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Math Help - Evaluating Integrals (inverse trig functions)

  1. #1
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    Evaluating Integrals (inverse trig functions)

    Original Problem:
    <br />
\int_{0}^{1}{4}/{(1+x^2)}dx<br />

    So I got tan^(-1)x ] 0 to 1<-- supposed to be inverse tanx evaluated from 0 to 1.

    Inverse tan of 1 is pi/4 minus inverse tan of 0 is 0, so I end up with pi/4 as the answer. The book says the answer is pi... What am I doing wrong?
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  2. #2
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    \int_0^1 \frac{4}{1+x^2} \, dx = 4\int_0^1 \frac{1}{1+x^2} \, dx = 4\left(\frac{\pi}{4}\right) = \pi

    looks like you forgot about the 4 in the original integral
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    Ah, thanks. I knew that the four would cancel somehow, I just didn't know that you could take it out like that.

    Also, what's a derivative of an integral? I know how to find it, but what exactly is it? If I understand correctly, an integral is the area between the x-axis and the function, so what information would the derivative of that yield?
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    Quote Originally Posted by tom ato View Post
    Also, what's a derivative of an integral? I know how to find it, but what exactly is it? If I understand correctly, an integral is the area between the x-axis and the function, so what information would the derivative of that yield?
    since a definite integral with constant limits of integration yields a constant, its derivative would be zero.

    a definite integral with at least one variable limit of integration is another story.

    look up the second fundamental theorem of calculus.
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