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Math Help - Area of a region

  1. #1
    Newbie ZosoPage's Avatar
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    Area of a region

    Find the exact area of the region bounded by ,
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  2. #2
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    Quote Originally Posted by ZosoPage View Post
    Find the exact area of the region bounded by ,
    Your post has an invalid image link. Please type the question out.
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  3. #3
    Newbie ZosoPage's Avatar
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    Sorry again.

    Here's the problem without any red x's.

    Find the exact area of the region bounded by y=\frac{x^3}{\sqrt{4-x^2}}, y=0, x=0, x=\sqrt{2}
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    Quote Originally Posted by ZosoPage View Post
    Sorry again.

    Here's the problem without any red x's.

    Find the exact area of the region bounded by y=\frac{x^3}{\sqrt{4-x^2}}, y=0, x=0, x=\sqrt{2}
    try the substitution ...

    u^2 = 4 - x^2

    you'll end up with an easier definite integral ...

    \int_{\sqrt{2}}^2 4 - u^2 \, du<br />
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  5. #5
    Newbie ZosoPage's Avatar
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    Quote Originally Posted by skeeter View Post
    try the substitution ...

    u^2 = 4 - x^2

    you'll end up with an easier definite integral ...

    \int_{\sqrt{2}}^2 4 - u^2 \, du<br />
    I'm confused. How did you use that substitution to come up with that integral? I need to take this integral but I don't know how:

     <br />
\int_{0}^{\sqrt{2}}\frac{x^3}{\sqrt{4-x^2}}dx<br />
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  6. #6
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    Quote Originally Posted by ZosoPage View Post
    I'm confused. How did you use that substitution to come up with that integral? I need to take this integral but I don't know how:

     <br />
\int_{0}^{\sqrt{2}}\frac{x^3}{\sqrt{4-x^2}}dx<br />
    u^2 = 4-x^2

    x^2 = 4-u^2

    2x \, dx = -2u \, du

    x \, dx = -u \, du

    \int \frac{x^3}{\sqrt{4-x^2}} \, dx = \int \frac{x^2}{\sqrt{4-x^2}} \cdot x \, dx

    substitute ...

    \int \frac{4-u^2}{u} \cdot (-u) \, du

    -\int 4 - u^2 \, du

    lower limit ... x = 0 , u = 2

    upper limit ... x = \sqrt{2} , u = \sqrt{2}

    -\int_2^{\sqrt{2}} 4 - u^2 \, du = \int_{\sqrt{2}}^2 4 - u^2 \, du
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