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Math Help - 2 Problems involving integrals

  1. #1
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    2 Problems involving integrals

    Suppose f(0)=0 and \int_0^2{f'(2t)e^{f(2t)}}dt=5. Find the value of f(4).

    So I first recognized that \frac{1}{2}e^{f(2t)} should be the antiderivative of the integral by FTC so that \frac{1}{2}(e^{f(4)}-e^{f(0)})=5.

    Since f(0)=0, the equation becomes: e^{f(4)}-1=10

    Isolating and then taking the natural log of both sides yields: f(4)=ln11
    Is that right?

    My other question is this: evaluate \lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?

    The first thing I notice is that the variable in the integrand isn't with respect to x. If I rewrote it through FTC, I'd put \lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?
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  2. #2
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    Quote Originally Posted by VectorRun View Post
    Suppose f(0)=0 and \int_0^2{f'(2t)e^{f(2t)}}dt=5. Find the value of f(4).

    So I first recognized that \frac{1}{2}e^{f(2t)} should be the antiderivative of the integral by FTC so that \frac{1}{2}(e^{f(4)}-e^{f(0)})=5.

    Since f(0)=0, the equation becomes: e^{f(4)}-1=10

    Isolating and then taking the natural log of both sides yields: f(4)=ln11
    Is that right?
    Yes, that is correct.

    My other question is this: evaluate \lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?
    Exactly right! That is \lim_{h\to 0}\frac{f(x+h)- f(x)}{h} where f(x)= \int_a^x e^{arctan(t)}dt. And, by the Fundamental Theorem of Calculus, \frac{d}{dx}\int_a^x f(t)dt= f(x).

    The first thing I notice is that the variable in the integrand isn't with respect to x.
    The variable in the integrand is a "dummy variable"- it could be anything without changing the integral.

    If I rewrote it through FTC, I'd put \lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?
    No, its not f that is being differentiated, it is the integral so the derivative of the integral is the integrand, f.
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  3. #3
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    Hello, VectorRun!

    Your first solution is correct . . . ecept for that \frac{1}{2}


    \displaystyle \text{Suppose }f(0)=0\:\text{ and }\:\int_0^2\!f'(2t)e^{f(2t)}}dt\:=\:5

    \text{Find the value of }f(4).


    \text{So I first recognized that }\frac{1}{2}e^{f(2t)}\text{ is the antiderivative of the integral.}
    Um ... not quite.

    \displaystyle \text{We have: }\:\int^2_0 u'\,e^u\,dt \;\;\text{ which equals: }\;e^u\,\bigg]^2_0

    \text{Hence: }\;e^{f(2t)} \,\bigg]^2_0 \;=\;5  \quad \hdots \;\text{ etc.}<br />

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