Expert: Paul Klarreich - 1/24/2007

Question
Hey Mr. Klarreich, just recently we started integration in my calculus class, my has gone on leave due to some difficulties and there is a test assigned in a few days. The review problem has a few questions I cannot answer or come close to figuring out, please help!

Suppose f is continuous and it is known that:
The integral from -1 to 1 f(x)dx=3

Evaluate the integral:
a) The integral from 0 to pi (sinx)f(cosx)dx
b) The integral from 2 to 3 xf(8-x^2)dx

I have an idea that I must use the substitution rule and set u=cosx for a, but i'm not sure on how to proceed. Thank you very much in advance!!

Answer
Questioner:   Joseph Malikay
Category:  Calculus
 
Subject:  Some integration problems
Question:  Hey Mr. Klarreich, just recently we started integration in my calculus class, my has gone on leave due to some difficulties

>> Your what has gone on leave?  Your dog?  [There is this old joke that the best player on the tennis team was reported to be the dog.  Why the dog?  Every time one of the players missed, the coach would yell out that the dog could make that shot.]

and there is a test assigned in a few days. The review problem has a few questions I cannot answer or come close to figuring out, please help!

Suppose f is continuous and it is known that:
The integral from -1 to 1 f(x)dx=3

Evaluate the integral:
a) The integral from 0 to pi (sinx)f(cosx)dx
b) The integral from 2 to 3 xf(8-x^2)dx

I have an idea that I must use the substitution rule and set u=cosx for a, but i'm not sure on how to proceed. Thank you very much in advance!!
...................................................
Hi, Joseph,
--------------------- Preliminary note.-----------
Since you are in a hurry I am trying to send a quick answer.  The first of your two is straightforward, the second is not.  Perhaps either (1) you mistyped something, or (2) your instructor has some really cute way to do part b.  If I come up with something in the next couple of days I'll send it along.
--------------------------------------------------
Yes, these are substitution integrals.  Your job will be to make the example integral match:

{1
|   f(x) dx
}-1
which, as you said, is equal to 3.

Keep in mind that the 'x' in that is a 'dummy variable' -- it disappears after you do the work.  So all of these:
{1
|   f(t) dt
}-1

{1
|   f(w) dw
}-1

etc.  are equal to 3.

Now when you do a substitution in a DEFINITE integral, you are supposed to change the boundaries of integration at the same time.  When you write:

{1
|   f(x) dx
}-1

you really mean:

{x=1
|   f(x) dx
}x=-1

So we will do that as part of the substitution process.
......................................

a) The integral from 0 to pi (sinx)f(cosx)dx

{x=pi
|   sin x f(cos x) dx
}x=0

Let  u = cos x.  Then we have:

du = - sin x dx  and  sin x dx = - du
At  x = 0, u = cos 0 = 1
At  x = pi, u = cos pi = -1.

Do the substitution:

{u=-1
|     - f(u) du
}u=1  

Now use the integration property that says if you switch the boundaries and integrate the 'other way', you get a minus sign (or you lose one).  So that becomes:

{u=1
|     + f(u) du
}u=-1  

which is equal to 3.

b) The integral from 2 to 3 of  x f(8-x^2)dx

{x=3
|    x f(8 - x^2) dx
}x=2

Let  u = 8 - x^2.
du = -2x dx,  and  x dx = - du/2
At x = 2,  u = 8 - 2^2 = 4
At x = 3,  u = 8 - 3^2 = -1

{u=-1    -1
|        --- f(u) du
}u=4      2

{u=4     1
|       --- f(u) du
}u=-1    2

Now this is not good.  The boundaries have to reduce to  -1 to 1 and they don't.
I'll see if I can come up with something soon.