<< Chapter < Page Chapter >> Page >

When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car travels away from its starting position in a straight line at a speed of 75 mph for 2 hours, then it is 150 mi away from its original position ( [link] ). Using integral notation, we have

0 2 75 d t = 150 .
A graph in quadrant 1 with the x-axis labeled as t (hours) and y-axis labeled as v (mi/hr). The area under the line v(t) = 75 is shaded blue over [0,2].
The area under the curve v ( t ) = 75 tells us how far the car is from its starting point at a given time.

In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at a speed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at a speed of 40 mph for 3 hours, it will be back at it starting position ( [link] ). Again, using integral notation, we have

0 2 60 d t + 2 5 −40 d t = 120 120 = 0.

In this case the displacement is zero.

A graph in quadrants one and four with the x-axis labeled as t (hours) and the y axis labeled as v (mi/hr). The first part of the graph is the line v(t) = 60 over [0,2], and the area under the line in quadrant one is shaded. The second part of the graph is the line v(t) = -40 over [2,5], and the area above the line in quadrant four is shaded.
The area above the axis and the area below the axis are equal, so the net signed area is zero.

Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area between the curve and the x -axis, regardless of whether that area is above or below the axis. This is called the total area    .

Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis (rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is

0 2 | 60 | d t + 2 5 | −40 | d t = 0 2 60 d t + 2 5 40 d t = 120 + 120 = 240.

Bringing these ideas together formally, we state the following definitions.

Definition

Let f ( x ) be an integrable function defined on an interval [ a , b ] . Let A 1 represent the area between f ( x ) and the x -axis that lies above the axis and let A 2 represent the area between f ( x ) and the x -axis that lies below the axis. Then, the net signed area between f ( x ) and the x -axis is given by

a b f ( x ) d x = A 1 A 2 .

The total area between f ( x ) and the x -axis is given by

a b | f ( x ) | d x = A 1 + A 2 .

Finding the total area

Find the total area between f ( x ) = x 2 and the x -axis over the interval [ 0 , 6 ] .

Calculate the x -intercept as ( 2 , 0 ) (set y = 0 , solve for x ). To find the total area, take the area below the x -axis over the subinterval [ 0 , 2 ] and add it to the area above the x -axis on the subinterval [ 2 , 6 ] ( [link] ).

A graph of a increasing line f(x) = x-2 going through the points (-2,-4), (0,2), (2,0), (4,2), and (6,4). The area under the line in quadrant one and to the left of the line x=6 is shaded and labeled A1. The area above the line in quadrant four is shaded and labeled A2.
The total area between the line and the x-axis over [ 0 , 6 ] is A 2 plus A 1 .

We have

0 6 | ( x 2 ) | d x = A 2 + A 1 .

Then, using the formula for the area of a triangle, we obtain

A 2 = 1 2 b h = 1 2 · 2 · 2 = 2
A 1 = 1 2 b h = 1 2 · 4 · 4 = 8 .

The total area, then, is

A 1 + A 2 = 8 + 2 = 10 .
Got questions? Get instant answers now!
Got questions? Get instant answers now!

Find the total area between the function f ( x ) = 2 x and the x -axis over the interval [ −3 , 3 ] .

18

Got questions? Get instant answers now!

Properties of the definite integral

The properties of indefinite integrals apply to definite integrals as well. Definite integrals also have properties that relate to the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals.

Rule: properties of the definite integral


  1. a a f ( x ) d x = 0

    If the limits of integration are the same, the integral is just a line and contains no area.

  2. b a f ( x ) d x = a b f ( x ) d x

    If the limits are reversed, then place a negative sign in front of the integral.

  3. a b [ f ( x ) + g ( x ) ] d x = a b f ( x ) d x + a b g ( x ) d x

    The integral of a sum is the sum of the integrals.

  4. a b [ f ( x ) g ( x ) ] d x = a b f ( x ) d x a b g ( x ) d x

    The integral of a difference is the difference of the integrals.

  5. a b c f ( x ) d x = c a b f ( x )

    for constant c . The integral of the product of a constant and a function is equal to the constant multiplied by the integral of the function.

  6. a b f ( x ) d x = a c f ( x ) d x + c b f ( x ) d x

    Although this formula normally applies when c is between a and b , the formula holds for all values of a , b , and c , provided f ( x ) is integrable on the largest interval.
Practice Key Terms 8

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask