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The Function for the Second Fundamental Theorem of Calculus for AP Calculus

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By McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus

The Function

The Second Fundamental Theorem of Calculus defines

and states that if f is continuous on [a, b], then F '(x) = f (x) for every point x in [a, b].

If f ≥ 0, then F ≥ 0. F (x) can be interpreted geometrically as the area under the curve of f from t = a to t = x. (See Figure 12.1-1.)

If f < 0, F < 0, F(x) can be treated as the negative value of the area between the curve of f and the t-axis from t = a to t = x. (See Figure 12.1-2.)

Example 1

If F (x) = 2 cos t dt for 0 ≤ x ≤ 2π, find the value(s) of x where f has a local minimum.

Method 1:

Method 2:   You can solve this problem geometrically by using area. See Figure 12.1-3.

The area "under the curve" is above the t-axis on [0, π/2] and below the x-axis on [π/2, 3π/2]. Thus the local minimum occurs at 3π/2.

Example 2

Let p(x) = f(t)dt and the graph of f is shown in Figure 12.1-4.

1. Evaluate: p(0), p(1), p(4).
2. Evaluate: p(5), p(7), p(8).
3. At what value of t does p have a maximum value?
4. On what interval(s) is p decreasing?
5. Draw a sketch of the graph of p.

Solution:

1. Since f ≥ 0 on the interval [0, 4], p attains a maximum at t = 4.
2. Since f(t) is below the x-axis from t = 4 to t = 8, if x > 4,
3. Thus, p is decreasing on the interval (4, 8).

4. p(x) = f(t)dt. See Figure 12.1-5 on page 262 for a sketch.

Example 3

The position function of a moving particle on a coordinate axis is:

s = f(x)dx, where t is in seconds and s is in feet.

The function f is a differentiable function and its graph is shown below in Figure 12.1-6.

1. What is the particle's velocity at t = 4?
2. What is the particle's position at t = 3?
3. When is the acceleration zero?
4. When is the particle moving to the right?
5. At t = 8, is the particle on the right side or left side of the origin?

Solution:

1. Since s = f(x)dx, then v(t)=s '(t) = f(t).
2. Thus, v(4) = –8 ft/sec.

3. .
4. a(t) = v '(t). Since v '(t) = f '(t), v '(t) = 0 at t = 4. Thus, a(4) = 0 ft/sec2.
5. The particle is moving to the right when v(t > 0. Thus, the particle is moving to the right on intervals (0, 2) and (7, 8).
6. The area of f below the x-axis from x = 2 to x = 7 is larger than the area of f above the x-axis from x = 0 to x = 2 and x = 7 to x = 8. Thus, f(x)dx < 0 and the particle is on the left side of the origin.

Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus

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