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Math Help - Three Intergral Problems

  1. #1
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    Three Intergral Problems

    1. Use the First Fundamental Theorem of Calculus to find the derivative of

    =???

    2. Consider the function .




    In this problem you will calculate by using the definition




    The summation inside the brackets is which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each subinterval. Calculate for on the interval and write your answer as a function of without any summation signs.
    ???
    ???

    ( So on this one I did: f(Xk)=f(x)= x^2/2+7) then plug in (2/n *k) for x and got 4/2n^2*k^2+7.
    then (4/2n^2*k^2+7) * 2/n and got (8/2n^3*k^2+14/n).
    8/2n^3*(2n^3+3n^2+n)= 16n^3+24n^2+8n/12n^3 + 14 answer. did i do something wrong?
    For the second part: i took the antiderivative of and evaluate from 0 to 3. the answer i got is (27/6+7). Help me out thanks.


    3.
    and


    ???
    The absolute maximum of occurs when ??? and is the value =????
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  2. #2
    Super Member Deadstar's Avatar
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    For the first on note that tan(t) = \frac{sin(t)}{cos(t)}, and since the derivative of cos(t) is -sin(t) we can say the derivative of \frac{sin(t)}{cos(t)} is -ln(cos(t)).

    So your integral becomes... \int_5^{\frac{1}{x}}8tan(t)dt = \int_5^{\frac{1}{x}}8\frac{sin(t)}{cos(t)}(t)dt = -8ln(cos(\frac{1}{x})) - (-8ln(cos(5))).
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  3. #3
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    Quote Originally Posted by Deadstar View Post
    For the first on note that tan(t) = \frac{sin(t)}{cos(t)}, and since the derivative of cos(t) is -sin(t) we can say the derivative of \frac{sin(t)}{cos(t)} is -ln(cos(t)).

    So your integral becomes... \int_5^{\frac{1}{x}}8tan(t)dt = \int_5^{\frac{1}{x}}8\frac{sin(t)}{cos(t)}(t)dt = -8ln(cos(\frac{1}{x})) - (-8ln(cos(5))).
    I think what's needed here (if it's to be done in the way the question instructs) is to first make the substitution u = \frac{1}{x}:


    y = \int_{5}^{1/x} 8 \tan (t) \, dt = \int_{5}^{u} 8 \tan (t) \, dt.


    Now use the chain rule: \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.


    From the Fundamental Theorem of Calculus \frac{dy}{du} = 8 \tan (u) and obviously \frac{du}{dx} = - \frac{1}{x^2}.


    Therefore h'(x) = \frac{dy}{dx} = - 8 \tan (u) \cdot \frac{1}{x^2} = - 8 \tan \left( \frac{1}{x} \right) \cdot \frac{1}{x^2} = - \frac{8}{x^2} \tan \left( \frac{1}{x} \right) .
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  4. #4
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    Can someone please look at number 2. Did i do something wrong or missing something? I cant figured out. Also i need help on 3 also. Thanks
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