Evaluate the integral

**Anonymous1** · November 13th 2009, 05:56 PM

Where k is a positive integer, evaluate:

$\int_{-1}^{1} \! (1 - x^2)^k \, dx.$

Possibly make a change of variables? Any help?

~Thanks

**Prove It** · November 13th 2009, 06:05 PM

Originally Posted by Anonymous1

Where k is a positive integer, evaluate:

$\int_{-1}^{1} \! (1 - x^2)^k \, dx.$

Possibly make a change of variables? Any help?

~Thanks

You can expand it using the Binomial Theorem...

**chengbin** · November 13th 2009, 06:11 PM

Sorry, wrong answer.

**redsoxfan325** · November 13th 2009, 06:18 PM

Originally Posted by chengbin

Look closer, the answer is 0 no matter what value of k.

$=\bigg [ \frac{1}{2k+2}(1-x^2)^{k+1} \bigg ]_{-1}^1$

$=\bigg ( \frac{1}{2k+2} (1-1)^{k+1} \bigg ) - \bigg ( \frac{1}{2k+2} (1-1)^{k+1} \bigg )$

$=0$

It can't be zero ever, because $(1-x^2)^k$ is an even function on a symmetric interval, so $\int_{-1}^1(1-x^2)^k\,dx=2\int_0^1(1-x^2)^k\,dx$ . Since $0\leq1-x^2\leq1$ on $[0,1]$ , this integral is always strictly positive.

Furthermore, the antiderivative is not $\frac{1}{2k+2}(1-x^2)^{k+1}$ . The derivative of that function is $-x(1-x^2)^k$ .

**chengbin** · November 13th 2009, 06:19 PM

Originally Posted by redsoxfan325

It can't be zero ever, because $(1-x^2)^k$ is an even function on a symmetric interval, so $\int_{-1}^1(1-x^2)^k\,dx=2\int_0^1(1-x^2)^k\,dx$ . Since $0\leq1-x^2\leq1$ on $[0,1]$ , this integral is always strictly positive.

Furthermore, the antiderivative is not $\frac{1}{2k+2}(1-x^2)^{k+1}$ . The derivative of that function is $-x(1-x^2)^k$ .

I realized that after I posted.

**redsoxfan325** · November 13th 2009, 06:21 PM

I realized that you realized that after I posted.

**Anonymous1** · November 13th 2009, 06:23 PM

Originally Posted by Prove It

You can expand it using the Binomial Theorem...

Really? The whole $x^2$ thing isn't an issue?

**redsoxfan325** · November 13th 2009, 06:25 PM

Nope. The binomial theorem works with anything.

**Prove It** · November 13th 2009, 06:28 PM

Originally Posted by Anonymous1

Where k is a positive integer, evaluate:

$\int_{-1}^{1} \! (1 - x^2)^k \, dx.$

Possibly make a change of variables? Any help?

~Thanks

Use the substitution $x = \sin{\theta}$ so that $dx = \cos{\theta}\,d\theta$ .

Also note that $1 - x^2 = 1 - \sin^2{\theta} = \cos^2{\theta}$ .

So $\int{(1 - x^2)^k\,dx} = \int{\cos^{2k}{\theta}\,\cos{\theta}\,d\theta}$

$= \int{\cos^{2k + 1}{\theta}\,d\theta}$ .

Thread: Evaluate the integral

LinkBack

Thread Tools

Display

Evaluate the integral

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Using a given property of an integral to evaluate and integral

Evaluate the integral using the FTC I

How do you evaluate this integral?

Search Tags