# Average value of a function problems

#### softballchick

Hi,

I just want to make sure I am doing this right.

Find the average value of the function n the given interval

f(x)=sec^2 (x/2) [0, pi/2]

So I integrated from 0 to pi/2. but I am stuck on what the anti derivative of sec^2 (x/2) is... I know the anti derivative of sec^2 (x) is tan (x)...

Thanks

#### pickslides

MHF Helper
$$\displaystyle \int \sec^2 \frac{x}{a}~dx = a \tan\frac{x}{a}+C ,\forall a \in \mathbb{R}$$

Hi, thank you!

#### softballchick

Can you help with this problem? I don't even know what it is asking.. It's the same type of problem

If f ave i[a,b] denotes the average value of f on the interval [a,b] and a<c<b, show that

f ave[a,b] = (c-a)/(b-a) f ave[a,b]+ (b-c)/(b-a) f ave[c,b]

ave means average.

Much thanks! I am stomped on that one.

#### pickslides

MHF Helper
Do you mean show that $$\displaystyle \frac{1}{b-a}\int_a^b f(x)~dx=\frac{c-a}{b-a}\int_a^c f(x)~dx +\frac{b-c}{b-a}\int_c^b f(x)~dx$$ ?

#### softballchick

I am not sure..... I wish I had a scanner. It basically says this:

If f ave[a,b] denotes the average value of f on the interval [a,b] and a<c<b, show that

f ave[a,b] = (c-a)/(b-a) f ave[a,b]+ (b-c)/(b-a) f ave[c,b]

ave means average.

So I am guessing i have to proof it?

#### Also sprach Zarathustra

$$\displaystyle f_{AVE[a,b]}=\frac{1}{b-a}\int^b_a f(x)dx$$

$$\displaystyle f_{AVE[a,c]}=\frac{1}{c-a}\int^c_a f(x)dx$$

$$\displaystyle f_{AVE[c,b]}=\frac{1}{b-c}\int^b_c f(x)dx$$

Now, put all these to you formula...

#### softballchick

i'm not sure how to put it in? I think it's obvious that it works, I just don't know how to show it.