Chapter 4.3: Area and Definite Integrals

Area and Definite Integrals

Objective

  • Find the area under the graph of a nonnegative function over a given closed interval.

  • Evaluate a definite integral.

  • Solve applied problems involving definite integrals.

Finding the Area Under a Nonnegative Function

To find the area under the graph of a nonnegative, continuous function f over the interval [a, b]:

  1. Find any antiderivative F(x) of f(x). The simplest one is typically where the constant of integration is 0.

  2. Evaluate F(x) using b and a, and compute F(b) - F(a). The result is the area under the graph over the interval [a, b].

Example 1: Area under y = x^2 + 1 over [-1, 2]

  1. Find an antiderivative F(x) of f(x) = x^2 + 1. The simplest one is:
    F(x) = \frac{x^3}{3} + x

  2. Substitute 2 and -1, and find the difference F(2) - F(-1):

    F(2) = \frac{2^3}{3} + 2 = \frac{8}{3} + 2 = \frac{14}{3}

    F(-1) = \frac{(-1)^3}{3} + (-1) = -\frac{1}{3} - 1 = -\frac{4}{3}

    F(2) - F(-1) = \frac{14}{3} - (-\frac{4}{3}) = \frac{18}{3} = 6

Therefore, the area under the graph of y = x^2 + 1 over the interval [-1, 2] is 6.

Quick Check 1

Refer to the function in Example 1, f(x) = x^2 + 1.

a) Calculate the area over the interval [0, 5].

Substitute 0 and 5, and find the difference:
F(5) = \frac{5^3}{3} + 5 = \frac{125}{3} + 5 = \frac{140}{3}
F(0) = \frac{0^3}{3} + 0 = 0
F(5) - F(0) = \frac{140}{3} - 0 = \frac{140}{3}

b) Calculate the area over the interval [-2, 2].

c) Suggest a shortcut for part (b)?

Integrate from 0 to 2, then double the results, because the graph of f is symmetric with the y-axis.

Note that f(-x) = f(x). Then,

\int{-2}^{2} f(x) dx = 2 \int{0}^{2} f(x) dx

Definition: Definite Integral

Let f be any continuous function over the interval [a, b] and F be any antiderivative of f. Then, the definite integral of f from a to b is:

\int_{a}^{b} f(x) dx = F(b) - F(a)

Example 2: Evaluate \int_{1}^{8} x^2 dx

Using the antiderivative F(x) = \frac{x^3}{3}, we have:

\int_{1}^{8} x^2 dx = F(8) - F(1) = \frac{8^3}{3} - \frac{1^3}{3} = \frac{512}{3} - \frac{1}{3} = \frac{511}{3}

It is convenient to use an intermediate notation:

\int{a}^{b} f(x) dx = F(x) \Big|{a}^{b} = F(b) - F(a)

where F(x) is an antiderivative of f(x).

Example 3: Evaluate

  1. \int_{-2}^{4} (x^3 + 3x^2 - 4) dx

    = (\frac{x^4}{4} + x^3 - 4x) \Big|_{-2}^{4}

    = (\frac{4^4}{4} + 4^3 - 4 \cdot 4) - (\frac{(-2)^4}{4} + (-2)^3 - 4 \cdot (-2))

    = (64 + 64 - 16) - (4 - 8 + 8)

    = 112 - 4 = 108

  2. \int_{1}^{4} 5\sqrt{x} dx

    = 5 \int_{1}^{4} x^{\frac{1}{2}} dx

    = 5 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \Big|_{1}^{4}

    = 5 \cdot \frac{2}{3} x^{\frac{3}{2}} \Big|_{1}^{4}

    = \frac{10}{3} (4^{\frac{3}{2}} - 1^{\frac{3}{2}})

    = \frac{10}{3} (8 - 1) = \frac{70}{3}

  3. \int_{0}^{1} e^{-0.02x} dx

    = \frac{e^{-0.02x}}{-0.02} \Big|_{0}^{1}

    = -50 e^{-0.02x} \Big|_{0}^{1}

    = -50 (e^{-0.02} - e^{0})

    = -50 (e^{-0.02} - 1)

    \approx 0.9901

Quick Check 2: Evaluate

a) \int_{0}^{5} 2x dx

= x^2 \Big|_{0}^{5} = 5^2 - 0^2 = 25

b) \int_{1}^{9} \frac{1}{x} dx

= ln(x) \Big|_{1}^{9} = ln(9) - ln(1) = ln(9)

c) \int_{0}^{4} 3 dx

= 3x \Big|_{0}^{4} = 3(4) - 3(0) = 12

The Fundamental Theorem of Integral Calculus

If a continuous function f has an antiderivative F over [a, b], then

\int_{a}^{b} f(x) dx = F(b) - F(a)

Example 4

Suppose that y is the profit per mile traveled and x is the number of miles traveled, in thousands. Find the area under y = \frac{1}{x} over the interval [1, 4] and interpret the significance of this area.

\int_{1}^{4} \frac{1}{x} dx

= ln(x) \Big|_{1}^{4} = ln(4) - ln(1) = ln(4) - 0 = ln(4) \approx 1.3863

The area represents a total profit of $1386.30 when the miles traveled increase from 1000 to 4000 miles.

Example 5

Predict the sign of the integral by examining the graph, and then evaluate the integral.

\int_{-1}^{2} (x^2 - 1) dx

From the graph, it appears that there is considerably more area below the x-axis than above. Thus, we expect that the sign of the integral will be negative.

Evaluating the integral, we have

\int_{-1}^{2} (x^2 - 1) dx

= (\frac{x^3}{3} - x) \Big|_{-1}^{2}

= (\frac{2^3}{3} - 2) - (\frac{(-1)^3}{3} - (-1))

= (\frac{8}{3} - 2) - (-\frac{1}{3} + 1)

= (\frac{8}{3} - \frac{6}{3}) - (-\frac{1}{3} + \frac{3}{3})

= \frac{2}{3} - \frac{2}{3} = -\frac{4}{3} \approx -1.333

Example 6

Northeast Airlines determines that the marginal profit resulting from the sale of x seats on a jet traveling from Atlanta to Kansas City, in hundreds of dollars, is given by

P'(x) = -x^2 + 25x - 300

Find the total profit when 60 seats are sold.

We integrate to find P(60).

\int_{0}^{60} (-x^2 + 25x - 300) dx

= (-\frac{x^3}{3} + \frac{25x^2}{2} - 300x) \Big|_{0}^{60}

= (-\frac{60^3}{3} + \frac{25(60)^2}{2} - 300(60)) - (0)

= -72000 + 45000 - 18000 = -45000

When 60 seats are sold, Northeast’s profit is –$5016.13. That is, the airline will lose $5016.13 on the flight.

Quick Check 3

Referring to Example 6, find the total profit of Northeast Airlines when 140 seats are sold.

From Example 6, we have

\int_{0}^{140} (-x^2 + 25x - 300) dx

= (-\frac{x^3}{3} + \frac{25x^2}{2} - 300x) \Big|_{0}^{140}

= (-\frac{140^3}{3} + \frac{25(140)^2}{2} - 300(140)) - (0)

When 140 seats are sold, Northeast Airlines makes

$132,666.67

Example 7

A particle starts out from some origin. Its velocity, in miles per hour, is given by

v(t) = t^2 + t

where t is the number of hours since the particle left the origin. How far does the particle travel during the second, third, and fourth hours (from t = 1 to t = 4)?

Recall that velocity, or speed, is the rate of change of distance with respect to time. In other words, velocity is the derivative of the distance function, and the distance function is an antiderivative of the velocity function. To find the total distance traveled from t = 1 to t = 4, we evaluate the integral

\int_{1}^{4} (t^2 + t) dt

= (\frac{t^3}{3} + \frac{t^2}{2}) \Big|_{1}^{4}

= (\frac{4^3}{3} + \frac{4^2}{2}) - (\frac{1^3}{3} + \frac{1^2}{2})

= (\frac{64}{3} + \frac{16}{2}) - (\frac{1}{3} + \frac{1}{2})

= (\frac{64}{3} + 8) - (\frac{1}{3} + \frac{1}{2})

= \frac{63}{3} + \frac{15}{2} = 21 + 7.5 = 28.5

Section Summary

  • The exact area between the x-axis and the graph of the nonnegative continuous function over the interval is found by evaluating the definite integral
    \int_{a}^{b} f(x) dx

    where F is an antiderivative of f.

  • If a function has areas both below and above the x-axis, the definite integral gives the net total area, or the difference between the sum of the areas above the x-axis and the sum of the areas below the x-axis.

  • If there is more area above the x-axis than below, the definite integral will be positive.

  • If there is more area below the x-axis than above, the definite integral will be negative.

  • If the areas above and below the x-axis are the same, the definite integral will be 0. Note that this does NOT mean that the function is zero. It means that the integral happens to resolve to zero.