Download Area under a Graph

Area under a Graph
In class on Wed Nov 12, we basically covered Sec 4.7 of Thomas/Finney,
though with an emphasis on area functions.
Let f (x) be a function with f (x) ≥ 0 for a ≤ x ≤ b. Let R be the region
under the graph y = f (x) and above the interval [a, b] on the x-axis. We
want to know Area(R).
Theorem Let F (x) = f (x) dx be an anti-derivative of f (x).
Then:
Area(R) = F (b) − F (a) .
The reason for this is as follows. Let the right wall of the region R vary,
rather than be fixed at x = b: that is, for any value of x, let Rx be the
region under the graph y = f (x) and above the x-axis interval [a, x]. (The
original region R = Rb .) Consider the function A(x) = Area(Rx ), so that
A(b) = Area(Rb ) = Area(R).
Then the derivative A (x) is equal to f (x). That is, the area under the
graph increases at a rate equal to the height of the graph. (See Fig 4.23, p. 334
of your book.) Now,
A (x) = f (x), so A(x) = f (x) dx, an anti-derivative
of f (x). If F (x) = f (x) dx is an arbitrary anti-derivative of f (x), we must
have, for some constant C:
A(x) = F (x) + C
We also know an intial value for A(x). The region Ra has its right wall
is equal to its left wall, meaning it has no width. We thus have A(a) =
Area(Ra ) = 0. Thus A(a) = F (a) + C = 0, and
C = −F (a) .
Therefore
A(x) = F (x) − F (a)
and
Area(R) = A(b) = F (b) − F (a) .
Definition We define the definite integral from a to b of f (x) to be area
under the region R. This is written:
b
f (x) dx := Area(R) .
a
The reason this notation is so similar to our previous notation for an antiderivative is the above Theorem. If we know an anti-derivative (or indefinite
integral)
F (x) =
f (x) dx
then the Theorem says that
b
f (x) dx := F (b) − F (a) .
a
Example Let
f (x) = x4 +
√
x,
which is positive on the interval [a, b] = [2,
√ 3]. We want to find the area of
the region R under the graph y = x4 + x and above the x-axis interval
[2, 3]. We know an anti-derivative
√
F (x) = x4 + x dx = 15 x5 + 23 x3/2 .
By the Theorem:
b
Area(R) =
a
f (x) dx
= F (b) − F (a)
= [ 15 (35 ) + 23 (33/2 ) ] − [ 15 (25 ) + 23 (23/2 ) ]
∼
= 43.78
Again, the reason for this is that if A(x) is the area of an arbitrary region
Rx under the graph with left wall at a = 2 and right wall at x, then the rate
of change of the area √
under the graph is equal to the height of the graph.
That is, A (x) = x4 + x, so we must have
√
A(x) = x4 + x dx = 15 x5 + 23 x3/2 + C
for some value of C. Since A(2) = 0, we have
2
C = −[ 15 (25 ) + (23/2 ) ]
3
and now Area(R) = A(3), given as above.
Homework: Ch. 4.7 #1, 2, 5, 7, 9, 11, 23, 25, 29, 31, 41, 42.
These problems ask you to compute definite integrals
b
f (x) dx
a
where f (x) ≥ 0 on the interval [a, b]. Remember that this means to compute
the area of the region under the graph and above the interval.
To do this,
use the Theorem. That is, find an anti-derivative F (x) = f (x) dx, and
compute
b
a
f (x) dx := F (b) − F (a) ,
√
as we did above for f (x) = x4 + x and F (x) = 15 x5 + 23 x3/2 .
If you have a graphing calculator, have it graph the curve y = f (x),
which will let you picture the region R. Try to estimate its area visually, and
compare with your calculus answer.