TBIL-Calc Area under curves (IN7)

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Section 4.7 Area under curves (IN7)

Learning Outcomes

Use definite integrals to find area under a curve.

Subsection 4.7.1 Activities

Remark 4.7.1.

A geometrical interpretation of

lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}) Δ x = \int_{a}^{b} f (x) d x

(Definition 4.5.3) defines

\int_{a}^{b} f (x) d x

as the net area between the graph of

y = f (x)

and the

x

-axis. By net area, we mean the area above the

x

-axis (when

f (x)

is positive) minus the area below the

x

-axis (when

f (x)

is negative).

As the number of subdivisions increases, the Riemann sum more closely appears to measure the net area between a curve and the x-axis. — Figure 100. Improving approximations of $\int_{0}^{5} (x - 2) (x - 4) d x$

Activity 4.7.2.

(a)

Write the net area between

f (x) = 6 x^{2} - 18 x

and the

x

-axis from

x = 2

to

x = 7

as a definite integral.

(b)

Evaluate this definite integral to verify the net area is equal to

265

square units.

Observation 4.7.3.

In order to find the total area between a curve and the

x

-axis, one must break up the definite integral at points where

f (x) = 0,

that is, wherever

f (x)

may change from positive to negative, or vice versa.

The total area is illustrated by breaking up the integral from 0 to 5 at 2 and 4 where (x-2) and (x-4) are equal to 0. — Figure 101. Partitioning $\int_{0}^{5} (x - 2) (x - 4) d x$ at $x = 2$ and $x = 4 .$

Since

f (x) = (x - 2) (x - 4)

is zero when

x = 2

and

x = 4,

we may compute the total area between

y = (x - 2) (x - 4)

and the

x

-axis using absolute values as follows:

Area = | \int_{0}^{2} (x - 2) (x - 4) d x | + | \int_{2}^{4} (x - 2) (x - 4) d x | + | \int_{4}^{5} (x - 2) (x - 4) d x |

Activity 4.7.4.

Follow these steps to find the total area between

f (x) = 6 x^{2} - 18 x

and the

x

-axis from

x = 2

to

x = 7 .

(a)

Find all values for

x

where

f (x) = 6 x^{2} - 18 x

is equal to

0 .

(b)

Only one such value is between

x = 2

and

x = 7 .

Use this value to fill in the

?

below, then verify that its value is

279

square units.

Area = | \int_{2}^{?} (6 x^{2} - 18 x) d x | + | \int_{?}^{7} (6 x^{2} - 18 x) d x |

Activity 4.7.5.

Answer the following questions concerning

f (x) = 6 x^{2} - 96 .

(a)

What is the total area between

f (x) = 6 x^{2} - 96

and the

x

-axis from

x = - 1

to

x = 9 ?

(b)

What is the net area between

f (x) = 6 x^{2} - 96

and the

x

-axis from

x = - 1

to

x = 9 ?

Subsection 4.7.2 Videos

Figure 102. Video for IN7