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# some integration questions!!! watch

1. There are five questions here so if anyone is able to show me how to do even just one of them I would be really grateful.

1. Use integration by parts to determine the exact value of ∫3xsin2x dx. between the limits of ½pi and 0. (answer = -0.75pi)

2. Use integration by parts to determine ∫x(e^2x) dx. between limits of 1/3 and 0. [answer = 1/4 - 1/12.(e^2/3)]

3. Use the substituition u=lnx to show that ∫1/(x√(lnx)) dx. between limits of e^2 and e [answer = 2√2 - 2]

4. Use the substituition u = 4 + x² to show that ∫x³/(√(4+x²)) dx.between limits of 1 and 0 [answer = 1/3(16-7√5)]

5. By using the substitution u = 3x + 1 show that ∫x/(3x+1)² dx. between limits of 1 and 0 [answer = (2/9)ln2 - (1/12)]

Thanks a billion!!!! xxxxxxxxx

(ok, the questions are right now. sorry for confusing people)
2. "1. Use integration by parts to determine the exact value of ∫3xsin2x dx. between the limits of ½pi and 0. (answer = -0.75pi)"

In Int[ u v'] = uv - Int[ u' v]

set u = 3x and v = sin2x

2. Use integration by parts to determine ∫x(e^2x) dx. between limits of 1/3pi and 0. [answer = 1/4 - 1/12.(e^2/3)]

This time u = x and v = e^(2x)

3. Use the substituition u=lnx to show that ∫1/(x√(lnx)) dx. between limits of e2 and e = 2√2 - 2

What is e2? Do you mean e^2 or 2e? Does e mean e=2.718... or is e=2(rt(2)-1)?

4. Use the substituition u = 4 + x² to show that ∫x³/(√(4+x²)) dx.
= 1/3(16-7√5)

Note du = 2x dx so that x^3 = (u-4) du/2

5. By using the substitution u = 3x + 1 show that ∫x/(3x+1)² dx. =
(2/9)ln2 - (1/12)

Where are you getting stuck?
3. Here are the first steps. Post back if you need further help.

1.
u = 3x, dv/dx = sin2x
v = 3, v = -(1/2)cos2x
=> ∫ 3x.sin2x dx = -(3/2)x.cos2x + (3/2) ∫ cos2x dx

2.
u = x, dv/dx = e^(2x)
du/dx = 1, v = (1/2)e^(2x)
∫ x.e^(2x) dx = (1/2)x.e^(2x) - (1/2) ∫ e^(2x) dx

3.
u = lnx, du/dx = 1/x
=> ∫ 1/(x . sqrt[lnx]) dx = ∫ 1/sqrt[u] du = ∫ u^(-1/2) du

4.
u = 4 + x², du/dx = 2x
=> ∫ (x².x)/(sqrt[4+x²]) dx = (1/2) ∫ (u-4)/sqrt[u] du = (1/2) ∫ u^(1/2) - 4u^(-1/2) du

5.
u = 3x + 1, du/dx = 3
∫ x/(3x+1)² dx = (1/3) ∫ (u-1)/u² du = (1/3) ∫ u^(-1) - u^(-2) du
4. Im not sure??!!!!

I know what i am meant to be doing but i get

1. 3pi/4 - 3

2. ¼ + (pi/6)e^(2pi/3) - ¼e^(2pi/3)

and the other three just didnt seem to be getting anywhere near what the actual answer was meant to be so i gave up.

I'll keep trying lol...

5. Are you sure that the answer for the first question is -3/4pi?

The book says it's 3/4pi, I think

The second ones not right as you are putting the wrong limits in!!!! it should be 1/3 and 0.
6. Q1 attached for clarification. Are you sure about the '-' sign though?
Attached Images

7. For question one the integrand is positive in that range so there's no way you could get a negative area
8. (Original post by fabz)
Are you sure that the answer for the first question is -3/4pi?

The book says it's 3/4pi, I think

The second ones not right as you are putting the wrong limits in!!!! it should be 1/3 and 0.
How can the area enclosed under a curve be negative? Areas are always positive!
If you end up with a negative value, treat the answer to be the positive of the negative answer you get, so:

-3/4pi >> 3/4pi square units
9. I just realised that! Doh!

Please could somebody post the working for the last question because I keep gatting (2/9) ln 2 - (27/4)???
10. See the attachment.
Attached Images

11. 1 and 0
12. Doh looks like my problem is just as much about reading the question properly as whether or not I can do the questions.

You are right about it being positive in the first one and the limit being just 1/3. Might explain why i couldnt get rid of pi =oS

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