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Math Help - [SOLVED] HELP: Evaluate the integral

  1. #1
    babyMT
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    [SOLVED] HELP: Evaluate the integral

    HELP: Evaluate the integral
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by babyMT View Post
    HELP: Evaluate the integral

    \int_0^1 (x^2+1)e^{-x} \; dx

    Use integration by parts twice.

    Or observe that there exists polynomials p(x) such that:

    \frac{d}{dx} p(x)e^{-x} = (x^2+1)e^{-x}

    and the polynomials must be quadratic. Now find such a
    polynomial and you will have the indefinite integral for your
    integral, so you can then apply the limits of integration to
    get your definite integral.

    RonL
    Last edited by CaptainBlack; February 1st 2008 at 11:03 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    and the polynomials must be quadratic. Now find such a
    polynomial and you will have the indefinite integral for your
    integral, so you can then apply the limits of integration to
    get your definite integral.

    RonL

    I am interested, can you explain how that is done / works please ?
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by babyMT View Post
    HELP: Evaluate the integral
    \int_0^1 (x^2+1)e^{-x}dx

    =\int_0^1 x^2e^{-x}dx + \int_0^1 e^{-x}dx

    =\int_0^1 x^2e^{-x}dx + \int_0^1 e^{-x}dx

    Integrate by parts:
    u=x^2
    du=2xdx
    dv=e^{-x}
    v=-e^{-x}

    =\left[-x^2e^{-x}+C\right]_0^1-(-2)\int_0^1 xe^{-x}dx + \int_0^1 e^{-x}dx

    Integrate by parts:
    u=x
    du=dx
    dv=e^{-x}
    v=-e^{-x}

    =\left[-x^2e^{-x}+C\right]_0^1+2\left([-xe^{-x}+C]_0^1 + \int_0^1 e^{-x}dx\right) + \int_0^1 e^{-x}dx

    =\left[-x^2e^{-x}+C\right]_0^1+[-2xe^{-x}+C]_0^1 + 3\int_0^1 e^{-x}dx

    =\left[-x^2e^{-x}-2xe^{-x}+C\right]_0^1 + 3\int_0^1 e^{-x}dx

    =\left[-x^2e^{-x}-2xe^{-x}+C\right]_0^1 + [-3e^{-x}+C]_0^1

    =\left[-x^2e^{-x}-2xe^{-x}-3e^{-x}+C\right]_0^1

    =(-1^2e^{-1}-2(1)e^{-1}-3e^{-1}+C) - (-0^2e^{-0}-2(0)e^{-0}-3e^{-0}+C)

    =-e^{-1}-2e^{-1}-3e^{-1}+C +3(1)-C

    =-6e^{-1} +3
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  5. #5
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    Quote Originally Posted by bobak View Post
    I am interested, can you explain how that is done / works please ?
    \frac{d}{dx} p(x)e^{-x} = (x^2+1)e^{-x}.

    Let p(x) = ax^2 + bx + c. Then:

    \frac{d}{dx} (ax^2 + bx + c) e^{-x} = (x^2+1)e^{-x}.

    From the product rule:

    (2ax + b) e^{-x} - (ax^2 + bx + c) e^{-x} = (x^2+1)e^{-x}

    Factorise, group like terms and simplify the left hand side:

    (-ax^2 + [2a - b]x + [b - c]) e^{-x} = (x^2+1) e^{-x}.

    Equate coefficients of the quadratic on each side:

    -a = 1 => a = -1.

    2a - b = 0 => -2 - b = 0 => b = -2.

    b - c = 1 => -2 - c = 1 => c = -3.

    So \frac{d}{dx} (-x^2 - 2x - 3) e^{-x} = (x^2+1)e^{-x}.

    Now integrate both sides, noting that the left hand side becomes (-x^2 - 2x - 3) e^{-x} .......
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    Super Member angel.white's Avatar
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    That was brilliant, do you have any others that can be done like this? I want to give it a try.
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    Quote Originally Posted by angel.white View Post
    That was brilliant, do you have any others that can be done like this? I want to give it a try.
    Let it first be said that the kudos goes to CaptainBlack (but I'll hold it in safe-keeping - for a while )

    It works for integrands of the form:

    (\text{polynomial of degree } n) e^{kx} (an example has already been seen).

    (\text{polynomial of degree } n) \sin (kx) or (\text{polynomial of degree } n) \cos (kx),

    where you would observe that there exists polynomials p(x) and q(x) of degree n such that \frac{d}{dx} [p(x) \cos (kx) + q(x) \sin (kx)] gives the integrand.

    Try integrating (x^2 + 1) \sin x this way. There's a subtle difference to the example just done ....

    If you think about it a bit, you'll think of more.

    The idea boils down to making a shrewd guess or observation as to the form of function which, when differentiated, gives the integrand. The technique is sometimes refered to as integration by recognition.
    Last edited by mr fantastic; February 2nd 2008 at 03:32 AM. Reason: Made a mistake - should've had p(x) AND q(x) in the sin and cos one .....
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    Quote Originally Posted by mr fantastic View Post
    Let it first be said that the kudos goes to CaptainBlack (but I'll hold it in safe-keeping - for a while )

    It works for integrands of the form:

    (\text{polynomial of degree } n) e^{kx} (an exmaple has already been seen).

    (\text{polynomial of degree } n) \sin (kx) or (\text{polynomial of degree } n) \cos (kx),

    where you would observe that there exists polynomials p(x) of degree n such that \frac{d}{dx} p(x) (\cos (kx) + \alpha \sin (kx) gives the integrand.

    Try integrating (x^2 + 1) \sin x this way. There's a subtle difference to the example just done ....

    If you think about it a bit, you'll think of more.

    The idea boils down to making a shrewd guess or observation as to the form of function which, when differentiated, gives the integrand. The technique is sometimes refered to as integration by recognition.
    I had a professor that solved his differential equations that way. He'd start with a proposed solution and modify it until it worked. I never have been able to figure out how he did it, other than perhaps a great deal of experience and intuition.

    -Dan
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  9. #9
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    topsquark beat my erratum edit by 1 minute, so I'll just add the corrected bit here so that there's no chance of a mixup when reading his quote of my previous reply.

    Quote Originally Posted by after editing by mr fantastic View Post
    [snip]
    (\text{polynomial of degree } n) \sin (kx) or (\text{polynomial of degree } n) \cos (kx),

    where you would observe that there exists polynomials p(x) and q(x) of degree n such that \frac{d}{dx} [p(x) \cos (kx) + q(x) \sin (kx)] gives the integrand.
    [snip]
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  10. #10
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Let it first be said that the kudos goes to CaptainBlack (but I'll hold it in safe-keeping - for a while )

    It works for integrands of the form:

    (\text{polynomial of degree } n) e^{kx} (an example has already been seen).

    (\text{polynomial of degree } n) \sin (kx) or (\text{polynomial of degree } n) \cos (kx),

    where you would observe that there exists polynomials p(x) and q(x) of degree n such that \frac{d}{dx} [p(x) \cos (kx) + q(x) \sin (kx)] gives the integrand.

    Try integrating (x^2 + 1) \sin x this way. There's a subtle difference to the example just done ....

    If you think about it a bit, you'll think of more.

    The idea boils down to making a shrewd guess or observation as to the form of function which, when differentiated, gives the integrand. The technique is sometimes refered to as integration by recognition.
    Well, here was my first attempt

    (x+1)sin(x)=\frac d{dx}\left[(a_1x^2+b_1x+c_1)sin(x) + (a_2x^2+b_2x+c_2)cos(x) \right]

    (x+1)sin(x)=(a_1x^2+b_1x+c_1)cos(x)+ (2a_1x+b_1)sin(x) + (2a_2x+b_2)cos(x)  - (a_2x^2+b_2x+c_2)sin(x)

    (x+1)sin(x)=[a_1x^2+x(b_1+2a_2)+(c_1+b_2)]cos(x)+ [-a_2x^2 +x(2a_1-b_2)+(b_1-c_2)]sin(x)

    I got really confused at this point and realized I had chosen n=2 which should have been n=1

    ------------------------------

    Here was my second attempt, I set n=1 and used only cosine
    (x+1)sin(x)=\frac d{dx}(ax+b)cos(x)

    (x+1)sin(x)=acos(x)-(ax+b)sin(x)

    Here I didn't know what to do with the acos(x) but I realized if I added -asin(x) to the initial formula, I could get it to go away, so I stuck that onto the initial formula:

    (x+1)sin(x)=\frac d{dx}[(ax+b)cos(x)-asin(x)]

    (x+1)sin(x)=(-ax-b)sin(x)

    x+1=-ax-b

    x+1=-ax-b

    a=-1
    b=-1

    Revising the Initial formula:

    (x+1)sin(x)=\frac d{dx}[(-x-1)cos(x)+sin(x)]

    \int (x+1)sin(x)=(-x-1)cos(x)+sin(x)

    ------------------------------

    Still didn't seem right, I tried it a third time using n=1 and both cos and sin

    (x+1)sin(x)=\frac d{dx}\left[(a_1x+b_1)sin(x) + (a_2x + b_2)cos(x) \right]

    (x+1)sin(x)=-(a_1x+b_1)cos(x)+a_1sin(x) + (a_2x + b_2)sin(x)+a_2cos(x)

    (x+1)sin(x)=[-a_1x+(a_2-b_1)]cos(x)+[a_2x+(a_1+b_2)]sin(x)

    ------------------------------

    Then I got frustrated, and checked my answer, and realized my second answer was correct, and I just differentiated it wrong when I checked it >.<

    So I guess It's still confusing to me to figure out how to set this one up. I think I can do the e^(kx) ones, though. They seem much more straightforward.
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by mr fantastic View Post
    Let it first be said that the kudos goes to CaptainBlack (but I'll hold it in safe-keeping - for a while )

    It works for integrands of the form:

    (\text{polynomial of degree } n) e^{kx} (an example has already been seen).

    (\text{polynomial of degree } n) \sin (kx) or (\text{polynomial of degree } n) \cos (kx),

    where you would observe that there exists polynomials p(x) and q(x) of degree n such that \frac{d}{dx} [p(x) \cos (kx) + q(x) \sin (kx)] gives the integrand.

    Try integrating (x^2 + 1) \sin x this way. There's a subtle difference to the example just done ....

    If you think about it a bit, you'll think of more.

    The idea boils down to making a shrewd guess or observation as to the form of function which, when differentiated, gives the integrand. The technique is sometimes refered to as integration by recognition.
    There is of course another trick to integrating (x^2 + 1) \sin x. That is to integrate (x^2 + 1) e^{ix}, x \in \mathbb{R} and keep the imaginary part of the integral.

    RonL
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