Integral Refresher Question

• Jan 18th 2011, 12:45 PM
Crell
Integral Refresher Question
Hi, I'm having a little bit of trouble with a integration question, here is the text from the question.

Find the antiderivitive of the given functions:

$\displaystyle F(x) = x(1-x^2)^7$

Basically I just have no idea how to expand $\displaystyle (1 - x^2)^7$
• Jan 18th 2011, 12:48 PM
dwsmith
Quote:

Originally Posted by Crell
Hi, I'm having a little bit of trouble with a integration question, here is the text from the question.

Find the antiderivitive of the given functions:

$\displaystyle F(x) = x(1-x^2)^7$

Basically I just have no idea how to expand $\displaystyle (1 - x^2)^7$

Dont expand!

Use substition.

u=1-x^2
• Jan 18th 2011, 12:53 PM
HallsofIvy
Well, you could expand it using the "binomial theorem": $\displaystyle (a+ b)^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix}a^ib^{n- i}$
For $\displaystyle (1- x^2)^7= 1- 7x^2+ 21x^4+ \cdot\cdot\cdot- 7x^{12}+ x^{14}$

But you don't want to do that. Instead, make the substituition $\displaystyle u= 1- x^2$ so that $\displaystyle (1- x^2)^7= u^7$ and $\displaystyle du= -2 xdx$.
• Jan 18th 2011, 01:20 PM
Crell
Quote:

Originally Posted by HallsofIvy
$\displaystyle du= -2 xdx$.

Thank you much, this is the part I was missing.

$\displaystyle \int x(1-x^2)^7dx$

Aside:

Let $\displaystyle u = 1-x^2$

Derive this equation

$\displaystyle \frac{du}{dx} = -2x$

Solve for $\displaystyle dx$

$\displaystyle dx = \frac{du}{-2x}$

Manipulate the first equation to make it appear like $\displaystyle -2x(u)^7dx$ so that when you substitute dx, the x is eliminated.

Therefore:

$\displaystyle -\frac{1}{2} \int -2x(u)^7dx$

Substitute $\displaystyle dx = \frac{du}{-2x}$ into first equation
$\displaystyle -\frac{1}{2} \int \frac {-2x(u)^7du}{-2x}$

$\displaystyle -\frac{1}{2} \int {(u)^7du}$

Integrate, sub in u and simplfy.

$\displaystyle - \frac{1}{2} [\frac {(u)^8}{8}]$

$\displaystyle -\frac{1}{2} [\frac {(1-x^2)^8}{8}]$

$\displaystyle -\frac {(1-x^2)^8}{16}$
• Jan 18th 2011, 01:26 PM
Plato
Did you differentiate that answer to cheek it for yourself?
• Jan 18th 2011, 01:35 PM
Crell
Have now, thanks.
• Jan 18th 2011, 01:37 PM
Quacky
This is a perfect place to use my favourite rule of integration:

$\displaystyle \displaystyle\int{f'(x)f(x)^n} dx=\frac{1}{n+1}f(x)^{n+1}+c$

So let $\displaystyle f(x)=(1-x^2)$
$\displaystyle f'(x)=-2x$
$\displaystyle n=7$

$\displaystyle f'(x)f(x)^n= -2x(1-x^2)^7$
You'll notice that it isn't quite a match to what you are trying to integrate. However, if you divide by $\displaystyle -2$, it is exactly what you are trying to integrate. Therefore, we can apply the rule as normal, but we must remember to divide our final answer by -2.

$\displaystyle \displaystyle\int x(1-x^2)^7 dx$

$\displaystyle =\displaystyle\frac{-1}{2}\times\frac{(1-x^2)^8}{8} + C$

$\displaystyle =\displaystyle -\frac{(1-x^2)^8}{16} + C$

I think in this case, this method is much nicer than substitution - if you spot the rule, it will take a few lines of working to obtain the full solution with no messy work required at all.
• Jan 18th 2011, 02:29 PM
TheCoffeeMachine
In speaking of the binomial theorem, you can find the anti-derivative
without performing an expansion, but in somewhat undesirable forms:

$\displaystyle \displaystyle \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+2}}{2k+2}\binom{7}{k}$ or $\displaystyle \displaystyle \sum_{k=0}^{7}\frac{(-1)^k(x)^{2k+2}}{2k+2}\binom{7}{k}$.

Plus the constant of integration. The u-sub does the job far better.

Spoiler:
1.

$\displaystyle \displaystyle x(1-x^2)^7 = x\left(\sum_{k=0}^{\infty}\binom{7}{k}(-1)^kx^{2k}\right) = \sum_{k=0}^{\infty}\binom{7}{k}(-1)^{k}x^{2k+1}.$

\displaystyle \begin{aligned}\displaystyle \therefore \int x(1-x^2)^7\;{dx} & = \int \left(\sum_{k=0}^{\infty}\binom{7}{k}(-1)^{k}x^{2k+1}\right)\;{dx} \\& = \sum_{k=0}^{\infty} \left(\int \binom{7}{k}(-1)^{k}x^{2k+1}\;{dx}\right) \\& = \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+2}}{2k+2}\binom{7}{k}+C.\end{aligned}

2.

\displaystyle \displaystyle \begin{aligned}x(1-x^2)^7 & = x\left(\sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k}\right) \\& = \sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k+1}.\end{aligned}

\displaystyle \displaystyle \begin{aligned} \therefore \int x(1-x^2)^7\;{dx} & = \int \left(\sum_{k=0}^{7}\binom{7}{k}(-1)^kx^{2k+1}\right)\;{dx} \\& = \sum_{k=0}^{7} \left(\int \binom{7}{k}(-1)^kx^{2k+1}\;{dx}\right) \\& = \sum_{k=0}^{7}\frac{(-1)^k(x)^{2k+2}}{2k+2}\binom{7}{k}+C.\end{aligned}