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Math Help - Definite integration help.

  1. #1
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    Definite integration help.

    My logic may be horrible.

    Differentiate g(x)= definite integral on [2, x^2+3x-1] sec(t)dt and then prove the theorem you applied.

    = antiderivative of sec(x^2+3x-1) - antiderivative of sec(2)

    d/dx(ntiderivative of sec(x^2+3x-1) - antiderivative of sec(2))

    = sec(x^2+3x-1) - sec(2)

    Which is exactly what the integral originally states, isn't it?
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  2. #2
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    Quote Originally Posted by pantsaregood View Post
    My logic may be horrible.

    Differentiate g(x)= definite integral on [2, x^2+3x-1] sec(t)dt and then prove the theorem you applied.

    = antiderivative of sec(x^2+3x-1) - antiderivative of sec(2)

    d/dx(ntiderivative of sec(x^2+3x-1) - antiderivative of sec(2))

    = sec(x^2+3x-1) - sec(2)

    Which is exactly what the integral originally states, isn't it?
    If I understand this correctly we have

    g(x)=\int_{2}^{x^2+3x-1}\sec(t)dt

    This can be viewed as a function composition. Let u=x^2+3x-1

    Then g(u)=\int_{2}^{u}\sec(t)dt

    Now by the chain rule

    \frac{d}{dx} g(x)=\frac{d}{dx}g(u)=g'(u)\cdot u'=g'(x^2+3x-1)(2x+3)

    Remember that the fundemental theorem of calculus says that(Notice the upper limit of integration) check you book for the hypothsis

    \frac{d}{dx} \int _{c}^{x}f(t)dt=f(x)

    Combining these two results we get

    \frac{d}{dx} g(x)=\frac{d}{dx}g(u)=g'(u)\cdot u'=g'(x^2+3x-1)(2x+3)=(2x+3)\sec(x^2+3x-1)

    Notice that if you integrate both sides of the last equaion with the same u substition that we did above you will get the same function g back. i.e u=x^2+3x-1

    I hope this helps
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