Integration

**Envy** · June 3rd 2009, 02:36 AM

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?

**CaptainBlack** · June 3rd 2009, 02:45 AM

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

**Prove It** · June 3rd 2009, 02:48 AM

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.

**Envy** · June 3rd 2009, 02:54 AM

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 $-4{x}^\frac{1}{2}$ !

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