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  1. #1
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    integral question

    I am studying for my final and I have come across two questions from an old exam that I can not figure out. Thanks to anyone who is willing to lend an ear!

    integral of f(x) dx + 2 equals integral of f(x) dx + 6 both from 2 to 5

    I am forgetting how to figure this out.


    integral of sin^2 t dt I have tried by substitution, by parts with no luck

    the answer on test is 1/2 integral of (1 - cos(2t)) dt How do you know this??????

    Thanks again, I LOVE and APPRECIATE this site more than I can say!
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  2. #2
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    Is this it: \int_{2}^{5} f(x) dx + 2 = \int_{2}^{5} f(x) dx + 6

    It's impossible. Let a = \int_{2}^{5} f(x)dx:
    a + 2 = a + 6 \: \: \Rightarrow \: \: 2 = 6 an absurd conclusion.

    --------------------------------------------

    \int \sin^{2}t \: dt

    Use the identity: \sin^{2} \theta = \frac{1 - \cos (2\theta)}{2}
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  3. #3
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    integral question

    the first integral should be

    from 2 to 5 of (f(x) + 2) dx

    while the second one is from 2 to 5 of f(x) dx + 6

    Sorry that I didn't make that clear before.
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  4. #4
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    \int_2^5 [f(x)+2]~dx

    \int_2^5 f(x)~dx~+~\int_2^5 2~dx

    \int_2^5 f(x)~dx~+~2x \bigg |_2^5

    \int_2^5 f(x)~dx + 6
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  5. #5
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    Quote Originally Posted by Frostking View Post
    the first integral should be

    from 2 to 5 of (f(x) + 2) dx

    while the second one is from 2 to 5 of f(x) dx + 6

    Sorry that I didn't make that clear before.
    So we have

    \int_2^5 \left(f(x) + 2\right)dx = \int_2^5f(x)\,dx + 6,

    correct? And I suppose you are trying to show that this statement is true? You can do:

    \int_2^5 \left(f(x) + 2\right)dx = \int_2^5f(x)\,dx + 6

    \Rightarrow\int_2^5f(x)\,dx + \int_2^52\,dx = \int_2^5f(x)\,dx + 6

    \Rightarrow\int_2^52\,dx = 6

    \Rightarrow2x\bigg|_2^5 = 6\Rightarrow2(5 - 2) = 6\Rightarrow6 = 6

    And the left side is equal to the right, provided that f(x) is indeed integrable over that interval.
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