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### Integration Questions

```
Date: 03/09/2001 at 07:53:16
From: Michelle
Subject: Integration

Hi there.

I'm currently doing integration, and blimey, I personally think it's
really hard. Anyway I'm stuck on a couple of questions and haven't a
clue how to do them.

1. Calculate the area of the finite region bounded by the curves
y = x^2 and y = sqrt(x)

2. The region R is bounded by the curve y = x^2+2, the x and y axes,
and the line joining point (2,6) to point (26,0). Find the area of
R. Also show that the line joining (2,6) to (26,0) is the normal
to the curve at (2,6).

3. Find the area of the finite region between the curve x = y^2 + 3y
and the line x = 4y.

I think this site is great. Keep it up :)
Thanks, Michelle.
```

```
Date: 03/09/2001 at 09:00:27
From: Doctor Jaffee
Subject: Re: Integration

Hi Michelle,

The first thing I always do when I try to solve problems like the ones
you presented is to draw a graph.

In problem 1 you should see that the region is a football-shaped
figure that intersects at (0,0) and (1,1). (I'm talking about American
football, by the way, not what we Yanks call Soccer.)  If you were to
draw a thin vertical strip anywhere in the region, you see that the
height of the strip is the y number on the curve y = sqrt(x) minus the
y number on the curve y = x^2. The thickness of the strip we can call
dx.

So the area of the region is the integral from 0 to 1 of
(sqrt(x) - x^2)dx.

In problem 2 you should see that the region can be partitioned into
two regions, the first, the region under the curve y = x^2 + 2 bounded
by the axes and the vertical line that goes through (2,6). Find the
area of that region using the same method as in problem number 1. The
second region is a triangle whose vertices are at (2,0), (2,6), and
(26,0). Since it is a right triangle, it should be easy to use
geometry to find the area of the triangle. Add these two areas
together and you will have the area of the entire region.

A normal line is the line perpendicular to the tangent line. Since
f'(x) = 2x, the slope of the tangent line at (2,6) will be f'(2).You
should be able to find the slope of the line that connects (2,6) and
(26,0). If the product of these two slopes is -1, you know that they
are perpendicular, which means that the line under consideration is
the normal line.

Do problem 3 the same as problem 1.  Hint: the region will be very
thin.

Give these problems a try and if you want to write back to check your
answers, or if you need more help understanding what to do, I'll be on

Good luck.

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

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