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Math Help - simple integration problem

  1. #1
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    simple integration problem

    Hi, new to this website. I am trying to teach myself integration out of a textbook and needs some help. The question is to find the area enclosed by the y axis, the curve y=1/(x^2) and the lines y=1 and y=4 in the first quadrant. I rearranged the equation to make x the subject {x= +or-sqrt(1/y)}, is that correct? Then, I don't know how to integrate +or-sqrt(1/y) to find the area between the limits of 4 and 1. If someone could explain this in beginners terms that is easy to follow I would greatly appreciate it. Thanks
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  2. #2
    Senior Member vincisonfire's Avatar
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    What you did is correct.
    Then you must compute \int_{1}^{4} \sqrt{\frac{1}{y}}dy =  \int_{1}^{4}\frac{1}{\sqrt{y}}dy
    Let  u = \sqrt{y} then  du = \frac{dy}{2\sqrt{y}}
    The integral becomes  \int_{y=1}^{y=4} 2du = 2u = 2\sqrt{y} = 2\sqrt{4} - 2\sqrt{1} = 2
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  3. #3
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    Quote Originally Posted by vincisonfire View Post
    What you did is correct.
    Then you must compute \int_{1}^{4} \sqrt{\frac{1}{y}}dy = \int_{1}^{4}\frac{1}{\sqrt{y}}dy
    Let  u = \sqrt{y} then  du = \frac{dy}{2\sqrt{y}}
    The integral becomes  \int_{y=1}^{y=4} 2du = 2u = 2\sqrt{y} = 2\sqrt{4} - 2\sqrt{1} = 2
    Please allow me to interject - there's no need for a substitution.

    \int_{1}^{4} \sqrt{\frac{1}{y}}dy = \int_{1}^{4} y^{-\frac{1}{2}}dy = \left. 2 y^{\frac{1}{2}} \right|_1^4 = 2 \left( \sqrt{4} - 1 \right) = 2
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    Senior Member vincisonfire's Avatar
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    I agree but for a beginner it might be harder to follow.
    In some sense there is never a need for a substitution. You could always see the hidden antiderivative. Substitutions are there to clarify things.
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  5. #5
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    Quote Originally Posted by vincisonfire View Post
    I agree but for a beginner it might be harder to follow.
    In some sense there is never a need for a substitution. You could always see the hidden antiderivative. Substitutions are there to clarify things.
    Your right, substitutions are there to simplify things. However, it is important to realize all of the standard integrals of which I believe one is

    \int x^n\, dx = \frac{x^{n+1}}{n+1} + c
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  6. #6
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    Quote Originally Posted by vincisonfire View Post
    I agree but for a beginner it might be harder to follow.
    In some sense there is never a need for a substitution. You could always see the hidden antiderivative. Substitutions are there to clarify things.
    In this case I think recognising \frac{1}{\sqrt{y}} = y^{-1/2} and applying the usual rule is a lot simpler than making a substitution.

    Note: There are many courses where integration and its mechanics are introduced but the technique of integration by substitution is not taught and a student would be required to deal with an integral like the one in this thread.
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