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# Integration. Watch

1. 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?

2. (Original post by boromir9111)
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?

3. (Original post by Clarity Incognito)
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?
4. (Original post by boromir9111)
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?

.
5. (Original post by james.h)

.
Cheers mate.
6. (Original post by boromir9111)
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!)
7. (Original post by Clarity Incognito)

(brackets, brackets, brackets!)
Oh dear me. What a silly error to make lol. Cheers for that mate, much appreciated
8. "Evaluate the double integral for x, from 1 to e and y, from 1 to e

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?
9. (Original post by boromir9111)
"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.
10. (Original post by Don John)
Draw a diagram for the function, and it'll become clear where to integrate across.
Yeah, I did draw a diagram but I am fishy on it tbh, is my limits wrong?
11. 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?
12. (Original post by Don John)
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.
13. (Original post by boromir9111)
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?
Attached Images

14. (Original post by Don John)
Here's a diagram. Does this help?
Actually that image worked perfect mate. I totally understand it now!!! thanks for that. + repped.
15. double integral for y, from 0 to pi/2 and x, from 0 to y

k is a constant.

So, (1-k^2*sin^2(x))^1/2. How do I integrate this?
16. (Original post by boromir9111)
double integral for y, from 0 to pi/2 and x, from 0 to y

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.
17. (Original post by Clarity Incognito)
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.

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?
18. (Original post by boromir9111)

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 and I'll let you do the rest.

so

Partial fractions and then integrate.
19. (Original post by Clarity Incognito)
I'm going to concentrate on getting you to the integral of and I'll let you do the rest.

so

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?
20. (Original post by boromir9111)
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?

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