- 6.3 1st Fundamental Theorem Of Calculus Ap Calculus Transcendentals
- 6.3 1st Fundamental Theorem Of Calculus Ap Calculus Multiple Choice
- 6.3 1st Fundamental Theorem Of Calculus Ap Calculus 14th Edition
- 6.3 1st Fundamental Theorem Of Calculus Ap Calculus Frq
Rolle’s Theorem; Fundamental Theorem of Calculus (two parts) Mean Value Theorem for Integrals; A Theorem by any other Name There are many other results and formulas in calculus that may not have the title of “Theorem” but are nevertheless important theorems. Every one of your derivative and antidifferentiation rules is actually a theorem. Cannot evaluate using the First Fundamental Theorem of Calculus since is discontinuous at x = 1. Using a graphing calculator, evaluate. Using a TI-89 graphing calculator, enter and obtain 2π. Second Fundamental Theorem of Calculus. If f is continuous on a, b and, then F ′ (x) = f (x) at every point x in a, b. This theorem, like the Fundamental Theorem of Calculus, says roughly that if we integrate a 'derivative-like function' (f ′ or ∇f) the result depends only on the values of the original function (f) at the endpoints. If a vector field F is the gradient of a function, F = ∇f, we say that F is a conservative vector field.
5 Steps to a 5: AP Calculus AB 2017 (2016)
STEP 4
Review the Knowledge You Need to Score High
CHAPTER 12
Big Idea 3: Integrals and the Fundamental Theorems of Calculus
Definite Integrals
IN THIS CHAPTER
Summary: In this chapter, you will be introduced to the summation notation, the concept of a Riemann Sum, the Fundamental Theorems of Calculus, and the properties of definite integrals. You will also be shown techniques for evaluating definite integrals involving algebraic, trigonometric, logarithmic, and exponential functions. In addition, you will learn how to work with improper integrals. The ability to evaluate integrals is a prerequisite to doing well on the AP Calculus AB exam.
Key Ideas
Summation Notation
Riemann Sums
Properties of Definite Integrals
The First Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus
Evaluating Definite Integrals
12.1 Riemann Sums and Definite Integrals
Main Concepts: Sigma Notation, Definition of a Riemann Sum, Definition of a Definite Integral, and Properties of Definite Integrals
Sigma Notation or Summation Notation
where i is the index of summation, l is the lower limit, and n is the upper limit of summation. (Note: The lower limit may be any non-negative integer ≤ n .)
Examples
Summation Formulas
If n is a positive integer, then:
1.
2.
3.
4.
5.
Example
.
(Note: This question has not appeared in an AP Calculus AB exam in recent years.)
• Remember: In exponential growth/decay problems, the formulas are and y = y0ekt.
Definition of a Riemann Sum
Let f be defined on [a , b ] and xibe points on [a , b ] such that x0 = a , xn= b , and a< x1< x2< x3 … < xn –1< b . The points a , x1 , x2 , x3 , … xn –1 , and b form a partition of f denoted as Δ on [a , b ]. Let Δxibe the length of the i th interval [xi –1 , xi] and cibe any point in the i th interval. Then the Riemann sum of f for the partition is .
Example 1
Let f be a continuous function defined on [0, 12] as shown below.
Find the Riemann sum for f (x ) over [0, 12] with 3 subdivisions of equal length and the midpoints of the intervals as ci.
Length of an interval . (See Figure 12.1-1 .)
Figure 12.1-1
The Riemann sum is 596.
Example 2
Find the Riemann sum for f (x ) = x3 + 1 over the interval [0, 4] using 4 subdivisions of equal length and the midpoints of the intervals as ci. (See Figure 12.1-2 .)
Figure 12.1-2
Length of an interval .
Enter Σ ((1 – 0.5)3 + 1, i , 1, 4) = 66.
The Riemann sum is 66.
Definition of a Definite Integral
Let f be defined on [a , b ] with the Riemann sum for f over [a , b ] written as .
If max Δxiis the length of the largest subinterval in the partition and the exists, then the limit is denoted by:
.
is the definite integral of f from a to b .
Example 1
Use a midpoint Riemann sum with three subdivisions of equal length to find the approximate value of .
midpoints are x = 1, 3, and 5.
Example 2
Using the limit of the Riemann sum, find .
Using n subintervals of equal lengths, the length of an interval
Let ci= xi; max Δxi→ 0 ⇒ n → ∞.
Thus,
(Note: This question has not appeared in an AP Calculus AB exam in recent years.)
Properties of Definite Integrals
1. If f is defined on [a , b ], and the exists, then f is integrable on [a , b ].
2. If f is continuous on [a , b ], then f is integrable on [a , b ].
If f (x ), g (x ), and h (x ) are integrable on [a , b ], then
3.
4.
5. when C is a constant.
6.
7. provided f (x ) ≥ 0 on [a , b ].
8. provided f (x ) ≥ g (x ) on [a , b ].
9.
10. ; provided g (x ) ≤ f (x ) ≤ h (x ) on [a , b ].
11. ; provided m ≤ f (x ) ≤ M on [a , b ].
12. ; provided f (x ) is integrable on an interval containing a , b , c .
Examples
1.
2.
3.
4.
5.
Note: Or
do not have to be arranged from smallest to largest.
The remaining properties are best illustrated in terms of the area under the curve of the function as discussed in the next section.
• Do not forget that .
12.2 Fundamental Theorems of Calculus
Main Concepts: First Fundamental Theorem of Calculus, Second Fundamental Theorem of Calculus
First Fundamental Theorem of Calculus
If f is continuous on [a , b ] and F is an antiderivative of f on [a , b ], then
.
Note: F (b ) – F (a ) is often denoted as .
Example 1
Evaluate .
.
Example 2
Evaluate .
Example 3
If , find k .
Example 4
If f ′ (x ) = g (x ), and g is a continuous function for all real values of x , express in terms of f .
Let u = 3x ; du = 3dx or .
Example 5
Evaluate .
Cannot evaluate using the First Fundamental Theorem of Calculus since is discontinuous at x = 1.
Example 6
Using a graphing calculator, evaluate .
Using a TI-89 graphing calculator, enter and obtain 2π .
Second Fundamental Theorem of Calculus
If f is continuous on [a , b ] and , then F ′ (x ) = f (x ) at every point x in [a , b ].
Example 1
Evaluate .
Let u = 2t ; du = 2dt or .
Example 2
Example 3
Find ; if y = .
Let u = 2x ; then .
Rewrite: .
Example 4
Find ; if .
Rewrite: .
Let u = x2 ; then .
Rewrite: .
Example 5
Find ; if .
Example 6
, integrate to find F (x ) and then differentiate to find f ′ (x ).
12.3 Evaluating Definite Integrals
Main Concepts: Definite Integrals Involving Algebraic Functions; Definite Integrals Involving Absolute Value; Definite Integrals Involving Trigonometric, Logarithmic, and Exponential Functions; Definite Integrals Involving Odd and Even Functions
• If the problem asks you to determine the concavity of f ′ (not f ), you need to know if f ″ is increasing or decreasing, or if f ′″ is positive or negative.
Definite Integrals Involving Algebraic Functions
Example 1
Evaluate .
Verify your result with a calculator.
Example 2
Evaluate .
Begin by evaluating the indefinite integral .
Let u = x2 – 1; du = 2x dx or .
Rewrite: .
Thus the definite integral
Verify your result with a calculator.
Example 3
Evaluate .
Rewrite:
Verify your result with a calculator.
• You may bring up to 2 (but no more than 2) approved graphing calculators to the exam.
Definite Integrals Involving Absolute Value
Example 1
Evaluate .
Set 3x – 6 = 0; x = 2; thus .
Rewrite integral:
Verify your result with a calculator.
Example 2
Evaluate .
Set x2 – 4 = 0; x = ± 2.
Thus .
.
Verify your result with a calculator.
• You are not required to clear the memories in your calculator for the exam.
Definite Integrals Involving Trigonometric, Logarithmic, and Exponential Functions
Example 1
Evaluate .
Rewrite: .
Verify your result with a calculator.
Example 2
Evaluate .
Let u = 3t ; du = 3dt or .
Verify your result with a calculator.
Example 3
.
Verify your result with a calculator.
Example 4
.
Verify your result with a calculator.
Definite Integrals Involving Odd and Even Functions
If f is an even function, that is, f (–x ) = f (x ), and is continuous on [–a , a ], then
.
If f is an odd function, that is, F (x ) = – f (–x ), and is continuous on [–a , a ], then
.
Example 1
Evaluate .
Since f (x ) = cos x is an even function,
Verify your result with a calculator.
Example 2
Evaluate .
Since f (x ) = x4 – x2 is an even function, i.e., f (–x ) = f (x ), thus
Verify your result with a calculator.
Example 3
Evaluate .
Since f (x ) = sin x is an odd function, i.e., f (–x ) = –f (x ), thus
.
Verify your result algebraically.
You can also verify the result with a calculator.
Example 4
If for all values of k , then which of the following could be the graph of f ? (See Figure 12.3-1 .)
Figure 12.3-1
Thus f is an even function. Choice (C).
12.4 Rapid Review
1. Evaluate .
Answer: .
2. Evaluate .
Answer: .
3. If , find G ′ (4).
Answer: G ′(x ) = (2x + 1)3/2 and G ′ (4) = 93/2 = 27.
4. If , find k .
Answer: .
5. If G (x ) is an antiderivative of (ex+ 1) and G (0) = 0, find G (1).
Answer: G (x ) = ex+ x + C
G (0) = e0 + 0 + C = 0 ⇒ C = –1.
G (1) = e1 + 1 – 1 = e .
6. If G ′ (x ) = g (x ), express in terms of G (x ).
Answer: Let .
. Thus .
12.5 Practice Problems
Part A—The use of a calculator is not allowed .
Evaluate the following definite integrals.
1 .
2 .
3 .
4 .
5 . If = 4, find k .
6 .
7 . If f ′ (x ) = g (x ) and g is a continuous function for all real values of x , express in terms of f .
8 .
9 .
10 . If (t )dt , find .
11 .
12 .
Part B—Calculators are allowed.
13 . Find k if .
14 . Evaluate to the nearest 100th.
15 . If , find .
16 . Use a midpoint Riemann sum with four subdivisions of equal length to find the approximate value of .
17 . Given
and , find
(a)
(b)
(c)
(d)
18 . Evaluate .
19 . Find if .
20 . Let f be a continuous function defined on [0, 30] with selected values as shown below:
Use a midpoint Riemann sum with three subdivisions of equal length to find the approximate value of .
12.6 Cumulative Review Problems
(Calculator) indicates that calculators are permitted.
21 . Evaluate .
22 . Find at x = 3 if y = |x2 – 4|.
6.3 1st Fundamental Theorem Of Calculus Ap Calculus Transcendentals
23 . The graph of f ′, the derivative of f , –6 ≤ x ≤ 8 is shown in Figure 12.6-1 .
Figure 12.6-1
(a) Find all values of x such that f attains a relative maximum or a relative minimum.
(b) Find all values of x such that f is concave upward.
(c) Find all values of x such that f has a change of concavity.
24 . (Calculator) Given the equation 9x2 + 4y2 – 18x + 16y = 11, find the points on the graph where the equation has a vertical or horizontal tangent.
25 . (Calculator) Two corridors, one 6 feet wide and another 10 feet wide meet at a corner. (See Figure 12.6-2 .) What is the maximum length of a pipe of negligible thickness that can be carried horizontally around the corner?
Figure 12.6-2
12.7 Solutions to Practice Problems
Part A—The use of a calculator is not allowed.
1 .
2 . Let u = x – 2 du = dx .
3 . Let u = t + 1; du = dt and t = u – 1.
4 . Set x – 3 = 0; x = 3.
5 .
Verify your results by evaluating
.
6 . Let u = 1 + cos x ; du = – sin x dx or – du = sin x dx .
7 . Let u = 4x ; du = 4 dx or .
8 .
9 . Let u = t + 3; du = dt .
10 .
11 .
Note that f (x ) = 4xex2is an odd function.
Thus, .
12 .
Note that f (x ) = cos x – x2 is an even function. Thus, you could have written and obtained the same result.
Part B—Calculators are allowed.
13 .
Set 4 + 2k = 10 and k = 3.
14 . Enter and obtain – 4.70208 ≈ – 4.702.
15 .
16 .
Midpoints are x = 1, 3, 5, and 7.
17 . (a)
(b)
(c)
(d)
18 .
19 .
20 .
Midpoints are x = 5, 15, and 25.
12.8 Solutions to Cumulative Review Problems
21 . As x → –∞, .
22 .
23 . (a) (See Figure 12.8-1 .)
Figure 12.8-1
The function f has a relative minimum at x = – 5 and x = 3, and f has a relative maximum at x = – 1 and x = 7.
(b) (See Figure 12.8-2 .)
Figure 12.8-2
The function f is concave upward on intervals (– 6, – 3) and (1, 5).
(c) A change of concavity occurs at x = – 3, x = 1, and x = 5.
24 . (Calculator) Differentiate both sides of 9x2 + 4y2 – 18x + 16y = 11.
Horizontal tangent .
Using a calculator, enter [Solve ]
(4y∧ 2 + 16y – 20 = 0, y ); obtaining y = – 5 or y = 1.
Thus at each of the points at (1, 1) and (1, – 5) the graph has a horizontal tangent.
Vertical tangent is undefined.
Set 8y + 16 = 0 ⇒ y = – 2.
At y = – 2, 9x2 + 16 – 18x – 32 = 11
⇒ 9x2 – 18x – 27 = 0.
Enter [Solve ] (9x2 – 18x – 27 = 0, x ) and obtain x = 3or x = – 1.
Thus, at each of the points (3, – 2) and (– 1, – 2), the graph has a vertical tangent. (See Figure 12.8-3 .)
Figure 12.8-3
25 . (Calculator)
Step 1: (See Figure 12.8-4 .) Let P = x + y where P is the length of the pipe and x and y are as shown. The minimum value of P is the maximum length of the pipe to be able to turn in the corner. By similar triangles, and thus,
.
Figure 12.8-4
Step 2: Find the minimum value of P .
Enter . Use the [Minimum ] function of the calculator and obtain the minimum point (9.306, 22.388).
Step 3: Verify with the First Derivative Test.
Enter y 2 = (y 1(x ), x ) and observe. (See Figure 12.8-5 .)
Figure 12.8-5
Step 4: Check endpoints.
The domain of x is (6, ∞). Since x = 9.306 is the only relative extremum, it is the absolute minimum. Thus the maximum length of the pipe is 22.388 feet.
In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function. Unfortunately, so far, the only tools we have available to calculate the value of a definite integral are geometric area formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section we look at some more powerful and useful techniques for evaluating definite integrals.
These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and explored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s, and it is codified in what we now call the Fundamental Theorem of Calculus, which has two parts that we examine in this section. Its very name indicates how central this theorem is to the entire development of calculus.
Isaac Newton’s contributions to mathematics and physics changed the way we look at the world. The relationships he discovered, codified as Newton’s laws and the law of universal gravitation, are still taught as foundational material in physics today, and his calculus has spawned entire fields of mathematics. To learn more, read a brief biography of Newton with multimedia clips.
Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives
The Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.
Fundamental Theorem of Calculus I
If (f(x)) is continuous over an interval ([a,b]), and the function (F(x)) is defined by
[F(x)=∫^x_af(t)dt,]
then (F′(x)=f(x)) over ([a,b]).
A couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, (F(x)), as the definite integral of another function, (f(t)), from the point a to the point x. At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of x, the definite integral is a number. So the function (F(x)) returns a number (the value of the definite integral) for each value of x.
Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function has an antiderivative.
Example (PageIndex{3}): Finding a Derivative with the Fundamental Theorem of Calculus
Use the Note to find the derivative of
(displaystyle g(x)=∫^x_1frac{1}{t^3+1}dt.)
Solution: According to the Fundamental Theorem of Calculus, the derivative is given by
(displaystyle g′(x)=frac{1}{x^3+1}.)
Exercise (PageIndex{3})
Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of (displaystyle g(r)=∫^r_0sqrt{x^2+4}dx).
Follow the procedures from Example to solve the problem.
(g′(r)=sqrt{r^2+4})
Example (PageIndex{4}): Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives
Let (displaystyle F(x)=∫^{sqrt{x}}_1sintdt.) Find (F′(x)).
Solution
Letting (u(x)=sqrt{x}), we have (displaystyle F(x)=∫^{u(x)}_1sintdt). Thus, by the Fundamental Theorem of Calculus and the chain rule,
[displaystyle F′(x)=sin(u(x))frac{du}{dx}=sin(u(x))⋅(frac{1}{2}x^{−1/2})=frac{sinsqrt{x}}{2sqrt{x}}.]
Exercise (PageIndex{4})
Let (displaystyle F(x)=∫^{x^3}_1costdt). Find (F′(x)).
Use the chain rule to solve the problem.
(F′(x)=3x^2cosx^3)
Example (PageIndex{5}): Using the Fundamental Theorem of Calculus with Two Variable Limits of Integration
Let (displaystyle F(x)=∫^{2x}_xt3dt). Find (F′(x)).
We have (displaystyle F(x)=∫^{2x}_xt^3dt). Both limits of integration are variable, so we need to split this into two integrals. We get
(displaystyle F(x)=∫^{2x}_xt^3dt=∫^0_xt^3dt+∫^{2x}_0t^3dt=−∫^x_0t^3dt+∫^{2x}_0t3dt.)
Differentiating the first term, we obtain
(displaystyle frac{d}{dx}[−∫^x_0t^3dt]=−x^3).
Differentiating the second term, we first let ((x)=2x.) Then,
(displaystyle frac{d}{dx}[∫^{2x}_0t^3dt]=frac{d}{dx}[∫^{u(x)}_0t^3dt]=(u(x))^3dudx=(2x)^3⋅2=16x^3.)
Thus,
(displaystyle F′(x)=frac{d}{dx}[−∫^x_0t^3dt]+frac{d}{dx}[∫^{2x}_0t^3dt]=−x^3+16x^3=15x^3)
Exercise (PageIndex{5})
Let (displaystyle F(x)=∫^{x2}_xcostdt.) Find (F′(x)).
Use the procedures from Example to solve the problem
(F′(x)=2xcosx^2−cosx)
Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem
The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of objects. Our view of the world was forever changed with calculus.
After finding approximate areas by adding the areas of n rectangles, the application of this theorem is straightforward by comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.
The Fundamental Theorem of Calculus, Part 2
If f is continuous over the interval ([a,b]) and (F(x)) is any antiderivative of (f(x),) then
[ ∫^b_af(x)dx=F(b)−F(a).]
We often see the notation (displaystyle F(x)|^b_a) to denote the expression (F(b)−F(a)). We use this vertical bar and associated limits a and b to indicate that we should evaluate the function (F(x)) at the upper limit (in this case, b), and subtract the value of the function (F(x)) evaluated at the lower limit (in this case, a).
The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting.
Proof
Let (P={x_i},i=0,1,…,n) be a regular partition of ([a,b].) Then, we can write
[ begin{align} F(b)−F(a) &=F(x_n)−F(x_0) nonumber &=[F(x_n)−F(x_{n−1})]+[F(x_{n−1})−F(x_{n−2})]+…+[F(x_1)−F(x_0)] nonumber &=sum^n_{i=1}[F(x_i)−F(x_{i−1})]. nonumber end{align}nonumber ]
Now, we know (F) is an antiderivative of (f) over ([a,b],) so by the Mean Value Theorem (see The Mean Value Theorem) for (i=0,1,…,n) we can find (c_i) in ([x_{i−1},x_i]) such that
[F(x_i)−F(x_{i−1})=F′(c_i)(x_i−x_{i−1})=f(c_i)Δx.]
Then, substituting into the previous equation, we have
[displaystyle F(b)−F(a)=sum_{i=1}^nf(c_i)Δx.]
Taking the limit of both sides as (n→∞,) we obtain
[displaystyle F(b)−F(a)=lim_{n→∞}sum_{i=1}^nf(c_i)Δx=∫^b_af(x)dx.]
□
Example (PageIndex{6}): Evaluating an Integral with the Fundamental Theorem of Calculus
Use Note to evaluate
[displaystyle ∫^2_{−2}(t^2−4)dt.]
Solution
Recall the power rule for Antiderivatives:
If
[displaystyle y=x^n,∫x^ndx=frac{x^{n+1}}{n+1}+C.]
Use this rule to find the antiderivative of the function and then apply the theorem. We have
(displaystyle ∫^2_{−2}(t^2−4)dt=frac{t^3}{3}−4t|^2−2)
(displaystyle =[frac{(2)^3}{3}−4(2)]−[frac{(−2)^3}{3}−4(−2)])
(displaystyle =(frac{8}{3}−8)−(−frac{8}{3}+8))
(displaystyle =frac{8}{3}−8+frac{8}{3}−8=frac{16}{3}−16=−frac{32}{3}.)
Analysis
6.3 1st Fundamental Theorem Of Calculus Ap Calculus Multiple Choice
Notice that we did not include the “+ C” term when we wrote the antiderivative. The reason is that, according to the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose the antiderivative with (C=0.) If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.
The region of the area we just calculated is depicted in Figure. Note that the region between the curve and the x-axis is all below the x-axis. Area is always positive, but a definite integral can still produce a negative number (a net signed area). For example, if this were a profit function, a negative number indicates the company is operating at a loss over the given interval.
Figure (PageIndex{3}): The evaluation of a definite integral can produce a negative value, even though area is always positive.
Example (PageIndex{7}): Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2
Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:
(displaystyle ∫^9_1frac{x−1}{sqrt{x}dx}.)
First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms by writing each one over the denominator:
(displaystyle ∫^9_1frac{x−1}{x^{1/2}}dx=∫^9_1(frac{x}{x^{1/2}}−frac{1}{x^{1/2}})dx.)
Use the properties of exponents to simplify:
(displaystyle ∫^9_1(frac{x}{x^{1/2}}−frac{1}{x^{1/2}})dx=∫^9_1(x^{1/2}−x^{−1/2})dx.)
Now, integrate using the power rule:
(displaystyle ∫^9_1(x^{1/2}−x^{−1/2})dx=(frac{x^{3/2}}{frac{3}{2}}−frac{x^{1/2}}{frac{1}{2}})∣^9_1)
(displaystyle =[frac{(9)^{3/2}}{frac{3}{2}}−frac{(9)^{1/2}}{frac{1}{2}}]−[frac{(1)^{3/2}}{frac{3}{2}}−frac{(1)^{1/2}}{frac{1}{2}}])
(displaystyle =[frac{2}{3}(27)−2(3)]−[frac{2}{3}(1)−2(1)]=18−6−frac{2}{3}+2=frac{40}{3}.)
See Figure.
.
Figure (PageIndex{4}): The area under the curve from (x=1) to (x=9) can be calculated by evaluating a definite integral.
Exercise (PageIndex{6})
Use Note to evaluate (displaystyle ∫^2_1x^{−4}dx.)
Use the power rule.
(frac{7}{24})
Example (PageIndex{8}): A Roller-Skating Race
James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the farthest after 5 sec wins a prize. If James can skate at a velocity of (f(t)=5+2t) ft/sec and Kathy can skate at a velocity of (g(t)=10+cos(frac{π}{2}t)) ft/sec, who is going to win the race?
Solution
We need to integrate both functions over the interval ([0,5]) and see which value is bigger. We are using (∫^5_0v(t)dt) to find the distance traveled over 5 seconds. For James, we want to calculate
[displaystyle ∫^5_0(5+2t)dt.]
Using the power rule, we have
[displaystyle ∫^5_0(5+2t)dt=(5t+t^2)∣^5_0=(25+25)=50.]
Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate
[displaystyle ∫^5_010+cos(frac{π}{2}t)dt.]
We know (sint) is an antiderivative of (cost), so it is reasonable to expect that an antiderivative of (cos(frac{π}{2}t)) would involve (sin(frac{π}{2}t)). However, when we differentiate (sin(π2t), we get π2cos(π2t) as a result of the chain rule, so we have to account for this additional coefficient when we integrate. We obtain
[displaystyle ∫^5_010+cos(frac{π}{2}t)dt=(10t+frac{2}{π}sin(frac{π}{2}t))∣^5_0]
6.3 1st Fundamental Theorem Of Calculus Ap Calculus 14th Edition
[=(50+frac{2}{π})−(0−frac{2}{π}sin0)≈50.6.]
Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!
Exercise (PageIndex{7})
Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec. Does this change the outcome?
Change the limits of integration from those in Example.
Kathy still wins, but by a much larger margin: James skates 24 ft in 3 sec, but Kathy skates 29.3634 ft in 3 sec.
A Parachutist in Free Fall
Julie is an avid skydiver. She has more than 300 jumps under her belt and has mastered the art of making adjustments to her body position in the air to control how fast she falls. If she arches her back and points her belly toward the ground, she reaches a terminal velocity of approximately 120 mph (176 ft/sec). If, instead, she orients her body with her head straight down, she falls faster, reaching a terminal velocity of 150 mph (220 ft/sec).
Figure (PageIndex{5}): Skydivers can adjust the velocity of their dive by changing the position of their body during the free fall. (credit: Jeremy T. Lock)
Since Julie will be moving (falling) in a downward direction, we assume the downward direction is positive to simplify our calculations. Julie executes her jumps from an altitude of 12,500 ft. After she exits the aircraft, she immediately starts falling at a velocity given by (v(t)=32t.)
She continues to accelerate according to this velocity function until she reaches terminal velocity. After she reaches terminal velocity, her speed remains constant until she pulls her ripcord and slows down to land.
On her first jump of the day, Julie orients herself in the slower “belly down” position (terminal velocity is 176 ft/sec). Using this information, answer the following questions.
- How long after she exits the aircraft does Julie reach terminal velocity?
- Based on your answer to question 1, set up an expression involving one or more integrals that represents the distance Julie falls after 30 sec.
- If Julie pulls her ripcord at an altitude of 3000 ft, how long does she spend in a free fall?
- Julie pulls her ripcord at 3000 ft. It takes 5 sec for her parachute to open completely and for her to slow down, during which time she falls another 400 ft. After her canopy is fully open, her speed is reduced to 16 ft/sec. Find the total time Julie spends in the air, from the time she leaves the airplane until the time her feet touch the ground.On Julie’s second jump of the day, she decides she wants to fall a little faster and orients herself in the “head down” position. Her terminal velocity in this position is 220 ft/sec. Answer these questions based on this velocity:
- How long does it take Julie to reach terminal velocity in this case?
- Before pulling her ripcord, Julie reorients her body in the “belly down” position so she is not moving quite as fast when her parachute opens. If she begins this maneuver at an altitude of 4000 ft, how long does she spend in a free fall before beginning the reorientation?
Some jumpers wear “wingsuits” (see Figure). These suits have fabric panels between the arms and legs and allow the wearer to glide around in a free fall, much like a flying squirrel. (Indeed, the suits are sometimes called “flying squirrel suits.”) When wearing these suits, terminal velocity can be reduced to about 30 mph (44 ft/sec), allowing the wearers a much longer time in the air. Wingsuit flyers still use parachutes to land; although the vertical velocities are within the margin of safety, horizontal velocities can exceed 70 mph, much too fast to land safely.
Figure (PageIndex{6}): The fabric panels on the arms and legs of a wingsuit work to reduce the vertical velocity of a skydiver’s fall. (credit: Richard Schneider)
Answer the following question based on the velocity in a wingsuit.
7. If Julie dons a wingsuit before her third jump of the day, and she pulls her ripcord at an altitude of 3000 ft, how long does she get to spend gliding around in the air
Key Concepts
- The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value c such that (f(c)) equals the average value of the function. See Note.
- The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. See Note.
- The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. The total area under a curve can be found using this formula. See Note.
Key Equations
- Mean Value Theorem for Integrals
If f(x)is continuous over an interval ([a,b]), then there is at least one point c∈[a,b] such that (displaystyle f(c)=frac{1}{b−a}∫^b_af(x)dx.)
- Fundamental Theorem of Calculus Part 1
6.3 1st Fundamental Theorem Of Calculus Ap Calculus Frq
If (f(x)) is continuous over an interval [a,b], and the function (F(x)) is defined by (displaystyle F(x)=∫^x_af(t)dt,) then (F′(x)=f(x).)
- Fundamental Theorem of Calculus Part 2
If f is continuous over the interval ([a,b]) and (F(x)) is any antiderivative of (f(x)), then (displaystyle ∫^b_af(x)dx=F(b)−F(a).)
Glossary
- fundamental theorem of calculus
- the theorem, central to the entire development of calculus, that establishes the relationship between differentiation and integration
- fundamental theorem of calculus, part 1
- uses a definite integral to define an antiderivative of a function
- fundamental theorem of calculus, part 2
- (also, evaluation theorem) we can evaluate a definite integral by evaluating the antiderivative of the integrand at the endpoints of the interval and subtracting
- mean value theorem for integrals
- guarantees that a point c exists such that (f(c)) is equal to the average value of the function
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.