Calculus Homework Help

[jai6638]

Hey

I was given certain homework probs on integration ( which I dont think I have a good grasp on .. ) .I'd appreciate it if you guys could confirm if the answers are correct.

...............

Thanks much..

 

[Kyteland]

If you just want to check your answers, try here: http://integrals.wolfram.com/index.jsp

Edit: Just looked at your answer again. You shouldn't be taking u=cos(e^(2x)). That's not what the substitution rule states.

? f(g(x))*g'(x) dx = ? f(u) du

You should be taking u to be what is inside of the function, ie u=e^(2x). That reduces the problem to ? sin(u)*cos(u)/2 du = ? sin(2*u)/4 du = -cos(2*u)/8 = -cos(2*e^(2*x))/8

 

[jai6638]

Originally posted by: Kyteland
If you just want to check your answers, try here: http://integrals.wolfram.com/index.jsp

Edit: Just looked at your answer again. You shouldn't be taking u=cos(e^(2x)). That's not what the substitution rule states.

? f(g(x))*g'(x) dx = ? f(u) du

You should be taking u to be what is inside of the function, ie u=e^(2x). That reduces the problem to ? sin(u)*cos(u)/2 du = ? sin(2*u)/4 du = -cos(2*u)/8 = -cos(2*e^(2*x))/8

Thanks for the link..

why did you divide by 2 in ? sin(u)*cos(u)/2 du ?

Could you please help me out with question 3? my answer's wrong and I dunno how to go about solving it.

Thanks much.

 

[Cal166]

Answer: 42

 

[jai6638]

Originally posted by: Cal166
Answer: 42

?

 

[Atheus]

Originally posted by: jai6638

Originally posted by: Cal166
Answer: 42

?

Hitchhiker's guide to the galaxy

 

[jai6638]

Originally posted by: Atheus

Originally posted by: jai6638

Originally posted by: Cal166
Answer: 42

?

Hitchhiker's guide to the galaxy

oh ok ..

anyhow, i'd appreciate some help with the stuff in my previous post..

Thanks

 

[MSCoder610]

4. Set u = the denominator instead, i.e. u = 3+ cos x, so -du = sin x which is the numerator, integral of 1/u du which you should know, then negate the whole thing (since it was -du)

 

[jai6638]

Originally posted by: MSCoder610
4. Set u = the denominator instead, i.e. u = 3+ cos x, so -du = sin x which is the numerator, integral of 1/u du which you should know, then negate the whole thing (since it was -du)

got the answer as -ln(cosx+3) which is correct ( from the link posted by Kyteland )


Thanks.

 

[QED]

Originally posted by: jai6638

3) find integarl of 1-x^n and show that it is equal to (n)/(n-1)

I am lost here since I do not know how to take the exponent into consideration when simplifying. I got x-ax^(n-1) which doesnt seem right.

The integral of 1-x^n is x - ( x ^ (n+1) ) / (n+1)...

I'm not sure where they get the n/(n-1) in there... the closest thing I can think of is if you try to take the definite integral of (1-x^n) over the interval [0,1] you end up with n/(n+1).

 

[George P Burdell]

http://www.at-wiki.com/wiki/index.php/I_have_a_math/chemistry/physics_question

 

[SpecialEd]

for #4, u=3+cos(x), du=-sin(x)dx,

 

[Kyteland]

Originally posted by: jai6638

Originally posted by: Kyteland
If you just want to check your answers, try here: http://integrals.wolfram.com/index.jsp

Edit: Just looked at your answer again. You shouldn't be taking u=cos(e^(2x)). That's not what the substitution rule states.

? f(g(x))*g'(x) dx = ? f(u) du

You should be taking u to be what is inside of the function, ie u=e^(2x). That reduces the problem to ? sin(u)*cos(u)/2 du = ? sin(2*u)/4 du = -cos(2*u)/8 = -cos(2*e^(2*x))/8

Thanks for the link..

why did you divide by 2 in ? sin(u)*cos(u)/2 du ?

Could you please help me out with question 3? my answer's wrong and I dunno how to go about solving it.

Thanks much.

if u=e^(2x) then du=2*e^(2x)
e^(2x) = (1/2) du. That's where that comes from.

for 3) are you sure you typed everything right? ? 1-x^n = x - x^(n+1)/(n+1). I don't see how that reduces to n/(n+1)

 

[jai6638]

I apologize. Didnt type the question right.

for question 3, it actually is: find the area when 0<=x<=1 ... I tried doing it and got 1/n+1 ( taking u=n ).

How do you guys solve such questions? Do you guys use "u-substitution" or is there some other method which i'm unaware of?

 

[QED]

Originally posted by: jai6638
I apologize. Didnt type the question right.

for question 3, it actually is: find the area when 0<=x<=1 ... I tried doing it and got 1/n+1 ( taking u=n ).

How do you guys solve such questions? Do you guys use "u-substitution" or is there some other method which i'm unaware of?

Area is just finding the integral of f(x) = 1 - x^n, and evaluating it at F[0] and F[1].

In this case, the intergral of 1-x^n is x - ( x^ (n+1) ) / (n+1), since n is just a constant.

Now F[0] is ( 0 - ( 0^(n+1) / (n+1))) = 0, and
F[1] = (1 - (1^(n+1)/(n+1))) = 1 - 1/(n+1) = ( (n+1) - 1 ) / (n+1) = n/(n+1).

 

[jai6638]

Originally posted by: MathMan

Originally posted by: jai6638
I apologize. Didnt type the question right.

for question 3, it actually is: find the area when 0<=x<=1 ... I tried doing it and got 1/n+1 ( taking u=n ).

How do you guys solve such questions? Do you guys use "u-substitution" or is there some other method which i'm unaware of?

Area is just finding the integral of f(x) = 1 - x^n, and evaluating it at F[0] and F[1].

In this case, the intergral of 1-x^n is x - ( x^ (n+1) ) / (n+1), since n is just a constant.

Now F[0] is ( 0 - ( 0^(n+1) / (n+1))) = 0, and
F[1] = (1 - (1^(n+1)/(n+1))) = 1 - 1/(n+1) = ( (n+1) - 1 ) / (n+1) = n/(n+1).

damn! thats what I did at first but then decided to use u substitution and got a different answer.. doh!

 

[ShadowBlade]

note: i know nothing about calculus

i need the "expr,var[,low,up]" and ill have your answer in under 5 secs

 

[jai6638]

Hey... Needed some help again..

Integral of ln[x]

Not sure how to go about it.. Cant use integration by parts since there is only one term and using u-substition will not help me get rid of the ln[x]

Integral of (x)(lnX)

I used U substituion and set u=x but I cant seem to get rid of the Ln X for some reason.. any ideas?

Integral of (lnx)^2

same prob

Any help is much appreciated.. Thanks

 

[hypn0tik]

Originally posted by: jai6638
Hey... Needed some help again..

Integral of ln[x]

Not sure how to go about it.. Cant use integration by parts since there is only one term and using u-substition will not help me get rid of the ln[x]

Integral of (x)(lnX)

I used U substituion and set u=x but I cant seem to get rid of the Ln X for some reason.. any ideas?

Integral of (lnx)^2

same prob

Any help is much appreciated.. Thanks

Integration by parts. u = ln(x), dv = dx

 

[jai6638]

Originally posted by: hypn0tik

Originally posted by: jai6638
Hey... Needed some help again..

Integral of ln[x]

Not sure how to go about it.. Cant use integration by parts since there is only one term and using u-substition will not help me get rid of the ln[x]

Integral of (x)(lnX)

I used U substituion and set u=x but I cant seem to get rid of the Ln X for some reason.. any ideas?

Integral of (lnx)^2

same prob

Any help is much appreciated.. Thanks

Integration by parts. u = ln(x), dv = dx

but dont I need a U and a V? In this case, if I assume dx to be V' , then what would V be?

 

[hypn0tik]

Originally posted by: jai6638

Originally posted by: hypn0tik

Originally posted by: jai6638
Hey... Needed some help again..

Integral of ln[x]

Not sure how to go about it.. Cant use integration by parts since there is only one term and using u-substition will not help me get rid of the ln[x]

Integral of (x)(lnX)

I used U substituion and set u=x but I cant seem to get rid of the Ln X for some reason.. any ideas?

Integral of (lnx)^2

same prob

Any help is much appreciated.. Thanks

Integration by parts. u = ln(x), dv = dx

isnt that u-substitution?

No. It's integration by parts.

int (u*dv) = u*v - int (v*du)

 

[jai6638]

whats V in this case though?

 

[hypn0tik]

Originally posted by: jai6638
whats V in this case though?

You have:
u = lnx
dv = dx

So:
du = 1/x dx
v = x

----------
Also, for the other two integrals, use integration by parts.

For xlnx. Set u = lnx and dv = x and try from there.

For (lnx)^2. u = lnx dv = lnx dx

 

[jai6638]

for ln[x]

I got xln(x)-x+c which seems correct

For xln[x]

I got 1/4 x^2(2lnx-1) which seems correct

However, (ln[x])^2 ... im not sure


I set u=ln x , u'=1/x dx
v'=lnxdx, v=lnx

Since its uv-int(lnx.1/x dx ) , how do i get antiderivatiev of lnx .. I suppose my substitutions are wrong..

 

[hypn0tik]

Originally posted by: jai6638
for ln[x]

I got xln(x)-x+c which seems correct

For xln[x]

I got 1/4 x^2(2lnx-1) which seems correct

However, (ln[x])^2 ... im not sure


I set u=ln x , u'=1/x dx
v'=lnxdx, v=lnx

Since its uv-int(lnx.1/x dx ) , how do i get antiderivatiev of lnx .. I suppose my substitutions are wrong..

Careful. Look at question 1 to find v from v'.

I have a meeting to run to. I'll be back in a few hours if you still need help.

Edit: I'd also be careful about your notation. u' indicates a derivative with respect to a variable. In this case, u' is the derivative of u with respect to x. So, you cannot have u' 1/x dx. You can have u' = du/dx = 1/x or you can have du = 1/x dx.