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Integral using Substitution

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Date: 4/11/96 at 4:3:39
From: Anonymous
Subject: integration by substitution

Dear Dr Math,

I am a year 12 student at high school in Sydney, Australia
and I have a math exam next week. I was wondering if you could
help me with this problem.

Using the given substitution evaluate the following correct to 3
decimal places.

xdx
--------
2        1/2
( x  +  1 )
2   2
given u = x  +  1

I don't know if you will be able to understand my question because
I don't have the right symbols.
2
u   this means u squared and the question is supposed to be
preceded by an integral sign with the limits of 3 and 0.
The question in written form is:

The integral of xdx over x squared plus one all square rooted.
I hope you can help.  Also do you know of any math pages located
in Australia?  Thanks a lot!

Bye,

Tracey Pearce
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Date: 4/12/96 at 23:47:40
From: Doctor Syd
Subject: Re: integration by substitution

Dear Tracey,

Your question is very clear, don't worry!  Ususally we notate
taking x to the second power by this: x^2, and that is the
notation I'll use since it is much shorter.

Okay, so to solve the problem, you let u = x^2 + 1.  Everything
inside the integral is in terms of x or dx, right?  So, all
we need to figure out before we begin the actual substitution
is what dx is.  If we differentiate both sides of the equation
u = x^2 + 1, we get du = 2xdx, right?  So, using these two
equations, let's make the substitution:

xdx = du/2 since du = 2xdx

(x^2 +1)^1/2 = u^1/2,

So we don't have to deal with a fraction, let's write what is in
the integral as:  (xdx)(x^2 + 1)^-1/2

Then, what is inside the integral ( I think it is called the
integrand if I am remembering correctly!) is:

(du/2)(u ^-1/2)

This is something that is much easier to integrate, right!

When evaluating the integral, you must be careful, since your
limits are in terms of x, but your integral is in terms of u.
You have two options:

1) As soon as you subsitute the u's for the x's, make similar
limits of the integral are x=0 and x=3, plug these values into the
equation relating x and u to get the corresponding values for u,
and use these new values as your limits in your new integral.  Or,

normally would when integrating something, except when you are at
the stage where you are ready to plug in limit numbers to what you
have found to be the antiderivative of the function you are
integrating, STOP!  At this stage, convert your antiderivative
that is in terms of u's back into something that is in terms of
x's using the original formula you used to convert from x's to

So, I think the rest of the problem is probably pretty doable.

You probably learned the chain rule for calculating derivatives.
What you just did in this problem was use an anti-chain rule of
sorts.  Have you talked about the anti-chain rule in class?  It is
a helpful tool, so think about what an anti-chain rule might say,
feel free to write us back!

-Doctor Sydney,  The Math Forum

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Associated Topics:
High School Calculus

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