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Math Help - Avg. Value

  1. #1
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    Avg. Value

    \int_{pi/4}^{pi/2} sinx dx
    I tried finding the average value of the function and came out with the answer .5554 sq. units. Is this correct?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by radioheadfan View Post
    \int_{pi/4}^{pi/2} sinx dx
    I tried finding the average value of the function and came out with the answer .5554 sq. units. Is this correct?
    No, I'm sorry

    I think you divided by \frac{4}{\pi}, whereas, since you flipped it, you should have multiplied.

    By the way, Radiohead is awesome.
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  3. #3
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    3 things fundamentally wrong with that.

    1) The minimum value of sin(x) on \left[\frac{\pi}{4},\frac{\pi}{2}\right] is in the neighborhood of 0.7 This definitely suggests that the AVERAGE value is unlikely to be less than 0.7.

    2) Square Units? Why would it be "square"? The sine function does not produce square units.

    3) You showed NONE of your work so now no one can help you fix whatever it was that caused you to wander off.

    Try this:
    A) What is the total area under the curve?
    B) If it were a rectangle with that width and that area, how tall would it be?
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    No, I'm sorry
    At what point did I mess up. I'll explain how I did the problem

    1/b-a, and got the anti derivative of sin x to be -cos x

    pi/4(-cos x)

    -pi/4(cos pi/2) -cos pi/3)
    -pi/4(0-.7071)
    +.5554 square units
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by radioheadfan View Post
    At what point did I mess up. I'll explain how I did the problem

    1/b-a, and got the anti derivative of sin x to be -cos x

    pi/4(-cos x)

    -pi/4(cos pi/2) -cos pi/3)
    -pi/4(0-.7071)
    +.5554 square units
    \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin{x}dx\appr  ox{.707}

    Now I've gotta divide by b-a=\frac{\pi}{4}. But dividing by \frac{\pi}{4} is the same as multiplying by \frac{4}{\pi}.

    So the answer:

    \approx\frac{4}{\pi}\cdot(.707)
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin{x}dx\appr  ox{.707}

    Now I've gotta divide by b-a=\frac{\pi}{4}. But dividing by \frac{\pi}{4} is the same as multiplying by \frac{4}{\pi}.

    So the answer:

    \approx\frac{4}{\pi}\cdot(.707)
    Wouldn't it be -.707? I say this because [Cos(pi/2) - Cos (pi/4)] = -.707
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by radioheadfan View Post
    Wouldn't it be -.707? I say this because [Cos(pi/2) - Cos (pi/4)] = -.707
    But recall \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\,dx=\left.\left[{\color{red}-}\cos x\right]\right|_{\frac{\pi}{4}}^{\frac{\pi}{2}}...

    Thus, the average value would be \frac{4}{\pi}\cdot\frac{\sqrt{2}}{2}=\frac{2\sqrt{  2}}{\pi} (Its usually best to leave the answer as an exact value)
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    (Its usually best to leave the answer as an exact value)
    I was gonna say something to that effect, but...

    I forgot. Thanks chris.
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  9. #9
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    Quote Originally Posted by Chris L T521 View Post
    But recall \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\,dx=\left.\left[{\color{red}-}\cos x\right]\right|_{\frac{\pi}{4}}^{\frac{\pi}{2}}...

    Thus, the average value would be \frac{4}{\pi}\cdot\frac{\sqrt{2}}{2}=\frac{2\sqrt{  2}}{\pi} (Its usually best to leave the answer as an exact value)
    ahhh thank you both
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  10. #10
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    is .9 the right answer?
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  11. #11
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    You still don't seem to know why it is NOT "square units".

    If you answer my two questions, you will KNOW and will not have to ask.
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  12. #12
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by radioheadfan View Post
    is .9 the right answer?
    The correct answer is \frac{2\sqrt{2}}{\pi}\approx 0.9003163158

    As I mentioned before, it is best that you leave the answer as an exact value, not an approximate value.
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  13. #13
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    But recall \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\,dx=\left.\left[{\color{red}-}\cos x\right]\right|_{\frac{\pi}{4}}^{\frac{\pi}{2}}...)
    Maybe if I finish this thought....


    \left[-cos{x}\right]^{\frac{\pi}{2}}_{\frac{\pi}{4}}=\overbrace{-\cos{\frac{\pi}{2}}-[-(\cos{\frac{\pi}{4})}]=-0+\frac{\sqrt{2}}{2}}^{\text{We know these values because these are special angles}}=\frac{\sqrt{2}}{2}

    Multiply by \frac{4}{\pi}
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  14. #14
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    Quote Originally Posted by VonNemo19 View Post
    Maybe if I finish this thought....


    \left[-cos{x}\right]^{\frac{\pi}{2}}_{\frac{\pi}{4}}=-\cos{\frac{\pi}{2}}-[-(\cos{\frac{\pi}{4})}]=-0+\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}

    Multiply by \frac{4}{\pi}
    Thank you but I know that you get 4/pi, but I don't understand how -cos (pi/4) = =\frac{\sqrt{2}}{2} My calculator spits out .707 and i multiply it buy 4/pi and get .9
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  15. #15
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by radioheadfan View Post
    Thank you but I know that you get 4/pi, but I don't understand how -cos (pi/4) = =\frac{\sqrt{2}}{2} My calculator spits out .707 and i multiply it buy 4/pi and get .9
    Have you never taken trig radioheadfan?

    Do you remember the special angles. Look at my last post again. And check this out
    http://library.thinkquest.org/C01102...io_special.htm

    Table for the 6 trigonometric functions for special angles
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