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Math Help - integral help

  1. #1
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    integral help

    so i was working with an integral and I have something like this

    \frac{x^2}{1+x^4} \int_{2}^{\sin{x}} dt + \frac{1}{1+x^4} \int_{2}^{\sin{x}} \cos{t^7} dt

    my question is does \int_{2}^{\sin{x}} dt = 0? because its like taking the area under the curve of nothing?
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  2. #2
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    It is the same as

    <br />
\frac{x^2}{1+x^4} \int_{2}^{\sin{x}} \; 1 \;  dt <br />

    <br />
\int \; f(x) \; dx \;= \int \; 1 \; dx.<br />

    It is the area under function f(x)=1.
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  3. #3
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    wow that entirely makes more sense. thanks
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by zzzoak View Post
    It is the same as

    <br />
\frac{x^2}{1+x^4} \int_{2}^{\sin{x}} \; 1 \; dt <br />

    <br />
\int \; f(x) \; dx \;= \int \; 1 \; dx.<br />

    It is the area under function f(x)=1.
    I am just curious on this one. The integral becomes

    \int_{2}^{\sin{x}} dt=|Sin x -2|

    Is this right?
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  5. #5
    Senior Member BAdhi's Avatar
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    why modulus?
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by BAdhi View Post
    why modulus?
    Sorry, that shouldnt be there.

    Just Sinx -2.

    So IM guessing thats right.
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