Mean Value Theorem and Average Value

[Scribendi]

Hello again, and thank you in advance to anyone who would so kindly offer my misdirected-mathsoul some guidance.

I apologize, as these problems seem ridiculously easy, but alas I was not graced with the beauty of being innately-numerical. If math "classes" literally existed, I would be the lowly peasant to your royal mind.

Numero Uno:

"Find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function over the given interval."

(*looks up at problem* I know, I know. It's simples, isn't it? Please kick my butt into gear.)

Numero Dos:

"Find the average value of the function over the given interval and all the values of x in the interval for which the function equals its average value."

*points down* I already got started on this problem and I have calculated the average value of the function. For some reason, I got stuck after I set the function equal to the average value. Can you clear up how to solve for x? Sorry for the MSPaintesque image.

One last problem:

I only really need assistance with Part A; just wanted to show the graph. I know this is going to be a lame question and my dignity is really going to regret asking this, but...how do you complete the chart? I'm not usually this mentally debilitated, I promise. I'm probably gonna slap a hand over my face once I figure it out.

Thank you once again for any help. Take care!

Sincerely,

Scribendi

 

[daon]

For 1, solve \(\displaystyle cos(c) = \frac{3}{2\pi}[sin(\frac{\pi}{3})-sin(-\frac{\pi}{3})]\)

For 2, on your last line, you wrote f(x). replace x with c then solve for c.

For 3, just plug in x. e.g.

x=1,

G(1) = \(\displaystyle \int_0^1f(t)dt\) = Area under f([0,1]) = 1.

Note that once f becomes negative, its hard to justify calling this area any longer, as area under the curve will start to cancel area lying above the curve. To gain intuition for this, look at the following:

\(\displaystyle \int_1^2f(t)dt = 1\)
\(\displaystyle \int_2^3f(t)dt = -2\)
\(\displaystyle \int_1^3f(t)dt = -1\)
\(\displaystyle \int_2^6f(t)dt = -8\)
\(\displaystyle \int_6^{10}f(t)dt = 12\)

 

[Scribendi]

Oh, thank you very much for your kindness, daon! I greatly appreciate your help. Sorry again for the trouble and take care.