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Math Help - Need help with a definite integral please

  1. #1
    Junior Member
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    Unhappy Need help with a definite integral please

    I would appreciate advice on solving the following problem.

    Evaluate the definite integral:

    integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx

    Here is my own attempt.

    Step 1: Solve for the generic (indefinite) integral via substitution

    Let u = x^2 + 1

    So then

    du / dx = 2x

    2x dx = du

    dx = du / 2x

    Returning to the original function:

    integral 2x (x^2 + 1) dx
    = integral 2x (u) dx
    = integral 2x (u) (du / 2x)
    = integral u du
    = integral x^2 + 1
    = (x^3 / 3 ) + x + C
    Step 2: solve for the definite integral given the stated limits

    At upper limit of 2:

    ((2)^3 / 3) + (2) +C
    = (8/3) +2 + C
    = (8/3) + (6/3) + C
    = 14/3 + C

    At lower limit of 1

    ((1)^3 / 3) + (1) +C
    =(1/3) +1 + C
    = (1/3) + (3/3) + C
    = 4/3 + C

    Subtracting the lower limit result from upper limit result:

    (14 /3 + C) - (4/3 + C)
    =14/3 + C - 4/3 - C
    = 10 / 3

    But my textbook and calculator both say the answer is (21/2)
    What am I doing wrong?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lingyai
    I would appreciate advice on solving the following problem.

    Evaluate the definite integral:

    integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx

    Here is my own attempt.

    Step 1: Solve for the generic (indefinite) integral via substitution

    Let u = x^2 + 1

    So then

    du / dx = 2x

    2x dx = du

    dx = du / 2x

    Returning to the original function:

    integral 2x (x^2 + 1) dx
    = integral 2x (u) dx
    = integral 2x (u) (du / 2x)
    = integral u du
    = integral x^2 + 1
    = (x^3 / 3 ) + x + C
    Step 2: solve for the definite integral given the stated limits

    At upper limit of 2:

    ((2)^3 / 3) + (2) +C
    = (8/3) +2 + C
    = (8/3) + (6/3) + C
    = 14/3 + C

    At lower limit of 1

    ((1)^3 / 3) + (1) +C
    =(1/3) +1 + C
    = (1/3) + (3/3) + C
    = 4/3 + C

    Subtracting the lower limit result from upper limit result:

    (14 /3 + C) - (4/3 + C)
    =14/3 + C - 4/3 - C
    = 10 / 3

    But my textbook and calculator both say the answer is (21/2)
    What am I doing wrong?
    Evaluate:

    \int_1^2 2x(x^2+1)dx

    Ssubstitute u=x^2+1 , dx=\frac{1}{2x}du:

    \int_{x=1}^2 2x(x^2+1)dx=\int_{u=2}^5 u\ du=\left[u^2/2\right]_{u=2}^5 =\frac{25}{2}-\frac{4}{2}=\frac{21}{2}

    RonL
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  3. #3
    MHF Contributor
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    Quote Originally Posted by lingyai
    I would appreciate advice on solving the following problem.

    Evaluate the definite integral:

    integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx

    Here is my own attempt.

    Step 1: Solve for the generic (indefinite) integral via substitution

    Let u = x^2 + 1

    So then

    du / dx = 2x

    2x dx = du

    dx = du / 2x

    Returning to the original function:

    integral 2x (x^2 + 1) dx
    = integral 2x (u) dx
    = integral 2x (u) (du / 2x)
    = integral u du
    = integral x^2 + 1
    = (x^3 / 3 ) + x + C
    Step 2: solve for the definite integral given the stated limits

    At upper limit of 2:

    ((2)^3 / 3) + (2) +C
    = (8/3) +2 + C
    = (8/3) + (6/3) + C
    = 14/3 + C

    At lower limit of 1

    ((1)^3 / 3) + (1) +C
    =(1/3) +1 + C
    = (1/3) + (3/3) + C
    = 4/3 + C

    Subtracting the lower limit result from upper limit result:

    (14 /3 + C) - (4/3 + C)
    =14/3 + C - 4/3 - C
    = 10 / 3

    But my textbook and calculator both say the answer is (21/2)
    What am I doing wrong?
    What you did wrong?
    That there after Integral u du.
    You continued with Integral x^2 +1.
    That is wrong.

    You were already in the substitution and then you went back to the original, without integrating yet. You should have continued the integration usin the substitution. That is why you chose to use substitution in the first place.

    Then in your going back to the original, from INT. u du, the switchback is wrong also.
    What is u?
    It is x^2 +1
    What is du?
    It is 2x dx.

    So why was your switchback
    INT. u du = INT. x^2 +1
    only?
    It should have been
    INT. u du = INT. (x^2 +1) *(2x dx)
    which is the same as the original integrand.
    Which would have put you back to square one or the beginning again.
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  4. #4
    Super Member

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    Lexington, MA (USA)
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    Hello, lingyai!

    Thank you for showing your work and reasoning . . .

    Evaluate: \:\int^2_1 2x(x^2 + 1)dx
    Did you (or anyone) notice that the integrand is a polynomial?

    \int^2_12x(x^2+1)dx \;= \;\int^2_1(2x^3 + 2x)\,dx \;= \;\frac{1}{2}x^4 + x^2\bigg|^2_1

    . . = \;\left(\frac{1}{2}\cdot2^4 + 2^2\right) - \left(\frac{1}{2}\cdot1^4 + 1^2\right) \;= \;(8 + 4) - \frac{3}{2}\;=\;\frac{21}{2}
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Soroban
    Hello, lingyai!

    Thank you for showing your work and reasoning . . .


    Did you (or anyone) notice that the integrand is a polynomial?

    \int^2_12x(x^2+1)dx \;= \;\int^2_1(2x^3 + 2x)\,dx \;= \;\frac{1}{2}x^4 + x^2\bigg|^2_1

    . . = \;\left(\frac{1}{2}\cdot2^4 + 2^2\right) - \left(\frac{1}{2}\cdot1^4 + 1^2\right) \;= \;(8 + 4) - \frac{3}{2}\;=\;\frac{21}{2}
    I was too busy noticing that:

    <br />
\:\int^2_1 2x(x^2 + 1)dx=\:\int^2_1 \frac{1}{2}\frac{d}{dx}(x^2 + 1)^2dx<br />
<br />
=\frac{1}{2}(x^2 + 1)^2 \bigg|_{x=1}^2=\frac{25-4}{2}=\frac{21}{2}

    but that would not help the OP spotting what they had done wrong

    RonL
    Last edited by CaptainBlack; May 28th 2006 at 07:09 AM.
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