1. ## Integration

I can't figure out why when integrating

$

-4{x}^\frac{3}{2}

$

you get

$

-\displaystyle{\frac{8}{3}{x}}^\frac{3}{2}

$

and not

$
-8{x}^\frac{3}{2}

$

Why does it become a fraction?

2. Originally Posted by Envy
I can't figure out why when integrating

$

-4{x} \frac{3}{2}

$

you get

$

-\displaystyle{\frac{8}{3}{x}}
\frac{3}{2}

$

and not

$
-8{x} \frac{3}{2}

$

Why does it become a fraction?
The rule is:

$\int k x^r\ dx= \frac{k}{r+1}x^{r+1} +C$

In your case $k=-4$, $r=3/2$ and so:

$\int -4 x^{3/2}\ dx= -\ \frac{4}{(5/2)}x^{5/2} +C=-\ \frac{8}{5}x^{5/2} + C$

CB

3. Originally Posted by Envy
I can't figure out why when integrating

$

-4{x}^\frac{3}{2}

$

you get

$

-\displaystyle{\frac{8}{3}{x}}^\frac{3}{2}

$

and not

$
-8{x}^\frac{3}{2}

$

Why does it become a fraction?
You shouldn't get either of those answers you've listed...

If $f(x) = ax^n$ then it's antiderivative is $F(x) = \int{ax^n\,dx} = \frac{ax^{n + 1}}{n + 1}+C$.

In other words, you add 1 to your power, and then divide by the new power.

In your case you have $-4x^{\frac{3}{2}}$.

Add 1 to the power and you get $\frac{5}{2}$.

Divide by your new power and you get

$-4\div\frac{5}{2}$

$= -4\times\frac{2}{5}$

$= -\frac{8}{5}$.

So your antiderivative is $-\frac{8x^{\frac{5}{2}}}{5} + C$ where C is a constant.

4. Originally Posted by CaptainBlack

$-\ \frac{4}{(5/2)}x^{5/2} +C=-\ \frac{8}{5}x^{5/2} + C$
Thanks, I forgot to flip the fraction and then times it

Originally Posted by Prove It
You shouldn't get either of those answers you've listed...
Totally correct, I wrote down the wrong quesion as $-4{x}^\frac{3}{2}$ which was my partially intergrated answer using ${ax^{n + 1}}$, the actual question was $-4{x}^\frac{1}{2}$!