$\displaystyle \int_{pi/4}^{pi/2} sinx dx$

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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- Aug 2nd 2009, 11:20 AM #1

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- Aug 2nd 2009, 11:34 AM #2

- Aug 2nd 2009, 11:37 AM #3

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3 things fundamentally wrong with that.

1) The minimum value of sin(x) on $\displaystyle \left[\frac{\pi}{4},\frac{\pi}{2}\right]$ 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?

- Aug 2nd 2009, 11:39 AM #4

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- Aug 2nd 2009, 11:44 AM #5
$\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin{x}dx\appr ox{.707}$

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

So the answer:

$\displaystyle \approx\frac{4}{\pi}\cdot(.707)$

- Aug 2nd 2009, 12:02 PM #6

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- Aug 2nd 2009, 12:05 PM #7
But recall $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\,dx=\left.\left[{\color{red}-}\cos x\right]\right|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$...

Thus, the average value would be $\displaystyle \frac{4}{\pi}\cdot\frac{\sqrt{2}}{2}=\frac{2\sqrt{ 2}}{\pi}$ (Its usually best to leave the answer as an exact value)

- Aug 2nd 2009, 12:11 PM #8

- Aug 2nd 2009, 12:13 PM #9

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- Aug 2nd 2009, 07:18 PM #10

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- Aug 2nd 2009, 07:28 PM #11

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- Aug 2nd 2009, 07:30 PM #12

- Aug 2nd 2009, 07:37 PM #13
Maybe if I finish this thought....

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

Multiply by $\displaystyle \frac{4}{\pi}$

- Aug 2nd 2009, 07:43 PM #14

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- Aug 2nd 2009, 07:46 PM #15
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