Integration.

Double integral for x limit, 0 to 1 and y limit x to root(x) with respect to dydx

Doing the inner integral gives 4(x^2)y + 4y^3/3, sub in the limits we get

(4x^5/2 + 4(x^3/2)/3) - (4x^3 + 4x^3/3)

Integrate again, we get (8(x^7/2)/7 + 8(x^5/2)/15) - (x^4 + 4x^4/12)

We forget the minus as that is related to the zero limit so focus on the upper limit 1 and yields an area of 88/105 and the answer is supposed to be 36/105. Am I wrong or is the sheet wrong?

Thanks in advance.

1^5/2 = 1/2, 1^7/2 = 1/2 thus giving me a fraction different to yours. How did you get 8/7, 8/15?

$1^{5/2}=\sqrt{1^5}=\sqrt{1}=1$.

1^5/2 = 1/2, 1^7/2 = 1/2 thus giving me a fraction different to yours. How did you get 8/7, 8/15?

(brackets, brackets, brackets!)

"Evaluate the double integral for x, from 1 to e and y, from 1 to y

y/x . dxdy

Reverse the order of integration and re-evaluate."

So from dxdy I can get the answer they want on the sheet which is, (1+e^2)/4 but the order reversed, dydx, I get (e^2 - 3)/4

So, limits are for y, from 1 to x and for x, 1 to e. Is this correct?

Draw a diagram for the function, and it'll become clear where to integrate across.

I've just read you've put integrating y from 1 to y, this makes no sense, do you mean integrating x from 1 to y?

Oh my bad, I meant 1 to e for my limits as y. Will change that now. That is what is given, I have to re-evaluate to integrate dydx.

Here's a diagram. Does this help?

double integral for y, from 0 to pi/2 and x, from 0 to y

$\cos(2y) \times \sqrt{1 - k^{2}sin^{2}x} dxdy$

k is a constant.

So, (1-k^2*sin^2(x))^1/2. How do I integrate this?

A substitution of u=(1-k^2sin^2(x)) works, rearranging to get sin(x) and using triangles to get cos(x) and then partial fractions, have yet to think of a more elegant method.

Thanks for your message!

I rearranged the integral to do it in the dxdy direction, I am left with

sin(2x)*(1-k^2*sin^2(x))^1/2 dx with the limits from 0 to pi/2 but that doesn't help at all.

The answer is supposed to be

((1-k^2)^3/2 -1)/3k^2

Did you get near that?

I'm going to concentrate on getting you to the integral of $(1-k^2sin^2(x))^{\frac{1}{2}}$ and I'll let you do the rest.

$\displaystyle \int (1-k^2sin^2(x))^{\frac{1}{2}} \ dx$

$u =1-k^2sin^2(x)$

$sin(x) = \dfrac{(1-u)^{\frac{1}{2}}}{k}$

so $cos(x) = \dfrac{(k^2-1+u)^{\frac{1}{2}}}{k}$

$\dfrac{du}{dx} = -2k^2sin(x)cos(x)$

$\displaystyle -\dfrac{1}{2} \int \dfrac{u^{\frac{1}{2}}}{k^2sin(x)cos(x)} \ du$

$\displaystyle -\dfrac{1}{2} \int \dfrac{u^{\frac{1}{2}}}{\sqrt{(k^2-1+u)(1-u)}} \ du$

Partial fractions and then integrate.

Thanks mate. Much appreciated!

Actually mate, isn't it supposed to be k -1+u and not k^2 -1+u? cos^2(x) = 1-sin^2(x),

So we have 1 - ((1-u)^1/2)k etc etc?