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Thread: Intergration by substitution 2

  1. #1
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    Intergration by substitution 2

    Find the area under the curve
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    Find the area under the curve
    $\displaystyle y = \sqrt {2x + 1}$

    The area under the curve between the limits $\displaystyle 0 \leq x \leq 1$ is given by:

    $\displaystyle A = \int_{0}^{1} \sqrt {2x + 1}~dx$

    We proceed by substitution

    Let $\displaystyle u = 2x + 1$

    $\displaystyle \Rightarrow du = 2 ~dx$

    $\displaystyle \Rightarrow \frac {1}{2} du = dx$

    Also, when $\displaystyle x = 1 \implies u = 3$

    and when $\displaystyle x = 0 \implies u = 1$ ..........we can use these to change the limits (I wouldn't, but that seems to be the preferred method around here, i always back-substitute to get the integral as a function of x and plug in the old limits)

    $\displaystyle \Rightarrow A = \frac {1}{2} \int_{1}^{3} \sqrt {u}~du$

    And i think you can take it from there
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  3. #3
    MHF Contributor red_dog's Avatar
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    Let $\displaystyle \displaystyle I(x)=\int\sqrt{2x+1}dx$.
    Substitution: $\displaystyle 2x+1=t\Rightarrow 2dx=dt$.
    Then $\displaystyle \displaystyle I(t)=\frac{1}{2}\int\sqrt{t}dt=\frac{1}{2}\frac{t^ {\frac{3}{2}}}{\frac{3}{2}}+C$.
    So $\displaystyle \displaystyle I(x)=\frac{1}{3}(2x+1)^{\frac{3}{2}}+C$
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