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### Evaluating Two Definite Integrals

```
Date: 02/28/98 at 16:52:35
From: Katie
Subject: calculus

Integral from 1 to -1,  x^3 + 1/x + 1 dx  =

answers:  12, 16, 18, 22, or none!
```

```
Date: 04/02/98 at 12:22:20
From: Doctor Sonya
Subject: Re: calculus

1) evaluate the indefinite integral
2) plug in upper and lower limits
3) simplify (being watchful of signs!)

In this case integrate the polynomial term-by-term. (Note: I'm
dropping the arbitrary constants, since we're ultimately finding a
definite integral.) Since

Int[x^3]dx = (1/4)x^4
Int[1/x]dx = ln|x|      (*)
Int[ 1 ]dx = x

We can write:

6Int[x^3 + 1/x + 1]dx =  6(Int[x^3]dx + Int[1/x]dx + Int[1]dx)

This will give you the general derivitive, and then you just have to
plug in your limits of integration to get the answer.  As a hint,
it is one of the numbers in your list of possible choices.

(*) Note that we're integrating 1/x from -1 to 1.  This bit of the
function is undefined at 0, which means we really should consider this
an improper integral. However, since both sides are symmetric (look at
the area under the curve on a graph) it comes out to 0 anyway. When
the function isn't symmetric, you must be more careful.

A quicker solution involves a bit more familiarity with the functions:
since x^3 and 1/x are odd functions, integrating them with symmetric
limits (from -a to a) gives 0, so we can drop those terms from the
original integral.

Think about why this works. Little tricks like that will help on the
AP Exam, where you're pressed for time.

-Doctor Sonya and Doctor Griffy, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 02/28/98 at 16:55:45
From: Katie
Subject: calculus

What is the area in the 1st quadrant that is enclosed by y = sin3x and
the x-axis from x = 0 to the first x-intercept on the positive x-axis?

I have no idea where to begin!

The possible answers are:  1/3, 2/3, 1, 2, or 6.
```

```
Date: 04/02/98 at 12:27:46
From: Doctor Sonya
Subject: Re: calculus

It sounds rather ugly, doesn't it? Fear not, that's what we're here
for. :-)

This problem is asking for the area under a particular curve in the
upper right quadrant of the graph.  This should scream "integration,"
because finding an integral actually computes the area from the curve
to the x axis. (Go ahead and plot the graph of this function.)

What we have to compute is a definite integral that looks like this:
Int(?,?){f(x)}dx. In this case, we're given

f(x) = y = sin(3x)

but we need to find the limits of the integral. The problem gives us
one limit, x = 0, but we need to find the other limit. It calls this
limit "the first x-intercept on the positive x-axis", which is asking
for the first point to the right of x=0 where the graph crosses the x-
axis.

To find the value, we set f(x) = 0, or sin(3x) = 0. Since the sine of
0, Pi, 2Pi, etc. is 0, then 3x must equal 0, Pi, 2Pi, etc. Setting 3x
= 0 gives us x = 0, which is the first limit we're given. And setting
3x = Pi gives us x  = Pi/3, which is the other limit.

So all this means we need to find Int(0,Pi/3){sin(3x)}dx.

You can do this with a change of variables. (Hint: let u = 3x)

-Doctor Sonya and Doctor Griffy, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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