Definite Integral

**Hardwork** · October 30th 2010, 09:38 AM

Find $\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}$ .

**Unknown008** · October 30th 2010, 09:44 AM

Is that what you are looking for?

$\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}\times \sqrt{\sqrt{15}+2x}}{\sqrt{(2-x)(2+x)}(4-x^2)^2}\ dx$

I only simplified the brackets and replace an 'x' by 'times', and simplified the denominator a little and added dx.

As you can see, the denominator and numerator can be further simplified.

$\displaystyle \int_{-1}^{1}\frac{\sqrt{(\sqrt{15}-2x)(\sqrt{15}+2x)}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$

$\displaystyle \int_{-1}^{1}\frac{\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$

Can you continue now?

EDIT: Since the 'x' is an actual x, this becomes:

$\displaystyle \int_{-1}^{1}\frac{x\sqrt{15 - 4x^2}}{\sqrt{4 - x^2}(4-x^2)^2}\ dx$

**Ackbeet** · October 30th 2010, 09:46 AM

$\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{\sqrt{4-x^{2}}(4-x^2)^{2}}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$

Do you notice anything about the integrand?

**Hardwork** · October 30th 2010, 09:53 AM

@Unknown - It was an actual $x$ . I'm sorry I should have made it clear.

Originally Posted by Ackbeet

$\displaystyle \int_{-1}^{1}\frac{\sqrt{\sqrt{15}-2x}x\sqrt{\sqrt{15}+2x}}{\sqrt{2-x}\sqrt{2+x}(4-x^2)^2}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{\sqrt{4-x^{2}}(4-x^2)^{2}}\,dx=\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$

Do you notice anything about the integrand?

Looks reasonable now. I'll try to take it from there.

**Ackbeet** · October 30th 2010, 09:54 AM

You should be able to write the answer down right now, with no further computation.

**Unknown008** · October 30th 2010, 09:55 AM

Ok, it just seemed strange that you'd have an x in the middle like that =P

**Hardwork** · October 30th 2010, 10:02 AM

Originally Posted by Ackbeet

You should be able to write the answer down right now, with no further computation.

You mean I don't need to find the antiderivative? That would be amazing, but I'm not sure how to do it. Is it to do with the domains of $\sqrt{15-4x^2}$ and $\sqrt{(4-x^2)^5}$ ?

**Ackbeet** · October 30th 2010, 10:09 AM

You mean I don't need to find the antiderivative?

Not in this case. That's because the integrand plus your particular limits have a certain property that allows you to use a fantastic shortcut. What happens to the integrand if you replace x with -x?

**Hardwork** · October 30th 2010, 10:15 AM

Originally Posted by Ackbeet

Not in this case. That's because the integrand plus your particular limits have a certain property that allows you to use a fantastic shortcut. What happens to the integrand if you replace x with -x?

Limits flip and the sign changes - i.e. we get $\displaystyle -\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.<br />$

**Ackbeet** · October 30th 2010, 10:33 AM

Actually, the limits flip, but the overall sign does NOT change. That's because you get one minus sign for the differential, and one minus sign for the lone x in the numerator. So you get

$\displaystyle \int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx=\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx.$

Now, what happens when you flip the right integral's limits back to what they are on the LHS?

**Hardwork** · October 30th 2010, 11:35 AM

You are right. I forgot to account for the minus from the differential.

Originally Posted by Ackbeet

Now, what happens when you flip the right integral's limits back to what they are on the LHS?

The integral on the right is

$\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

Flipping the limits to what they were

$\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

I get the integral in the left.

**Ackbeet** · October 30th 2010, 11:47 AM

No, no. What usually happens when you flip limits? (Scroll down the link to just below the Conventions section).

**Hardwork** · October 30th 2010, 11:59 AM

Originally Posted by Ackbeet

No, no. What usually happens when you flip limits? (Scroll down the link to just below the Conventions section).

Oh, I knew I was missing something.

So $\int_{1}^{-1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

So $\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx = -\int_{-1}^{1}\frac{\sqrt{15-4x^{2}}x}{(4-x^2)^{5/2}}\,dx$

**Ackbeet** · October 30th 2010, 12:25 PM

So, if y = -y, what is y?

**Hardwork** · October 30th 2010, 12:45 PM

Originally Posted by Ackbeet

So, if y = -y, what is y?

$0$

That's it?

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