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Math Help - Finding the area under a curve using integration of trig functions.

  1. #1
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    Finding the area under a curve using integration of trig functions.

    Hello!!

    I had an integration exam today and I'm desperate to know if I answered this question correctly. As follows....

    Find the area under the curve when Y= 5xsin(2x) when x=0, x=PI/2

    I got 452, which I'm 99% sure is wrong but you never know!!

    Thanks for your help and time in advance.

    Chris. x
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  2. #2
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    Re: Finding the area under a curve using integration of trig functions.

    Bad luck. int 5x sin(2x) - Wolfram|Alpha

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,



    ... is lazy integration by parts, doing without u and v.




    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  3. #3
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    Re: Finding the area under a curve using integration of trig functions.

    Hello, Purn!

    Find the area under the curve: . y \:=\: 5x\sin(2x)\:\text{ for }x \in [0,\tfrac{\pi}{2}]

    You need to integrate by parts . . .

    We have: . I \;=\;\int 5x\sin(2x)\,dx

    . \begin{Bmatrix}u &=& 5x && dv &=& \sin2x\,dx \\ du &=& 5\,dx && v &=& \text{-}\frac{1}{2}\cos(2x) \end{Bmatrix}

    \text{Then: }\:I \;=\;\text{-}\tfrac{5}{2}x\cos(2x) + \tfrac{5}{2}\int\cos(2x)\,dx

    n . . . . . =\;\text{-}\tfrac{5}{2}x\cos(2x) + \tfrac{5}{4}\sin(2x)\,\bigg]^{\frac{\pi}{2}}_0

    n . . . . . =\;\bigg[\text{-}\tfrac{5}{2}\left(\tfrac{\pi}{2}\right)\cos\pi + \tfrac{5}{4}\sin\pi\bigg] - \bigg[\text{-}\tfrac{5}{2}(0)\cos0 + \tfrac{5}{4}\sin0\bigg]

    n . . . . . = \; \text{-}\tfrac{5}{4}\pi(\text{-}1)

    n . . . . . =\;\frac{5\pi}{4}

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  4. #4
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    Re: Finding the area under a curve using integration of trig functions.

    Quote Originally Posted by Purn View Post
    Hello!!

    I had an integration exam today and I'm desperate to know if I answered this question correctly. As follows....

    Find the area under the curve when Y= 5xsin(2x) when x=0, x=PI/2

    I got 452, which I'm 99% sure is wrong but you never know!!

    Thanks for your help and time in advance.

    Chris. x
    You might, by the way, have notice that since sine is never larger than 1, "5x sin(x/2)", between x= 0 and x= \pi/2, cannot be larger than 5(\pi/2) and so its integral cannot be larger than 5(\pi/x)^2 which is about 12.33. I can't imagine how you got "452"!
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