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This is actually a part of a bigger problem, which involves using the Mean Value Theorem for Integrals. The question is to find, $f_{ave}$:

$f(x) = 5 \sin 4x$ for $x\in [−π, π].$

Using the theorem, I have gotten it down to:

$$\frac{1}{2\pi} \int_{-\pi}^\pi5\sin(4x)dx = 5\sin(4c)$$

I know that I have to find the antiderivative and then solve for c, but I don't know how to find the antiderivative. Any help will be appreciated.

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HINT: Do a substitution; set $u=4x$, and solve the integral. – Arturo Magidin Feb 20 '12 at 3:31
Am I overall on the right track? – user754950 Feb 20 '12 at 3:34
@user754950: Yes, what you are doing will lead to an answer, once you "solve" the integral. And the hint I gave you should help you solve that integral. – Arturo Magidin Feb 20 '12 at 3:38
A solution method surprisingly not listed below is integration by parts. I know,it would initially have resulted in a more complicated integral then the other methods here-but it probably would result in the same solution more directly then the methods below. I suggest it just to make sure the list of proposed solutions is as complete as possible. – Mathemagician1234 Feb 20 '12 at 7:40
up vote 1 down vote accepted

1st you can pull the $5$ out in front of the integral sign to give you $\frac{5}{2\pi}$. Then the antiderivative of $\sin(4x)$ is $-\frac{1}{4}\cos(4x)$ because the antiderivative of $\sin(x)$ is $-\cos(x)$ [recall that the derivative of $\cos(x)$ is $-\sin(x)$] then you multiply that by the reciprocal of the constant associated with $x$ [meaning $1$ over $4$ or $\frac{1}{4}$]. Make sure you evaluate the problem from $-\pi$ to $\pi$.

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alright, I did as you said and I got: -1/(4pi) = sin(4c). Is this correct so far? If so, how do I solve for c? does it involve sine inverse? – user754950 Feb 20 '12 at 3:50
I am not really sure where you get the 5sin(4c) from. Can you explain that so I can better understand your goal? – Jared Feb 20 '12 at 4:09
Take a look at the graph of your equation on a graphing calculator. Note that the region you are integrating cancels out, but if you change your limits so you are integrating from 0 to pi/4, you can multiply the integral by 8 to get a non-zero, non-negative value. – Jared Feb 20 '12 at 4:23
I'm not sure how to put integral symbols in the text; they will be written out in words with curly brackets. Now that you have redefined your limits, your integral problem should read (8/2pi){integral from 0 to pi/4}5sin(4x)dx = (4/pi){integral from 0 to pi/4}5sin(4x)dx. Now since the 5 is a constant, move it outside the integral: (20/pi){integral from 0 to pi/4}sin(4x)dx. Use u-substitution so u=4x and du=4dx, which means (1/4)du=dx. Since you changed from 4x to u, also change your interval: from 4*0=0 to4*(pi/4)= pi. So now you have (1/4)(20/pi){integral from 0 to pi}sin(u)dx. – Jared Feb 20 '12 at 4:31
Now use the antiderivative of sin(x) I mentioned earlier to integrate. You have F(u)=-(5/pi)cos(u){from 0 to pi}. Plug in your intervals so you get F(pi)-F(0)=(final answer). That is [-(5/pi)(-1)]-[-(5/pi)(1)]=[5/pi]-[-5/pi]=(5/pi)+(5/pi)=(10/pi). I hope this is the final answer you are looking for. Remember to add units if your teacher requires that. – Jared Feb 20 '12 at 4:37

For what it's worth:

You can use the "guess and check" method as follows.

The derivative of $-\cos x$ is $\sin x$. So, perhaps the antiderivative of $5\sin(4x)$ is $$ -5\cos(4x).$$

Does this work? Let's check:

The derivative of our guess has to be $5\sin(4x)$; but, $$ {d\over dx} \bigl(-5\cos (4x)\bigr)=5\sin(4x)\cdot 4= 5\cdot 4\sin(4x). $$ Hmm, it's not quite right, we do not want that "4" there on the right hand side, that arose from the chain rule. But, if we introduced a multiplicative factor of $1\over4$ in our guess for the antiderivative, things would work out: $$ {d\over dx} \bigl(-{5\over4}\cos (4x)\bigr)={5\over4} \sin(4x)\cdot 4= 5 \sin(4x). $$

So, indeed, an antiderivative of $5\sin(4x)$ is $-{5\over4}\cos(4x)$.

More generally:

If $F(x)$ is an antiderivative of $f(x)$, then

$\ \ \ $1) $cF(x)$ is an antiderivative of $cf$ (since $(cf)'=c f'$)

$\ \ \ $2) For $c\ne0$, ${1\over c}F(cx)$ is an antiderivative of $f(cx)$ (by the chain rule).

Of course, you can use the "substitution method" for integrals for your problem.

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I think "guess and check",although useful in some differential equations problems,leads to sloppy habits. But it does work in this case,so you can't argue with success......... – Mathemagician1234 Feb 20 '12 at 7:42
I would view adjusting constants at the end as a quite efficient method. – André Nicolas Feb 20 '12 at 21:53

From the basic theory of primitives you can check that

$$\int {f\left( {ax} \right)dx = \frac{1}{a}F\left( {ax} \right) + C} $$

So you can use this and put

$$5\int {\sin \left( {4x} \right)dx = - \frac{5}{4}\cos \left( {4x} \right) + C} $$

Alternatively $\sin x$ is odd, you will have that the integral over any symmertric interval around the origin will be zero, that is

$$\int\limits_{ - \pi }^\pi {\sin \left( {4x} \right)dx} = 0$$

So you problem ultimately is finding $c\in[-\pi,\pi]$ for

$$5\sin \left( {4c} \right) = 0$$

which has solutions. $(0,\pm\pi/4,\pm\pi/2,\pm 3\pi/4,\pm\pi)$

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You may want to add "Alternatively" before "Since $\sin x$ is odd..." since what follows there is really an alternate way of solving the problem that does not require finding an antiderivative for $5\sin(4x)$. – Arturo Magidin Feb 20 '12 at 4:03
@ArturoMagidin Indeed. Thanks. – Pedro Tamaroff Feb 20 '12 at 4:06

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