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Thread: Simple Integral with U substitution

  1. #1
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    Simple Integral with U substitution

    Hi,

    I'm trying to solve this integral

    $\displaystyle \int_{0}^{0.4}12.5x(1-x)^4 dx$

    I'm letting u = 1-x, so that I have

    $\displaystyle \int_{0.6}^{1}12.5(1-u)u^4 du$

    Then

    $\displaystyle 2.5u^5 - \frac{25}{12}u^6 |^{1}_{0.6} =0.3195$

    But if I replace u with 1-x then

    $\displaystyle 2.5(1-x)^5 - \frac{25}{12}(1-x)^6 |^{0.4}_{0}$

    I don't get the correct answer.

    Why is this? Can I not substitute (1-x) for u?

    Thanks.


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  2. #2
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    Re: Simple Integral with U substitution

    Quote Originally Posted by downthesun01 View Post
    Hi,

    I'm trying to solve this integral

    $\displaystyle \int_{0}^{0.4}12.5x(1-x)^4 dx$

    I'm letting u = 1-x, so that I have

    $\displaystyle \int_{0.6}^{1}12.5(1-u)u^4 du$

    Then

    $\displaystyle 2.5u^5 - \frac{25}{12}u^6 |^{1}_{0.6} =0.3195$

    But if I replace u with 1-x then

    $\displaystyle 2.5(1-x)^5 - \frac{25}{12}(1-x)^6 |^{0.4}_{0}$

    I don't get the correct answer.

    Why is this? Can I not substitute (1-x) for u?

    Thanks.


    Let's go through the substitution a little more carefully:

    $$\int_0^{0.4} 12.5x(1-x)^4dx$$

    $$u = 1-x, du = -dx$$

    So:

    $$\int_0^{0.4} 12.5x(1-x)^4dx = -\int_{1}^{0.6} 12.5(1-u)u^4du$$

    You used the fact that:

    $$-\int_1^{0.6}f(u)du = \int_{0.6}^1 f(u)du$$

    But, when you switch back, you have to remember you reversed the bounds of integration. So, you need:

    $$\left. 2.5(1-x)^5 - \dfrac{25}{12}(1-x)^6 \right|_{0.4}^0$$

    This will give you the same answer.
    Thanks from downthesun01 and topsquark
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  3. #3
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    Re: Simple Integral with U substitution

    I knew it was something obvious that I just wasn't seeing. Thank you.
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