Integrals and Trig Functions

Date: 4/6/96 at 16:2:52
From: Anonymous
Subject: Calculus: integral from a to b of trig function 

Hey there!

This is a problem for Advanced Placement Calculus, and if I solve it, 
I'll get extra credit.

Evaluate EXACTLY: the integral from pi to zero of the square root 
of (1-sinX).

I don't even know where to start and what should I use to solve
this problem, so I would apreciate it if you would be able to give 
me some hints and starting points. That'll be really great. 

Thanks.  Bye,   Ksenya.   8)

Date: 4/7/96 at 19:49:1
From: Doctor Aaron
Subject: Re: Calculus: integral from a to b of trig function 

Hi, Ksenya,

It's hard to imagine how someone in a  pre-calc class would do an 
integral and understand what they were doing.  I'll give you some 
things to think about, and I hope they'll be helpful.

First I'll define an integral.

Let f(x) be a function that takes the domain (a,b) to the range 
(c,d).

Let a<x1<x2<.....<x(n-1)<b. Now we're going to let x0 = a and 
xn = b. Here the 1,2,n-1,n should be subscript and denote indices.  
We could just use different variable names, but we're going to 
want to take a limit as n goes to infinity so unless we use 
indices, we'll run out of variable names.

Then we can think about the partition of (a,b) as all of the 
subintervals:

(x0,x1), (x1,x2), ... (x(n-2),x(n-1)), (x(n-1),xn).  

Then we can call an arbitrary interval (xi,x(i+1)).  We can think 
of the group of subintervals as this arbitrary one where i ranges 
from 0 to n-1.

The next thing is to pick a random point xij that is inside of 
(xi,x(i+1)).

Now we get back to the function.  We're going to look at f(xij).

We multiply f(xij) by (x(i+1) - xi).

Now we're ready for the definition of the integral:

The _Integral_ of f from a to b is equal to the sum over all i of
f(xij) times (x(i+1) -xi).  

You may be thinking that this is completely unintelligible and 
even if it is, why would anyone want to know about that awful sum.

Well it might help to think of this geomoetrically.  Get a piece 
of graph paper and draw the graph of your favorite function.  

Now pick an interval on the x-axis.  An interval of length 5 with 
endpoints that are integers is good.  Draw vertical lines at the 
endpoints of this interval.  Only look at the piece of the graph 
that is contained within these vertical lines.

Now pick a bunch of points inside of your interval on the x-axis.  
It's much easier if you pick integers.  Now draw vertical lines at 
these points. Your graph should be partitioned into lots of little 
pieces. Pick a random point in each of these pieces.  Now multiply 
the function value at that point by the length of the interval 
which contains it.

Now the last step is to add all of these guys up. We call this a 
Riemann sum.

Geometric Interpretation:

We remember that the product of 2 numbers is the area of the 
rectangle whose sides have length equal to the numbers.  Now draw 
a vertical line down from the function value at the random point.  
We're multiplying by the subinterval length, so in each 
subinterval on the graph of the function, we can draw a rectangle 
whose base is the subinterval on the x-axis, and whose height is 
the value of the function at the random point.  If we do this in 
each subinterval, we get something that approximates the area 
under the graph of the function.

An integral is defined to be the limit as n goes to infinity of 
this Riemann sum.  Then we can say that the integral of a function 
from a to b is exactly the area under the graph of f from a to b.

Your function is sqrt(1-Sin[x]).  It sure would be a pain to have 
to do this whole process again, especially for such an annoying 
function.

There is something called the fundamental theorem of the Calculus 
that gives a shortcut because life is too short to do Riemann 
sums, especially if we have to take the limit as n goes to 
infinity. The proof is hard, and even the statement of the theorem 
requires a knowledge of derivatives, but here it is:

**Caution: this is a deep mathematical statement so don't use it 
without thinking about what it means.**

If f is the derivative of F, then the integral of f from a to b is 
equal to F(b) - F(a).  

I think that if you think about the Riemann sum a lot, you will 
learn a lot of math, and even if you can't evaluate the integral 
exactly I'm sure your teacher will give you partial credit (no 
promises - I don't know him or her). 

Good luck. If you understand what's going on by your 5th reading 
you're doing great.

-Doctor Aaron,  The Math Forum