Chapter 6 The Definite Integral

Chapter Outline Antidifferentiation The Definite Integral and Net Change of a Function The Definite Integral and Area under a Graph Areas in the xy-Plane Applications of the Definite Integral

§ 6.1 Antidifferentiation

Section Outline Antidifferentiation Finding Antiderivatives Theorems of Antidifferentiation The Indefinite Integral Rules of Integration Antiderivatives in Application

Antidifferentiation - Antiderivative Definition Example Antidifferentiation: The process of determining f (x) given f ΄(x) If , then

Finding Antiderivatives EXAMPLE Find all antiderivatives of the given function. SOLUTION The derivative of x9 is exactly 9x8. Therefore, x9 is an antiderivative of 9x8. So is x9 + 5 and x9 -17.2. It turns out that all antiderivatives of f (x) are of the form x9 + C (where C is any constant*) as we will see next. *The derivative of a constant function is zero.

Theorems of Antidifferentiation

The Indefinite Integral

Rules of Integration Power Rule: Exponential Rule: Log Rule:

Rules of Integration

Finding Antiderivatives EXAMPLE Determine the following. SOLUTION Using the rules of indefinite integrals, we have

Finding Antiderivatives EXAMPLE Find the function f (x) for which and f (1) = 3. SOLUTION The unknown function f (x) is an antiderivative of . One antiderivative is . Therefore, by Theorem I, Now, we want the function f (x) for which f (1) = 3. So, we must use that information in our antiderivative to determine C. This is done below.

Finding Antiderivatives CONTINUED So, 3 = 1 + C and therefore, C = 2. Therefore, our function is

Antiderivatives in Application EXAMPLE A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v(t) = -32t feet per second. (a) Find s(t), the height of the rock above the ground at time t. (b) How long will the rock take to reach the ground? (c) What will be its velocity when it hits the ground? SOLUTION (a) We know that s΄(t) = v(t) = -32t and we also know that s(0) = 400. We can now use this information to find an antiderivative of v(t) for which s(0) = 400. The antiderivative of v(t) is To determine C, Therefore, C = 400. So, our antiderivative is s(t) = -16t2 + 400.

Antiderivatives in Application CONTINUED (b) To determine how long it will take for the rock to reach the ground, we simply need to find the value of t for which the position of the rock is at height 0. In other words, we will find t for when s(t) = 0. s(t) = -16t2 + 400 This is the function s(t). 0 = -16t2 + 400 Replace s(t) with 0. -400 = -16t2 Subtract. 25 = t2 Divide. 5 = t Take the positive square root since t ≥ 0. So, it will take 5 seconds for the rock to reach the ground.

Antiderivatives in Application CONTINUED (c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5). v(t) = -32t This is the function v(t). v(5) = -32(5) = -160 Replace t with 5 and solve. So, the velocity of the rock, as it hits the ground, is 160 feet per second in the downward direction (because of the minus sign).

§ 6.2 Definite Integral and Net Change of a Function

Section Outline The Definite Integral Properties of the Definite Integral The Definite Integral as a Net Change of a Function

The Definite Integral

The Definite Integral Computing a Definite Integral Evaluate EXAMPLE Computing a Definite Integral Evaluate SOLUTION An antiderivative of f(x) = x is

Properties of the Definite Integral

Using Properties of the Integral EXAMPLE Given that SOLUTION By (7), we have Using the given values of the integrals, we find

The Definite Integral as a Net Change of a Function In the definition of the definite integral (1), since f = F, we can rewrite (1) as (8) Recall that the derivative is the rate of change of F. Thus, (8) states that the integral of the rate of change of F is its net change as x varies from a to b.

§ 6.3 Definite Integral and Area under a Graph

Section Outline Area Under a Graph Calculating Definite Integrals The Fundamental Theorem of Calculus Area Under a Curve as an Antiderivative

Area Under a Graph Definition Example Area Under the Graph of f (x) from a to b: An example of this is shown to the right

Theorem I: Area under a Graph

Area Under the Graph of a Function EXAMPLE Use Theorem 1 to compute the area of the shaded region under the graph of f(x) = 3x2 + ex, from x = −1 to x = 1. SOLUTION

Riemann Sums In this section we will learn to estimate the area under the graph of f (x) from x = a to x = b by dividing up the interval into partitions (or subintervals), each one having width where n = the number of partitions that will be constructed. In the example below, n = 4. A Riemann Sum is the sum of the areas of the rectangles generated above.

Riemann Sums to Approximate Areas EXAMPLE Use a Riemann sum to approximate the area under the graph f (x) on the given interval using midpoints of the subintervals SOLUTION The partition of -2 ≤ x ≤ 2 with n = 4 is shown below. The length of each subinterval is x1 x2 x3 x4 -2 2

Riemann Sums to Approximate Areas CONTINUED Observe the first midpoint is units from the left endpoint, and the midpoints themselves are units apart. The first midpoint is x1 = -2 + = -2 + .5 = -1.5. Subsequent midpoints are found by successively adding midpoints: -1.5, -0.5, 0.5, 1.5 The corresponding estimate for the area under the graph of f (x) is So, we estimate the area to be 5 (square units).

Approximating Area With Midpoints of Intervals CONTINUED

Riemann Sums to Approximate Areas EXAMPLE Use a Riemann sum to approximate the area under the graph f (x) on the given interval using left endpoints of the subintervals SOLUTION The partition of 1 ≤ x ≤ 3 with n = 5 is shown below. The length of each subinterval is 1 1.4 1.8 2.2 2.6 3 x1 x2 x3 x4 x5

Riemann Sums to Approximate Areas CONTINUED The corresponding Riemann sum is So, we estimate the area to be 15.12 (square units).

Approximating Area Using Left Endpoints CONTINUED

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus EXAMPLE Use the Fundamental Theorem of Calculus to calculate the following integral. SOLUTION An antiderivative of 3x1/3 – 1 – e0.5x is . Therefore, by the fundamental theorem,

The Fundamental Theorem of Calculus EXAMPLE (Heat Diffusion) Some food is placed in a freezer. After t hours the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where (a) Compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2. (b) What does the area in part (a) represent? SOLUTION (a) To compute the area under the graph of y = r(t) over the interval 0 ≤ t ≤ 2, we evaluate the following.

The Fundamental Theorem of Calculus CONTINUED (b) Since the area under a graph can represent the amount of change in a quantity, the area in part (a) represents the amount of change in the temperature between hour t = 0 and hour t = 2. That change is 24.533 degrees Fahrenheit.

§ 6.4 Areas in the xy-Plane

Section Outline Properties of Definite Integrals Area Between Two Curves Finding the Area Between Two Curves

The Definite Integral Geometric interpretation of the definite integral:

Area Between Two Curves

Finding the Area Between Two Curves EXAMPLE Find the area of the region between y = 2x2 – 4x+6 and y = – x2 + 2x + 1 from x = 1 to x = 2. SOLUTION Upon sketching the two graphs (Fig. 7), we see that f(x) = 2x2 − 4x + 6 lies above g(x) = −x2 +2x+1 for 1 ≤ x ≤ 2. Therefore, our formula gives the area of the shaded region as

Finding the Area Between Two Curves EXAMPLE Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure. SOLUTION Since the two shaded regions are (1) disjoint and (2) have different functions on top, we will need a separate integral for each. Therefore

Finding the Area Between Two Curves CONTINUED Therefore, to represent all the shaded regions, we have

Finding the Area Between Two Curves EXAMPLE Two rockets are fired simultaneously straight up into the air. Their velocities (in meters per second) are v1(t) and v2(t), respectively, and v1(t) ≥ v2(t) for t ≥ 0. Let A denote the area of the region between the graphs of y = v1(t) and y = v2(t) for 0 ≤ t ≤ 10. What physical interpretation may be given to the value of A? SOLUTION Since v1(t) ≥ v2(t) for t ≥ 0, this suggests that the first rocket is always traveling at least as fast as the second rocket. Therefore, we have

Finding the Area Between Two Curves CONTINUED But again, since v1(t) ≥ v2(t) for t ≥ 0, we know that . So, this implies that . This means that the position of the first rocket is always at least as high (up in the air) as that of the second rocket. That is, the first rocket is always higher up than the second rocket (or at the same height).

§ 6.5 Applications of the Definite Integral

Section Outline Average Value of a Function Over an Interval Consumers’ Surplus Future Value of an Income Stream Volume of a Solid of Revolution

Average Value of a Function Over an Interval

Average Value of a Function Over an Interval EXAMPLE Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1. SOLUTION Using (2) with a = -1 and b = 1, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is equal to An antiderivative of 1 – x is . Therefore, So, the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1 is 1.

Average Value of a Function Over an Interval EXAMPLE (Average Temperature) During a certain 12-hour period the temperature at time t (measured in hours from the start of the period) was degrees. What was the average temperature during that period? SOLUTION The average temperature during the 12-hour period from t = 0 to t = 12 is

Definition: Consumers’ Surplus

Consumers’ Surplus (1 of 2) EXAMPLE Find the consumers’ surplus for the following demand curve at the given sales level x. SOLUTION Since 20 units are sold, the price must be Therefore, the consumers’ surplus is

Consumers’ Surplus (2 of 2) CONTINUED That is, the consumers’ surplus is $20.

Future Value of an Income Stream

Future Value of an Income Stream EXAMPLE (Future Value) Suppose that money is deposited daily into a savings account at an annual rate of $2000. If the account pays 6% interest compounded continuously, approximately how much will be in the account at the end of 2 years? SOLUTION Divide the time interval from 0 to 2 years into daily subintervals. Each subinterval is then of duration years. Let t1, t2, ..., tn be points chosen from these subintervals. Since we deposit money at an annual rate of $2000, the amount deposited during one of the subintervals is dollars. If this amount is deposited at time ti, the dollars will earn interest for the remaining 2 – ti years. The total amount resulting from this one deposit at time ti is then

Future Value of an Income Stream CONTINUED Add the effects of the deposits at times t1, t2, ..., tn to arrive at the total balance in the account: This is a Riemann sum for the function on the interval 0 ≤ t ≤ 2. Since is very small when compared with the interval, the total amount in the account, A, is approximately That is, the approximate balance in the account at the end of 2 years is $4250.

Volume of a Solid of Revolution

Volume of a Solid of Revolution EXAMPLE Find the volume of a solid of revolution generated by revolving about the x-axis the region under the following curve. SOLUTION Here g(x) = x2, and