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Math Help - intigration! some1 please cheack my work and help me with a problem!

  1. #1
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    intigration! some1 please cheack my work and help me with a problem!

    how do i do this?

    ∫ (x-1)^2 x dx

    are these right?
    (1)


    3π/4
    ∫ sins x dx
    π/2

    3π/4
    ∫ sins x dx = -cos x
    π/2


    (-cos 3
    π/4) - (-cos π/2 ) = 0.71


    (2)

    2
    ∫ (2+x) dx
    0

    2
    ∫ (2+x) dx =
    2x+ (1/2)x^2
    0

    (2(2)+(1/2)(2)^2) - ( 2(0) + (1/2)(0)^2)

    =6
    are these right?
    Last edited by arsenal12345; September 12th 2012 at 12:22 AM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: intigration! some1 please cheack my work and help me with a problem!

    For problem 0) expand the binomial, then distribute the x to get a polynomial integrand.

    1) Yes, but I would write the result as \frac{1}{\sqrt{2}} instead.

    2.) I would use a substitution to make the computation simpler: u=x+2,\:du=dx and write:

    \int_2^4 u\,du=\frac{1}{2}\left(4^2-2^2 \right)=6
    Last edited by MarkFL; September 12th 2012 at 12:36 AM.
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  3. #3
    Member kalyanram's Avatar
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    Re: intigration! some1 please cheack my work and help me with a problem!

    Quote Originally Posted by arsenal12345 View Post
    (1)


    [/B]3π/4
    ∫ sins x dx
    π/2

    3π/4
    ∫ sins x dx = -cos x
    π/2


    (-cos 3
    π/4) - (-cos π/2 ) = 0.71


    (2)

    2
    ∫ (2+x) dx
    0

    2
    ∫ (2+x) dx =
    2x+ (1/2)x^2
    0

    (2(2)+(1/2)(2)^2) - ( 2(0) + (1/2)(0)^2)

    =6
    are these right?
    Your solution is correct. However your notation of writing  \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin x dx = -\cos x should be written as  \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin x dx = -\cos x \vert_{\frac{\pi}{2}} ^{\frac{3\pi}{4}} , which is read as ' -\cos x evaluated at the lower limit and higher limit of integration'. Use the same notation in the second integral as well.

    Quote Originally Posted by arsenal12345 View Post
    how do i do this?
    ∫ (x-1)^2 x dx
    You can expand (x-1)^2 using binomial expansion and evaluated the integral as in (2)
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  4. #4
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    Re: intigration! some1 please cheack my work and help me with a problem!

    Quote Originally Posted by MarkFL2 View Post
    For problem 0) expand the binomial, then distribute the x to get a polynomial integrand.

    1) Yes, but I would write the result as \frac{1}{\sqrt{2}} instead.

    2.) I would use a substitution to make the computation simpler: u=x+2\:du=dx and write:

    \int_2^4 u\,du=\frac{1}{2}\left(4^2-2^2 \right)=6
    is my working right though? would get full marks for what i did without making the changes mentioned?

    also regarding the first one is it

    (x^2-2x+1)x = x^3-2x^2+3

    ∫x^3-2x^2+3

    (1/4)x^4 -(2/3)x^3+(1/2)(x^2)+c

    is this right? with proper working?
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: intigration! some1 please cheack my work and help me with a problem!

    You expanded the binomial correctly, but did not distribute the x correctly, although your final answer is correct...

    I am not a teacher, but I would not give full credit for 0.71 when it may easily be given exactly.
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