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Math Help - check evaluation of indefinate integral

  1. #1
    Junior Member shepherdm1270's Avatar
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    check evaluation of indefinate integral

    Integral from 0 to 1: x^2 (4x^3+2)^3 dx = x^3/3(x^4)^3 = F(1) - F(0)

    F(1) = 1/3
    F(0) = 0

    = 1/3
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  2. #2
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    Quote Originally Posted by shepherdm1270 View Post
    Integral from 0 to 1: x^2 (4x^3+2)^3 dx = x^3/3(x^4)^3 = F(1) - F(0)

    F(1) = 1/3
    F(0) = 0

    = 1/3
    \int_0^1\left(x^2 (4x^3+2)^3\right)dx = \frac43 \int_0^1\left(\frac34 \cdot 8x^2 (2x^3+1)^3\right) dx = \left. \frac43 \cdot \frac14 \cdot (2x^3+1)^4 \right|_0^1

    I'll leave the rest for you. (Btw the result is \frac{80}3 )
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  3. #3
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    Or you could think of it as \int af\prime(x)~f(x)^n~=~\frac{af(x)^{n+1}}{n+1}

    Hence  \int_0^1 x^2(4x^3+2)^3 \, dx

     = \left[ \frac{1}{48}(4x^3+2)^4 \right]_0^1
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  4. #4
    Junior Member shepherdm1270's Avatar
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    I'm not understanding where you got the 1/48
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  5. #5
    Junior Member shepherdm1270's Avatar
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    Quote Originally Posted by earboth View Post
    \int_0^1\left(x^2 (4x^3+2)^3\right)dx = \frac43 \int_0^1\left(\frac34 \cdot 8x^2 (2x^3+1)^3\right) dx = \left. \frac43 \cdot \frac14 \cdot (2x^3+1)^4 \right|_0^1

    I'll leave the rest for you. (Btw the result is \frac{80}3 )


    I'm not sure how you found the anti-derivative, but I'll try to go back and figure that one out on my own...

    i got F(1) = 1/3(2x^3+1)^4 = 1/3 (3)^4 = 81/3

    and i got F(0) = 0

    so I got 81/3..... ?
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  6. #6
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    Hello,

    F(0)=\frac{1}{3} (2*0+1)^4=\frac{1}{3}*1=\frac{1}{3}
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  7. #7
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    Quote Originally Posted by shepherdm1270 View Post
    I'm not understanding where you got the 1/48
    If you think of it as \int af\prime(x)~f(x)^n~=~\frac{af(x)^{n+1}}{n+1}

    Then f(x)=4x^3+2 \implies f\prime(x)=12x^2

    Hence a must be \frac{1}{12}

    Given the general result we have \frac{\frac{1}{12}(4x^3+2)^4}{4}

     <br />
=~\frac{1}{48}(4x^3+2)^4<br />
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  8. #8
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    Quote Originally Posted by shepherdm1270 View Post
    I'm not sure how you found the anti-derivative, but I'll try to go back and figure that one out on my own...

    ...
    Probably you have seen what I've done. Only in case you need some confirmation I'm going to show you how I got the anit-derivative:

    \int_0^1\left(x^2 (4x^3+2)^3\right) dx = \int_0^1\left(x^2\cdot 2^3 \cdot (2x^3+1)^3\right) dx

    Let u = 2x^3+1~\implies~\frac{du}{dx}=6x^2~\implies~du = 6x^2 dx

    That means:

    du = \frac43 \cdot \underbrace{\frac34 \cdot 2^3 \cdot x^2}_{= 6x^2} \cdot dx

    The integral becomes:

    \int_0^1\left(\frac43 \underbrace{\cdot \underbrace{\frac34 \cdot 2^3 \cdot x^2}_{= 6x^2}}_{\frac{du}{dx}} \cdot (\underbrace{2x^3+1}_{u})^3)\right) dx = \frac43 \cdot \int_{u(0)}^{u(1)} (u^3) du = \frac43 \cdot \left[\frac14 \cdot u^4\right]_{u(0)}^{u(1)}

    Now re-substitute and you'll get:
    <br />
\int_0^1\left(x^2 (4x^3+2)^3\right) dx =\left. \frac{(2x^3+1)^4}{3}\right|_0^1
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  9. #9
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    Introduce a new variable:
    u = 4x^3 + 2

    They you can get dx=du/12x^2. You insert both of these two into you initial equation and get:

    x^2*u^3*du/12x^2

    Due to new variable you must change limits into 2 to 6, getting final equation

    u^4/48
    Inserting the limits you obtain the result 26.67 or 80/3.
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