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### Integrals and Trig Functions

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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)
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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

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

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Associated Topics:
High School Calculus

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