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Integration Formulas and the Net Change Theorem
Apply the basic integration formulas.
Basic Integration Formulas
Recall the integration formulas given in the table in Antiderivatives and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.
Example
Try it.
Use the power rule to integrate the function \({\int }_{1}^{4}\sqrt{t}(1+t)dt.\)
Solution
The first step is to rewrite the function and simplify it so we can apply the power rule:
\[\begin{array}{ll}{\int }_{1}^{4}\sqrt{t}(1+t)dt & ={\int }_{1}^{4}{t}^{1\text{/}2}(1+t)dt \\ \\ & ={\int }_{1}^{4}({t}^{1\text{/}2}+{t}^{3\text{/}2})dt.\end{array}\]Now apply the power rule:
\[\begin{array}{ll}{\int }_{1}^{4}({t}^{1\text{/}2}+{t}^{3\text{/}2})dt & ={(\frac{2}{3}{t}^{3\text{/}2}+\frac{2}{5}{t}^{5\text{/}2})|}_{1}^{4} \\ & =[\frac{2}{3}{(4)}^{3\text{/}2}+\frac{2}{5}{(4)}^{5\text{/}2}]-[\frac{2}{3}{(1)}^{3\text{/}2}+\frac{2}{5}{(1)}^{5\text{/}2}] \\ & =\frac{256}{15}.\end{array}\]The Net Change Theorem
The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.
Subtracting \(F(a)\) from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.
We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in .
Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by
\[\begin{array}{ll}{\int }_{2}^{5}v(t)dt & ={\int }_{2}^{4}40dt+{\int }_{4}^{5}-30dt \\ & =80-30 \\ & =50.\end{array}\]Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by
\[\begin{array}{ll} \\ \\ {\int }_{2}^{5}|v(t)|dt & ={\int }_{2}^{4}40dt+{\int }_{4}^{5}30dt \\ & =80+30 \\ & =110.\end{array}\]Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
Example
Try it.
Given a velocity function \(v(t)=3t-5\) (in meters per second) for a particle in motion from time \(t=0\) to time \(t=3,\) find the net displacement of the particle.
Solution
Applying the net change theorem, we have
\[\begin{array}{ll}{\int }_{0}^{3}(3t-5)dt & =\frac{3{t}^{2}}{2}-5t{|}_{0}^{3} \\ \\ & =[\frac{3{(3)}^{2}}{2}-5(3)]-0 \\ & =\frac{27}{2}-15 \\ & =\frac{27}{2}-\frac{30}{2} \\ & =-\frac{3}{2}.\end{array}\]The net displacement is \(-\frac{3}{2}\) m ().
Condensed — the full section is in OpenStax Calculus Volume 2.
Applying the Net Change Theorem
The net change theorem can be applied to the flow and consumption of fluids, as shown in .
Example
Try it.
If the motor on a motorboat is started at \(t=0\) and the boat consumes gasoline at the rate of \(5-0.1{t}^{3}\) gal/hr, how much gasoline is used in the first 2 hours?
Solution
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval \([0,2].\) We have
\[{\int }_{0}^{2}(5-0.1{t}^{3})dt=(5t-0.1\frac{{t}^{4}}{4}){{|}^{2}}_{0}\ =[5(2)-0.1\frac{{(2)}^{4}}{4}]-0\ =10-0.4\ =9.6\]Thus, the motorboat uses 9.6 gal of gas in 2 hours.
Example
Try it.
As we saw at the beginning of the chapter, top iceboat racers () can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed. Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function \(v(t)=20t+5.\) For the second half hour of Andrew’s outing, the wind remains steady at 15 mph. In other words, the wind speed is given by
\[v(t)=\{\begin{array}{lll}20t+5 & \text{for} & 0\le t\le \frac{1}{2} \\ 15 & \text{for} & \frac{1}{2}\le t\le 1.\end{array}\]Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?
Solution
To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then
Distance \(={\int }_{0}^{1}2v(t)dt.\)
Substituting the expressions we were given for \(v(t),\) we get
\[\begin{array}{ll}{\int }_{0}^{1}2v(t)dt & ={\int }_{0}^{1\text{/}2}2v(t)dt+{\int }_{1\text{/}2}^{1}2v(t)dt \\ & ={\int }_{0}^{1\text{/}2}2(20t+5)dt+{\int }_{1\text{/}2}^{1}2(15)dt \\ & ={\int }_{0}^{1\text{/}2}(40t+10)dt+{\int }_{1\text{/}2}^{1}30dt \\ & =[20{t}^{2}+10t]{|}_{0}^{1\text{/}2}+[30t]{|}_{1\text{/}2}^{1} \\ & =(\frac{20}{4}+5)-0+(30-15) \\ & =25.\end{array}\]Andrew is 25 mi from his starting point after 1 hour.
Condensed — the full section is in OpenStax Calculus Volume 2.
Integrating Even and Odd Functions
We saw in Functions and Graphs that an even function is a function in which \(f(\text{-}x)=f(x)\) for all x in the domain—that is, the graph of the curve is unchanged when x is replaced with −x. The graphs of even functions are symmetric about the y-axis. An odd function is one in which \(f(\text{-}x)=\text{-}f(x)\) for all x in the domain, and the graph of the function is symmetric about the origin.
Integrals of even functions, when the limits of integration are from −a to a, involve two equal areas, because they are symmetric about the y-axis. Integrals of odd functions, when the limits of integration are similarly \([\text{-}a,a],\) evaluate to zero because the areas above and below the x-axis are equal.
Example
Try it.
Integrate the even function \({\int }_{-2}^{2}(3{x}^{8}-2)dx\) and verify that the integration formula for even functions holds.
Solution
The symmetry appears in the graphs in . Graph (a) shows the region below the curve and above the x-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetry about the y-axis of an even function. We have
\[\begin{array}{ll}{\int }_{-2}^{2}(3{x}^{8}-2)dx & =(\frac{{x}^{9}}{3}-2x){|}_{-2}^{2} \\ \\ \\ & =[\frac{{(2)}^{9}}{3}-2(2)]-[\frac{{(-2)}^{9}}{3}-2(-2)] \\ & =(\frac{512}{3}-4)-(-\frac{512}{3}+4) \\ & =\frac{1000}{3}.\end{array}\]To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then check to make sure we get the same answer.
\[\begin{array}{ll}{\int }_{0}^{2}(3{x}^{8}-2)dx & =(\frac{{x}^{9}}{3}-2x){|}_{0}^{2} \\ \\ & =\frac{512}{3}-4 \\ & =\frac{500}{3}\end{array}\]Since \(2\cdot \frac{500}{3}=\frac{1000}{3},\) we have verified the formula for even functions in this particular example.
Condensed — the full section is in OpenStax Calculus Volume 2.
Key Concepts
- The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
- The area under an even function over a symmetric interval can be calculated by doubling the area over the positive x-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.
Integration Formulas and the Net Change Theorem
Use basic integration formulas to compute the following antiderivatives or definite integrals.
For the next two exercises use the data in the following table, which displays bald eagle populations from 1963 to 2000 in the continental United States.
| Year | Population of Breeding Pairs of Bald Eagles |
| 1963 | 487 |
| 1974 | 791 |
| 1981 | 1188 |
| 1986 | 1875 |
| 1992 | 3749 |
| 1996 | 5094 |
| 2000 | 6471 |
As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.
| Time (sec) | Acceleration (mph/sec) |
| 1 | 11.2 |
| 2 | 10.6 |
| 3 | 8.1 |
| 4 | 5.4 |
| 5 | 0 |
As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the next three exercises use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.
| Time (sec) | Acceleration (mph/sec) |
| 1 | 11.2 |
| 2 | 10.6 |
| 3 | 8.1 |
| 4 | 5.4 |
| 5 | 0 |
Basic Integration Formulas
Recall the integration formulas given in the table in Antiderivatives and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.
Example
Try it.
Use the power rule to integrate the function \({\int }_{1}^{4}\sqrt{t}(1+t)dt.\)
Solution
The first step is to rewrite the function and simplify it so we can apply the power rule:
\[\begin{array}{ll}{\int }_{1}^{4}\sqrt{t}(1+t)dt & ={\int }_{1}^{4}{t}^{1\text{/}2}(1+t)dt \\ \\ & ={\int }_{1}^{4}({t}^{1\text{/}2}+{t}^{3\text{/}2})dt.\end{array}\]Now apply the power rule:
\[\begin{array}{ll}{\int }_{1}^{4}({t}^{1\text{/}2}+{t}^{3\text{/}2})dt & ={(\frac{2}{3}{t}^{3\text{/}2}+\frac{2}{5}{t}^{5\text{/}2})|}_{1}^{4} \\ & =[\frac{2}{3}{(4)}^{3\text{/}2}+\frac{2}{5}{(4)}^{5\text{/}2}]-[\frac{2}{3}{(1)}^{3\text{/}2}+\frac{2}{5}{(1)}^{5\text{/}2}] \\ & =\frac{256}{15}.\end{array}\]The Net Change Theorem
The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.
Subtracting \(F(a)\) from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.
We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in .
Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by
\[\begin{array}{ll}{\int }_{2}^{5}v(t)dt & ={\int }_{2}^{4}40dt+{\int }_{4}^{5}-30dt \\ & =80-30 \\ & =50.\end{array}\]Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by
\[\begin{array}{ll} \\ \\ {\int }_{2}^{5}|v(t)|dt & ={\int }_{2}^{4}40dt+{\int }_{4}^{5}30dt \\ & =80+30 \\ & =110.\end{array}\]Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
Example
Try it.
Given a velocity function \(v(t)=3t-5\) (in meters per second) for a particle in motion from time \(t=0\) to time \(t=3,\) find the net displacement of the particle.
Solution
Applying the net change theorem, we have
\[\begin{array}{ll}{\int }_{0}^{3}(3t-5)dt & =\frac{3{t}^{2}}{2}-5t{|}_{0}^{3} \\ \\ & =[\frac{3{(3)}^{2}}{2}-5(3)]-0 \\ & =\frac{27}{2}-15 \\ & =\frac{27}{2}-\frac{30}{2} \\ & =-\frac{3}{2}.\end{array}\]The net displacement is \(-\frac{3}{2}\) m ().
Condensed — the full section is in OpenStax Calculus Volume 1.
Applying the Net Change Theorem
The net change theorem can be applied to the flow and consumption of fluids, as shown in .
Example
Try it.
If the motor on a motorboat is started at \(t=0\) and the boat consumes gasoline at the rate of \(5-0.1{t}^{3}\) gal/hr, how much gasoline is used in the first 2 hours?
Solution
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval \([0,2].\) We have
\[{\int }_{0}^{2}(5-0.1{t}^{3})dt=(5t-0.1\frac{{t}^{4}}{4}){{|}^{2}}_{0}\ =[5(2)-0.1\frac{{(2)}^{4}}{4}]-0\ =10-0.4\ =9.6\]Thus, the motorboat uses 9.6 gal of gas in 2 hours.
Example
Try it.
As we saw at the beginning of the chapter, top iceboat racers () can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed. Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function \(v(t)=20t+5.\) For the second half hour of Andrew’s outing, the wind remains steady at 15 mph. In other words, the wind speed is given by
\[v(t)=\{\begin{array}{lll}20t+5 & \text{for} & 0\le t\le \frac{1}{2} \\ 15 & \text{for} & \frac{1}{2}\le t\le 1.\end{array}\]Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?
Solution
To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then
Distance \(={\int }_{0}^{1}2v(t)dt.\)
Substituting the expressions we were given for \(v(t),\) we get
\[\begin{array}{ll}{\int }_{0}^{1}2v(t)dt & ={\int }_{0}^{1\text{/}2}2v(t)dt+{\int }_{1\text{/}2}^{1}2v(t)dt \\ & ={\int }_{0}^{1\text{/}2}2(20t+5)dt+{\int }_{1\text{/}2}^{1}2(15)dt \\ & ={\int }_{0}^{1\text{/}2}(40t+10)dt+{\int }_{1\text{/}2}^{1}30dt \\ & =[20{t}^{2}+10t]{|}_{0}^{1\text{/}2}+[30t]{|}_{1\text{/}2}^{1} \\ & =(\frac{20}{4}+5)-0+(30-15) \\ & =25.\end{array}\]Andrew is 25 mi from his starting point after 1 hour.
Condensed — the full section is in OpenStax Calculus Volume 1.
Integrating Even and Odd Functions
We saw in Functions and Graphs that an even function is a function in which \(f(\text{-}x)=f(x)\) for all x in the domain—that is, the graph of the curve is unchanged when x is replaced with −x. The graphs of even functions are symmetric about the y-axis. An odd function is one in which \(f(\text{-}x)=\text{-}f(x)\) for all x in the domain, and the graph of the function is symmetric about the origin.
Integrals of even functions, when the limits of integration are from −a to a, involve two equal areas, because they are symmetric about the y-axis. Integrals of odd functions, when the limits of integration are similarly \([\text{-}a,a],\) evaluate to zero because the areas above and below the x-axis are equal.
Example
Try it.
Integrate the even function \({\int }_{-2}^{2}(3{x}^{8}-2)dx\) and verify that the integration formula for even functions holds.
Solution
The symmetry appears in the graphs in . Graph (a) shows the region below the curve and above the x-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetry about the y-axis of an even function. We have
\[\begin{array}{ll}{\int }_{-2}^{2}(3{x}^{8}-2)dx & =(\frac{{x}^{9}}{3}-2x){|}_{-2}^{2} \\ \\ \\ & =[\frac{{(2)}^{9}}{3}-2(2)]-[\frac{{(-2)}^{9}}{3}-2(-2)] \\ & =(\frac{512}{3}-4)-(-\frac{512}{3}+4) \\ & =\frac{1000}{3}.\end{array}\]To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then check to make sure we get the same answer.
\[\begin{array}{ll}{\int }_{0}^{2}(3{x}^{8}-2)dx & =(\frac{{x}^{9}}{3}-2x){|}_{0}^{2} \\ \\ & =\frac{512}{3}-4 \\ & =\frac{500}{3}\end{array}\]Since \(2\cdot \frac{500}{3}=\frac{1000}{3},\) we have verified the formula for even functions in this particular example.
Condensed — the full section is in OpenStax Calculus Volume 1.
Key Concepts
- The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
- The area under an even function over a symmetric interval can be calculated by doubling the area over the positive x-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.
Integration Formulas and the Net Change Theorem
Use basic integration formulas to compute the following antiderivatives or definite integrals.
For the next two exercises use the data in the following table, which displays bald eagle populations from 1963 to 2000 in the continental United States.
| Year | Population of Breeding Pairs of Bald Eagles |
| 1963 | 487 |
| 1974 | 791 |
| 1981 | 1188 |
| 1986 | 1875 |
| 1992 | 3749 |
| 1996 | 5094 |
| 2000 | 6471 |
As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.
| Time (sec) | Acceleration (mph/sec) |
| 1 | 11.2 |
| 2 | 10.6 |
| 3 | 8.1 |
| 4 | 5.4 |
| 5 | 0 |
As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the next three exercises use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.
| Time (sec) | Acceleration (mph/sec) |
| 1 | 11.2 |
| 2 | 10.6 |
| 3 | 8.1 |
| 4 | 5.4 |
| 5 | 0 |
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Use the power rule to integrate the function \({\int }_{1}^{4}\sqrt{t}(1+t)dt.\)
Fi àwọn àgbèwọlé hàn
The first step is to rewrite the function and simplify it so we can apply the power rule:
\[\begin{array}{ll}{\int }_{1}^{4}\sqrt{t}(1+t)dt & ={\int }_{1}^{4}{t}^{1\text{/}2}(1+t)dt \\ \\ & ={\int }_{1}^{4}({t}^{1\text{/}2}+{t}^{3\text{/}2})dt.\end{array}\]Now apply the power rule:
\[\begin{array}{ll}{\int }_{1}^{4}({t}^{1\text{/}2}+{t}^{3\text{/}2})dt & ={(\frac{2}{3}{t}^{3\text{/}2}+\frac{2}{5}{t}^{5\text{/}2})|}_{1}^{4} \\ & =[\frac{2}{3}{(4)}^{3\text{/}2}+\frac{2}{5}{(4)}^{5\text{/}2}]-[\frac{2}{3}{(1)}^{3\text{/}2}+\frac{2}{5}{(1)}^{5\text{/}2}] \\ & =\frac{256}{15}.\end{array}\] -
Find the definite integral of \(f(x)={x}^{2}-3x\) over the interval \([1,3].\)
Fi àwọn àgbèwọlé hàn
\(-\frac{10}{3}\)
-
Given a velocity function \(v(t)=3t-5\) (in meters per second) for a particle in motion from time \(t=0\) to time \(t=3,\) find the net displacement of the particle.
Fi àwọn àgbèwọlé hàn
Applying the net change theorem, we have
\[\begin{array}{ll}{\int }_{0}^{3}(3t-5)dt & =\frac{3{t}^{2}}{2}-5t{|}_{0}^{3} \\ \\ & =[\frac{3{(3)}^{2}}{2}-5(3)]-0 \\ & =\frac{27}{2}-15 \\ & =\frac{27}{2}-\frac{30}{2} \\ & =-\frac{3}{2}.\end{array}\]The net displacement is \(-\frac{3}{2}\) m ().
-
Use to find the total distance traveled by a particle according to the velocity function \(v(t)=3t-5\) m/sec over a time interval \([0,3].\)
Fi àwọn àgbèwọlé hàn
The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.
To continue with the example, use two integrals to find the total distance. First, find the t-intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for t. Thus,
\[\begin{array}{lll}3t-5 & = & 0 \\ 3t & = & 5 \\ t & = & \frac{5}{3}.\end{array}\]The two subintervals are \([0,\frac{5}{3}]\) and \([\frac{5}{3},3].\) To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval \([0,\frac{5}{3}],\) we have \(|v(t)|=\text{-}v(t)\) over that interval. Over \([\frac{5}{3},3],\) the function is positive, so \(|v(t)|=v(t).\) Thus, we have
\[\begin{array}{ll} \\ \\ {\int }_{0}^{3}|v(t)|dt & ={\int }_{0}^{5\text{/}3}\text{-}v(t)dt+{\int }_{5\text{/}3}^{3}v(t)dt \\ \\ & ={\int }_{0}^{5\text{/}3}5-3tdt+{\int }_{5\text{/}3}^{3}3t-5dt \\ & ={(5t-\frac{3{t}^{2}}{2})|}_{0}^{5\text{/}3}+{(\frac{3{t}^{2}}{2}-5t)|}_{5\text{/}3}^{3} \\ & =[5(\frac{5}{3})-\frac{3{(5\text{/}3)}^{2}}{2}]-0+[\frac{27}{2}-15]-[\frac{3{(5\text{/}3)}^{2}}{2}-\frac{25}{3}] \\ & =\frac{25}{3}-\frac{25}{6}+\frac{27}{2}-15-\frac{25}{6}+\frac{25}{3} \\ & =\frac{41}{6}.\end{array}\]So, the total distance traveled is \(\frac{41}{6}\) m.
-
Find the net displacement and total distance traveled in meters given the velocity function \(f(t)=\frac{1}{2}{e}^{t}-2\) over the interval \([0,2].\)
Fi àwọn àgbèwọlé hàn
Net displacement: \(\frac{{e}^{2}-9}{2}\approx -0.8055\ \text{m;}\) total distance traveled: \(4\ \text{ln}\ 4-7.5+\frac{{e}^{2}}{2}\approx 1.740\) m
-
If the motor on a motorboat is started at \(t=0\) and the boat consumes gasoline at the rate of \(5-0.1{t}^{3}\) gal/hr, how much gasoline is used in the first 2 hours?
Fi àwọn àgbèwọlé hàn
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval \([0,2].\) We have
\[{\int }_{0}^{2}(5-0.1{t}^{3})dt=(5t-0.1\frac{{t}^{4}}{4}){{|}^{2}}_{0}\ =[5(2)-0.1\frac{{(2)}^{4}}{4}]-0\ =10-0.4\ =9.6\]Thus, the motorboat uses 9.6 gal of gas in 2 hours.
-
As we saw at the beginning of the chapter, top iceboat racers () can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed. Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function \(v(t)=20t+5.\) For the second half hour of Andrew’s outing, the wind remains steady at 15 mph. In other words, the wind speed is given by
\[v(t)=\{\begin{array}{lll}20t+5 & \text{for} & 0\le t\le \frac{1}{2} \\ 15 & \text{for} & \frac{1}{2}\le t\le 1.\end{array}\]Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?
Fi àwọn àgbèwọlé hàn
To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then
Distance \(={\int }_{0}^{1}2v(t)dt.\)
Substituting the expressions we were given for \(v(t),\) we get
\[\begin{array}{ll}{\int }_{0}^{1}2v(t)dt & ={\int }_{0}^{1\text{/}2}2v(t)dt+{\int }_{1\text{/}2}^{1}2v(t)dt \\ & ={\int }_{0}^{1\text{/}2}2(20t+5)dt+{\int }_{1\text{/}2}^{1}2(15)dt \\ & ={\int }_{0}^{1\text{/}2}(40t+10)dt+{\int }_{1\text{/}2}^{1}30dt \\ & =[20{t}^{2}+10t]{|}_{0}^{1\text{/}2}+[30t]{|}_{1\text{/}2}^{1} \\ & =(\frac{20}{4}+5)-0+(30-15) \\ & =25.\end{array}\]Andrew is 25 mi from his starting point after 1 hour.
-
Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function \(v(t)=-10t+20.\) In other words, the wind speed is given by
\[v(t)=\{\begin{array}{lll}20t+5 & \text{for} & 0\le t\le \frac{1}{2} \\ -10t+15 & \text{for} & \frac{1}{2}\le t\le 1.\end{array}\]Under these conditions, how far from his starting point is Andrew after 1 hour?
Fi àwọn àgbèwọlé hàn
17.5 mi
-
Integrate the even function \({\int }_{-2}^{2}(3{x}^{8}-2)dx\) and verify that the integration formula for even functions holds.
Fi àwọn àgbèwọlé hàn
The symmetry appears in the graphs in . Graph (a) shows the region below the curve and above the x-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetry about the y-axis of an even function. We have
\[\begin{array}{ll}{\int }_{-2}^{2}(3{x}^{8}-2)dx & =(\frac{{x}^{9}}{3}-2x){|}_{-2}^{2} \\ \\ \\ & =[\frac{{(2)}^{9}}{3}-2(2)]-[\frac{{(-2)}^{9}}{3}-2(-2)] \\ & =(\frac{512}{3}-4)-(-\frac{512}{3}+4) \\ & =\frac{1000}{3}.\end{array}\]To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then check to make sure we get the same answer.
\[\begin{array}{ll}{\int }_{0}^{2}(3{x}^{8}-2)dx & =(\frac{{x}^{9}}{3}-2x){|}_{0}^{2} \\ \\ & =\frac{512}{3}-4 \\ & =\frac{500}{3}\end{array}\]Since \(2\cdot \frac{500}{3}=\frac{1000}{3},\) we have verified the formula for even functions in this particular example.
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Evaluate the definite integral of the odd function \(-5\ \text{sin}\ x\) over the interval \([\text{-}\pi ,\pi ].\)
Fi àwọn àgbèwọlé hàn
The graph is shown in . We can see the symmetry about the origin by the positive area above the x-axis over \([\text{-}\pi ,0],\) and the negative area below the x-axis over \([0,\pi ].\) We have
\[\begin{array}{ll}{\int }_{\text{-}\pi }^{\pi }-5\ \text{sin}\ xdx & =-5(\text{-}\text{cos}\ x){|}_{\text{-}\pi }^{\pi } \\ \\ \\ & =5\ \text{cos}\ x{|}_{\text{-}\pi }^{\pi } \\ & =[5\ \text{cos}\ \pi ]-[5\ \text{cos}(\text{-}\pi )] \\ & =-5-(-5) \\ & =0.\end{array}\] -
Integrate the function \({\int }_{-2}^{2}{x}^{4}dx.\)
Fi àwọn àgbèwọlé hàn
\(\frac{64}{5}\)
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\(\int (\sqrt{x}-\frac{1}{\sqrt{x}})dx\)
Fi àwọn àgbèwọlé hàn
\(\int (\sqrt{x}-\frac{1}{\sqrt{x}})dx=\int {x}^{1\text{/}2}dx-\int {x}^{-1\text{/}2}dx=\frac{2}{3}{x}^{3\text{/}2}+{C}_{1}-2{x}^{1\text{/}2}+{C}_{2}=\frac{2}{3}{x}^{3\text{/}2}-2{x}^{1\text{/}2}+C\)
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\(\int ({e}^{2x}-\frac{1}{2}{e}^{x\text{/}2})dx\)
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\(\int \frac{dx}{2x}\)
Fi àwọn àgbèwọlé hàn
\(\int \frac{dx}{2x}=\frac{1}{2}\text{ln}|x|+C\)
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\(\int \frac{x-1}{{x}^{2}}dx\)
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\({\int }_{0}^{\pi }(\text{sin}\ x-\text{cos}\ x)dx\)
Fi àwọn àgbèwọlé hàn
\({\int }_{0}^{\pi }\text{sin}\ xdx-{\int }_{0}^{\pi }\text{cos}\ xdx=\text{-}\text{cos}\ x{|}_{0}^{\pi }-(\text{sin}\ x){|}_{0}^{\pi }=(\text{-}(-1)+1)-(0-0)=2\)
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\({\int }_{0}^{\pi \text{/}2}(x-\text{sin}\ x)dx\)
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Write an integral that expresses the increase in the perimeter \(P(s)\) of a square when its side length s increases from 2 units to 4 units and evaluate the integral.
Fi àwọn àgbèwọlé hàn
\(P(s)=4s,\) so \(\frac{dP}{ds}=4\) and \({\int }_{2}^{4}4ds=8.\)
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Write an integral that quantifies the change in the area \(A(s)={s}^{2}\) of a square when the side length doubles from S units to 2S units and evaluate the integral.
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A regular N-gon (an N-sided polygon with sides that have equal length s, such as a pentagon or hexagon) has perimeter Ns. Write an integral that expresses the increase in perimeter of a regular N-gon when the length of each side increases from 1 unit to 2 units and evaluate the integral.
Fi àwọn àgbèwọlé hàn
\({\int }_{1}^{2}Nds=N\)
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The area of a regular pentagon with side length \(a>0\) is pa2 with \(p=\frac{1}{4}\sqrt{5\left(5+2\sqrt{5}\right)}.\) The Pentagon in Washington, DC, has inner sides of length 360 ft and outer sides of length 920 ft. Write an integral to express the area of the roof of the Pentagon according to these dimensions and evaluate this area.
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A dodecahedron is a Platonic solid with a surface that consists of 12 pentagons, each of equal area. By how much does the surface area of a dodecahedron increase as the side length of each pentagon doubles from 1 unit to 2 units?
Fi àwọn àgbèwọlé hàn
With p as in the previous exercise, each of the 12 pentagons increases in area from 2p to 4p units so the net increase in the area of the dodecahedron is 36p units.
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An icosahedron is a Platonic solid with a surface that consists of 20 equilateral triangles. By how much does the surface area of an icosahedron increase as the side length of each triangle doubles from a unit to 2a units?
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Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from s unit to 2s units and evaluate the integral.
Fi àwọn àgbèwọlé hàn
\(18{s}^{2}=6{\int }_{s}^{2s}2xdx\)
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Write an integral that quantifies the increase in the volume of a cube when the side length doubles from s unit to 2s units and evaluate the integral.
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Write an integral that quantifies the increase in the surface area of a sphere as its radius doubles from R unit to 2R units and evaluate the integral.
Fi àwọn àgbèwọlé hàn
\(12\pi {R}^{2}=8\pi {\int }_{R}^{2R}rdr\)
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Write an integral that quantifies the increase in the volume of a sphere as its radius doubles from R unit to 2R units and evaluate the integral.
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Suppose that a particle moves along a straight line with velocity \(v(t)=4-2t,\) where \(0\le t\le 2\) (in meters per second). Find the displacement at time t and the total distance traveled up to \(t=2.\)
Fi àwọn àgbèwọlé hàn
\(d(t)={\int }_{0}^{t}v(s)ds=4t-{t}^{2}.\) The total distance is \(d(2)=4\ \text{m}\text{.}\)
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Suppose that a particle moves along a straight line with velocity defined by \(v(t)={t}^{2}-3t-18,\) where \(0\le t\le 6\) (in meters per second). Find the displacement at time t and the total distance traveled up to \(t=6.\)
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Suppose that a particle moves along a straight line with velocity defined by \(v(t)=|2t-6|,\) where \(0\le t\le 6\) (in meters per second). Find the displacement at time t and the total distance traveled up to \(t=6.\)
Fi àwọn àgbèwọlé hàn
\(d(t)={\int }_{0}^{t}v(s)ds.\) For \(t<3,d(t)={\int }_{0}^{t}(6-2t)dt=6t-{t}^{2}.\) For \(t>3,d(t)=d(3)+{\int }_{3}^{t}(2t-6)dt=9+({t}^{2}-6t){|}_{3}^{6}.\) The total distance is \(d(6)=18\ \text{m}\text{.}\)
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Suppose that a particle moves along a straight line with acceleration defined by \(a(t)=t-3,\) where \(0\le t\le 6\) (in meters per second). Find the velocity and displacement at time t and the total distance traveled up to \(t=6\) if \(v(0)=3\) and \(d(0)=0.\)
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A ball is thrown upward from a height of 1.5 m at an initial speed of 40 m/sec. Acceleration resulting from gravity is −9.8 m/sec2. Neglecting air resistance, solve for the velocity \(v(t)\) and the height \(h(t)\) of the ball t seconds after it is thrown and before it returns to the ground.
Fi àwọn àgbèwọlé hàn
\(v(t)=40-9.8t\text{m/sec};h(t)=1.5+40t-4.9{t}^{2}\) m/s
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A ball is thrown upward from a height of 3 m at an initial speed of 60 m/sec. Acceleration resulting from gravity is −9.8 m/sec2. Neglecting air resistance, solve for the velocity \(v(t)\) and the height \(h(t)\) of the ball t seconds after it is thrown and before it returns to the ground.
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The area \(A(t)\) of a circular shape is growing at a constant rate. If the area increases from 4π units to 9π units between times \(t=2\) and \(t=3,\) find the net change in the radius during that time.
Fi àwọn àgbèwọlé hàn
The net increase is 1 unit.
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A spherical balloon is being inflated at a constant rate. If the volume of the balloon changes from 36π in.3 to 288π in.3 between time \(t=30\) and \(t=60\) seconds, find the net change in the radius of the balloon during that time.
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Water flows into a conical tank with cross-sectional area πx2 at height x and volume \(\frac{\pi {x}^{3}}{3}\) up to height x. If water flows into the tank at a rate of 1 m3/min, find the height of water in the tank after 5 min. Find the change in height between 5 min and 10 min.
Fi àwọn àgbèwọlé hàn
At \(t=5,\) the height of water is \(x={(\frac{15}{\pi })}^{1\text{/}3}\ \text{m}\text{.}.\) The net change in height from \(t=5\) to \(t=10\) is \({(\frac{30}{\pi })}^{1\text{/}3}-{(\frac{15}{\pi })}^{1\text{/}3}\) m.
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A horizontal cylindrical tank has cross-sectional area \(A(x)=4(6x-{x}^{2}){m}^{2}\) at height x meters above the bottom when \(x\le 3.\)
- The volume V between heights a and b is \({\int }_{a}^{b}A(x)dx.\) Find the volume at heights between 2 m and 3 m.
- Suppose that oil is being pumped into the tank at a rate of 50 L/min. Using the chain rule, \(\frac{dx}{dt}=\frac{dx}{dV}\ \frac{dV}{dt},\) at how many meters per minute is the height of oil in the tank changing, expressed in terms of x, when the height is at x meters?
- How long does it take to fill the tank to 3 m starting from a fill level of 2 m?
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The following table lists the electrical power in gigawatts—the rate at which energy is consumed—used in a certain city for different hours of the day, in a typical 24-hour period, with hour 1 corresponding to midnight to 1 a.m.
Hour Power Hour Power 1 28 13 48 2 25 14 49 3 24 15 49 4 23 16 50 5 24 17 50 6 27 18 50 7 29 19 46 8 32 20 43 9 34 21 42 10 39 22 40 11 42 23 37 12 46 24 34 Find the total amount of energy in gigawatt-hours (gW-h) consumed by the city in a typical 24-hour period.
Fi àwọn àgbèwọlé hàn
The total daily power consumption is estimated as the sum of the hourly power rates, or 911 gW-h.
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The average residential electrical power use (in hundreds of watts) per hour is given in the following table.
Hour Power Hour Power 1 8 13 12 2 6 14 13 3 5 15 14 4 4 16 15 5 5 17 17 6 6 18 19 7 7 19 18 8 8 20 17 9 9 21 16 10 10 22 16 11 10 23 13 12 11 24 11 - Compute the average total energy used in a day in kilowatt-hours (kWh).
- If a ton of coal generates 1842 kWh, how long does it take for an average residence to burn a ton of coal?
- Explain why the data might fit a plot of the form \(p(t)=11.5-7.5\ \text{sin}(\frac{\pi t}{12}).\)
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The data in the following table are used to estimate the average power output produced by Peter Sagan for each of the last 18 sec of Stage 1 of the 2012 Tour de France.
Second Watts Second Watts 1 600 10 1200 2 500 11 1170 3 575 12 1125 4 1050 13 1100 5 925 14 1075 6 950 15 1000 7 1050 16 950 8 950 17 900 9 1100 18 780 Estimate the net energy used in kilojoules (kJ), noting that 1W = 1 J/s, and the average power output by Sagan during this time interval.
Fi àwọn àgbèwọlé hàn
17 kJ
Symbols used here
Antiderivative (indefinite) or signed area from a to b (definite).
The non-negative number whose square (n-th power) is x.
Ratio of a circle's circumference to its diameter, 3.14159…
Chance of A; chance of A given that B happened.
Instantaneous rate of change; slope of the graph.
Inequalities that allow equality; < and > exclude it.
2.71828…, the base whose exponential is its own derivative.
Not a number: "grows without bound" in limits and intervals.
Ratios of sides in a right triangle; coordinates on the unit circle.
The exponent b must be raised to for x; ln uses base e.
Add a_k for k = 1 up to n.
The value f(x) approaches as x approaches a.
Prime notation for derivatives with respect to x (or t).
Constants of integration fixed by initial conditions.
How to: Integration Formulas and the Net Change Theorem
- Apply the basic integration formulas.
- Explain the significance of the net change theorem.
- Use the net change theorem to solve applied problems.
- Apply the integrals of odd and even functions.
- The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
- The area under an even function over a symmetric interval can be calculated by doubling the area over the positive
- The volume
- Suppose that oil is being pumped into the tank at a rate of 50 L/min. Using the chain rule,
Questions people ask
What is a derivative in one sentence?
The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.
Wárá
Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
Diẹ̀ nínú Calculus
LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests