1 Areas Between Curves

Lecture 9: Some Basic Applications of The Integral (Stewart Chapter 6) Rahul Krishna

1

Areas Between Curves

This is easy, so we’re just going to blow through it. Suppose I have two curves, one given by y = f (x) ; and the other given by y = g (x) : How can I compute the area between the curves? [Draw picture, nice and big of the two curves.] Well, for a given partition P : a = x0 < x1 < ::: < xn = b, the upper and lower Riemann sums S (P) =

n X

(f

g) (xmax ) xk k

(f

g) xmin k

k=1

and S (P) =

n X

xk

k=1

bound the area A between the curves. Note that here, xmax is chosen so that f (xmax ) g (xmax ) = k k k max maxx2[xk 1 ;xk ] (f (x) g (x)): That is, xk is chosen with respect to the function f g: Similarly for xmin k : So we have S (P) A S (P) and thus

Z

Z

b

f (x)

g (x) dx

A

a

b

f (x)

g (x) dx

a

so if f g is integrable, we have that the area between the curves y = f (x) and y = g (x) is given by Rb f (x) g (x) dx: Note that this is the signed area between the curves; if we want the real area a between the curves, then it should be clear that we should integrate Z

b

a

jf (x)

g (x)j dx

OK, I think this is all completely intuitively clear. Let’s do an example and move on. Example 1 Find the area bounded above by y = 2x x2 and below by y = x: First we should …nd the intersection points. [Draw picture–zeros at 0; 2; max at 1 with value 1:] So we calculate: 2x x2 = x has solution at x = 0; x = 1: Therefore, our integral should go from 0 to 1: We write Z 1 A= 2x x2 xdx 0

=

Z

1

x2 dx

x

0

=

x3 3

x2 2

1

= 0

1 6

Bounds of integration are important to keep track of. They are sometimes less than obvious. Sometimes we should also change coordinates to make our job easier. 1

Example 2 Find the area bounded by the curves y = x and 4x + y 2 = 12: [Draw picture.] Well, in this case we should reverse the roles of x and y; and write x = y and x = 12 y 2 =4: First let’s …nd the points of intersection. They occur at ( 6; 6) and (2; 2) : How did we get this? Well, substituting, we are solving 0 = x2 + 4x 12 = (x + 6) (x 2) so the points of intersection are at x = 6 and x = 2: Since the points lie on the line y = x; they are ( 6; 6) and (2; 2) : OK, instead of viewing y as a function of f (which it ISN’T!) we should view x as a function of y: That is, x = f (y) and x = g (y) are our equations. So we get Z 2 2 12 y3 y2 12 y 2 ydy = y 4 4 12 2 6 6 =

8 12

6

2

( 18

( 18)

(18))

= 64=3 If computing a real area, break it up into chunks.

2

Average Value of a Function

How would you compute the average value of a function? For example, if I asked for the average value of f (x) = x on the interval, how should you proceed? Well, this one is easy–[draw f (x) = x] since this is just a line, we know that the average value of f (x) on an interval should be the value it assumes at its midpoint. But what do I really mean by average? One example gives the following intuitive answer. Suppose we had a function f (x) on an interval [a; b] which gave the water level of a tank of water at a position x: So if you like, we would have something like [draw picture of rectangular tank, draw function, draw the other dimension of length l]. How would I try to measure the average water level? Well if we let the water level sort of settle down, it should tend towards the average water level, right? But the volume is going to be preserved, so we …nd that, if fave represents the …nal water level which is the same as the initial average water level, then Z b fave (b a) l = l f (x) dx a

So we can solve this to write fave =

1 b

a

Z

b

f (x) dx

a

Rb OK, so this gives one motivation for why we might want to call b 1 a a f (x) dx the average value of f in the interval [a; b] : Let’s try another motivation. Suppose f (t) measured the temperature at time t; and we want to know the average temperature over the interval [a; b] : Well, we can approximate this by taking sample points and taking a weighted average of these. To ensure that we have some variation in our sample points, let’s divide up the interval [a; b] into n pieces. That is, take a partition P : a = x0 < x1 < x2 < ::: < xn = b: Now we should weight the temperatures in each interval [xk 1 ; xk ] by the length of the interval [xk 1 ; xk ] : If we choose the maximum temperature in each interval, we get a Riemann sum of the form Pn

1

k=1

xk

n X

f (xmax ) xk = k

k=1

2

1 b

a

S (P)

1 b aS

If we choose the minumum temperature, we get gives us Z b 1 f (x) dx b a a

(P) : Taking minima and maxima over all partitions 1

fave

b

a

Z

b

f (x) dx

a

so we …nd that, if f is integrable, then 1

fave =

b

a

Z

b

f (x) dx

a

Pn If you’re uncomfortable with the step where I put k=1 xk in the denominator, we can think about the case where P is just xk = a + k x; where x = b na : Then the maximal average for this partition is n

1X f (xmax ) k n k=1

and the minimum is

n

1X f xmin k n k=1

But if we write this as n

n

1X 1 b aX f (xmax )= f (xmax ) k k n b a n k=1

k=1

=

1

b

x

a

n X

f (xmax )= k

k=1

1 b

a

n X

f (xmax ) xk k

k=1

where xk = x = b na for all k; then it becomes apparent that this is, indeed, an average! OK, enough motivation. I think you understand now why we should de…ne, for a continuous function f , the average value of f on the interval [a; b] to be 1

fave =

b

a

Z

b

f (x) dx

a

Example 3 Let v (t) = 50 ln (t + 1) be the velocity of a car from time t = 0 to t = 1; with t measured in hours and v (t) measured in miles per hour. Then what is the average velocity of the car? We do this in two di¤ erent ways. First, we …nd the average value of v (t) using the formula above. We compute vave =

1 b

=1

Z

a 1

Z

b

v (t) dt

a

50 ln (t + 1) dt

0 1

= 50 ((t + 1) ln (t + 1) (t + 1))0 = 50 (2 ln 2 2 + 1) = 100 ln 2 50 which is about 19 mph. There must have been bad tra¢ c. We can also do this a more standard way–we …nd how far the car went, then divide by the time. BUT THIS IS THE EXACT SAME CALCULATION. Of course you can show that the two methods will always agree. See Example 3 in 6.5. 3

3

Volumes

Approximate volumes by cylindrical disks of area A (x) : Z

V =

b

V (x) dx

a

Example 4 (Sphere) Find the volume of the region enclosed by x2 + y 2 + z 2 = r2 ; i.e. the volume of a p 2 sphere. Well, how do we do this? Let’s write what A (z) is: it should be r2 z 2 : So in total V =

Z

r

r2

z 2 dz

r r

z3 3

r2 z

=

r

3

= 2 r3

2

r 3

4 3 r 3

= Hurray!

Example 5 Find volume of solid obtained by rotating the region bounded by y = x3 ; y = 8; and x = 0 2 around the y-axis. [Draw picture.] [Draw rotated solid.] How do we compute this? Well A (y) = r (y) ; 1=3 where r (y) = y ; so V =

Z

8

y 2=3 dy

0 8

=

3 5=3 y 5

=

3 5=3 8 = 5

0

3 5 2 = 5

96 5

Example 6 Region enclosed by y = x and y = x2 rotated around the x-axis. A (x) = x2 x2

=

2

x2 x4

so V =

Z

1

A (x) dx

0

=

Z

1

x2

x4 dx

0

= =

x3 3 2 15

Volumes by shells.

4

x5 5

1 0

Example 7 Consider the same example as before. [Draw picture again.] Z

1

p 2 y( y

y) dy =

0

Z

1

2 y 3=2

0

2 = 2 5 2 = 15

5

1 2 3

2 y 2 dy