}{2}$

5. Originally Posted by FailCalculus
ok thanks just to clarify things. so i need to use u=|t|+1 and use the general integration formula?
No, like I said, you have to break it into two integrals.

So for the first, use $u = t + 1$ and in the second use $v = -t + 1$.