Area under the Graph of a Function

Area under the Graph of a Function

How to find the area under the graph of a function: formula, proof, examples, and their solutions.

Formula

(area under the graph) = [inteval, from a to b, of |f(x)| dx]

If y = f(x) is above the x-axis, (in [c, b])
then the area is the integral of f(x).

But if y = f(x) is below the x-axis, (in [a, c])
then the area is the integral of -f(x).

So the area under y = f(x)
is the integral of |f(x)|.

Definite integration of absolute value functions

Proof

Area under the Graph of a Function: Proof of the Formula

If y = f(x) is below the x-axis, (in [a, c])
f(x) is (-).

So the area of the sliced rectangle, f(x) dx, is (-).
[ (-)⋅(+) = (-) ]

So the integral of f(x) dx is (-).

But the sign of the area should be (+).

So, in order to change this (-) sign,
change the integrand f(x) into -f(x).

So the area below the x-axis is ∫ac -f(x) dx.

If y = f(x) is above the x-axis, (in [c, b])
the area above the x-axis is ∫cb f(x) dx.

Because of the change of the integrands, [-f(x), f(x)]
the integrand is always (+).

So the absolute value sign can be used
to express the area under the graph of y = f(x).

Example 1

Find the area bounded by y = x^2 - 2x, x = 3, and the x-axis.

First find the bounded region.

y = x2 - 2x
= x(x - 2)

So the zeros of y = x2 - 2x are 0 and 2.

Quadratic functions - factored form

In [0, 2],
y = x2 - 2x is below the x-axis.

So the area in [0, 2] is
02 -(x2 - 2x) dx.

In [2, 3],
y = x2 - 2x is above the x-axis.

So the area in [2, 3] is
23 (x2 - 2x) dx.

Solve the integrals.

Definite integration of polynomials

To make the denominators the same,
multiply 3/3 by 8.

Example 2

Find the area bounded by y = sin x (in [0, 2pi]) and the x-axis.

First find the bounded region.

Graphing sine functions

In [0, π],
y = sin x is above the x-axis.

So the area in [0, π] is
0π (sin x) dx.

In [π, 2π],
y = sin x is below the x-axis.

So the area in [π, 2π] is
π2π (-sin x) dx.

Indefinite integration of sin x