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## Section4.4The Fundamental Theorem of Calculus

###### Motivating Questions
• How can we find the exact value of a definite integral without taking the limit of a Riemann sum?

• What is the statement of the Fundamental Theorem of Calculus, and how do antiderivatives of functions play a key role in applying the theorem?

• What is the meaning of the definite integral of a rate of change in contexts other than when the rate of change represents velocity?

Much of our work in Chapter4 has been motivated by the velocity-distance problem: if we know the instantaneous velocity function, $v(t)\text{,}$ for a moving object on a given time interval $[a,b]\text{,}$ can we determine the distance it traveled on $[a,b]\text{?}$ If the velocity function is nonnegative on $[a,b]\text{,}$ the area bounded by $y = v(t)$ and the $t$-axis on $[a,b]$ is equal to the distance traveled. This area is also the value of the definite integral $\int_a^b v(t) \, dt\text{.}$ If the velocity is sometimes negative, the total area bounded by the velocity function still tells us distance traveled, while the net signed area tells us the object's change in position.

For instance, for the velocity function in Figure4.54, the total distance $D$ traveled by the moving object on $[a,b]$ is

\begin{equation*} D = A_1 + A_2 + A_3\text{,} \end{equation*}

and the total change in the object's position is

\begin{equation*} s(b) - s(a) = A_1 - A_2 + A_3\text{.} \end{equation*}

The areas $A_1\text{,}$ $A_2\text{,}$ and $A_3$ are each given by definite integrals, which may be computed by limits of Riemann sums (and in special circumstances by geometric formulas).

We turn our attention to an alternate approach.

###### Example4.55

A student with a third floor dormitory window 32 feet off the ground tosses a water balloon straight up in the air with an initial velocity of 16 feet per second. It turns out that the instantaneous velocity of the water balloon is given by $v(t) = -32t + 16\text{,}$ where $v$ is measured in feet per second and $t$ is measured in seconds.

1. Let $s(t)$ represent the height of the water balloon above the ground at time $t\text{,}$ and note that $s$ is an antiderivative of $v\text{.}$ That is, $v$ is the derivative of $s\text{:}$ $s'(t) = v(t)\text{.}$ Find a formula for $s(t)$ that satisfies the initial condition that the balloon is tossed from 32 feet above ground. In other words, make your formula for $s$ satisfy $s(0) = 32\text{.}$

2. When does the water balloon reach its maximum height? When does it land?

3. Compute $s(\frac{1}{2}) - s(0)\text{,}$ $s(2) - s(\frac{1}{2})\text{,}$ and $s(2) - s(0)\text{.}$ What do these represent?

4. What is the total vertical distance traveled by the water balloon from the time it is tossed until the time it lands?

5. Sketch a graph of the velocity function $y = v(t)$ on the time interval $[0,2]\text{.}$ What is the total net signed area bounded by $y = v(t)$ and the $t$-axis on $[0,2]\text{?}$

Hint
1. Try to find a function whose derivative is $v(t)\text{.}$ Shift this function to satisfy the initial condition.

2. Remember what the derivative tells us about the maximum height, and use the function from above to find the time it hits the ground.

3. Use the function to find the given values.

4. Try finding the maximum height.

5. Graph the function.

1. $s(t)=-16t^2+16t+32$

2. The water balloon reaches its maximum height at 0.5 seconds. It hits the ground at 2 seconds.

3. $s(\frac{1}{2}) - s(0)=36-32=4\text{,}$ $s(2) - s(\frac{1}{2})=0-36=-36\text{,}$ and $s(2) - s(0)=0-32=-32\text{.}$

4. The total vertical distance traveled by the water balloon is 40 feet.

The total net signed area is -32.

Solution
1. Starting with the velocity function given, $v(t)=-32t+16\text{,}$ we want to think about what function could give us $v(t)$ as its derivative. Using the power rule, we know that $\frac{d}{dt}t^k=kt^{k-1}\text{.}$ In this case, we are starting with some $at^{k-1}\text{,}$ so we must add 1 to the exponent and divide by that new exponent to undo the differentiation. So, we get that $s(t)=\frac{-32t^2}{2}+\frac{16t}{1}\text{.}$ However, adding a constant to $s(t)$ keeps the derivative $v(t)$ (remember that constants have derivative 0). So, we have most generally that $s(t)=-16t^2+16t+c\text{,}$ where $c$ is some constant.

Now we want to find $c$ given the condition that $s(0)=32\text{.}$ So,

\begin{equation*} 32=s(0)=-16(0)^2+16(0)+c. \end{equation*}

Therefore, $c=32\text{.}$ Then $s(t)=-16t^2+16t+32$

2. Recall that we have a critical point where the derivative of a function is 0, which will result in a maximum point. Since the position function is a quadratic function, this will tell us the time that the water balloon reaches its maximum height. So, let's set the derivative equal to 0:

\begin{equation*} -32t+16=0, \end{equation*}

so $t=0.5\text{.}$ Therefore, after 0.5 seconds the water balloon reaches its maximum height.

To find the time it takes for the water balloon to land we set the position function equal to 0:

\begin{equation*} -16t^2+16t+32=0 \end{equation*}
\begin{equation*} -16(t-2)(t+1)=0 \end{equation*}

Thus, it takes 2 seconds for the water balloon to hit the groups.

3. We just need to plug the values into the position function: $s(\frac{1}{2}) - s(0)=36-32=4\text{,}$ $s(2) - s(\frac{1}{2})=0-36=-36\text{,}$ and $s(2) - s(0)=0-32=-32\text{.}$

4. We know from part (c) that the total vertical distance from time 0 to the maximum height is 4 feet. Again, from part (c), we know that from the maximum height to when the water balloon hits the ground is a vertical distance of 36 feet. The total vertical distance traveled by the water balloon is 40 feet.

We can calculate the area under the curve by breaking this into two triangles. The first triangle has height 16 and width 0.5, so the area is $16\cdot 0.5\cdot 0.5=4\text{.}$ The second triangle has a negative height of -48 and width of 1.5, so the area is $-48\cdot 1.5\cdot 0.5=-36\text{.}$ Notice that we want to keep the area under the $x$-axis negative. Thus, the total net signed area is -32.

### SubsectionThe Fundamental Theorem of Calculus

Suppose we know the position function $s(t)$ and the velocity function $v(t)$ of an object moving in a straight line, and for the moment let us assume that $v(t)$ is positive on $[a,b]\text{.}$ Then, as shown in Figure4.58, we know two different ways to compute the distance, $D\text{,}$ the object travels: one is that $D = s(b) - s(a)\text{,}$ the object's change in position. The other is the area under the velocity curve, which is given by the definite integral, so $D = \int_a^b v(t) \, dt\text{.}$ Figure4.58Finding distance traveled when we know a velocity function $v\text{.}$

Since both of these expressions tell us the distance traveled, it follows that they are equal, so

\begin{equation} s(b) - s(a) = \int_a^b v(t) \, dt\text{.}\label{E-FTCVel}\tag{4.3} \end{equation}

Equation(4.3) holds even when velocity is sometimes negative, because $s(b) - s(a)\text{,}$the object's change in position, is also measured by the net signed area on $[a,b]$ which is given by $\int_a^b v(t) \, dt\text{.}$

Perhaps the most powerful fact Equation(4.3) reveals is that we can compute the integral's value if we can find a formula for $s\text{.}$ Remember, $s$ and $v$ are related by the fact that $v$ is the derivative of $s\text{,}$ or equivalently that $s$ is an antiderivative of $v\text{.}$

###### Example4.59

Determine the exact distance traveled on $[1,5]$ by an object with velocity function $v(t) = 3t^2 + 40$ feet per second. The distance traveled on the interval $[1,5]$ is given by

\begin{equation*} D = \int_1^5 v(t) \,dt = \int_1^5 (3t^2 + 40) \, dt = s(5) - s(1)\text{,} \end{equation*}

where $s$ is an antiderivative of $v\text{.}$ Now, the derivative of $t^3$ is $3t^2$ and the derivative of $40t$ is $40\text{,}$ so it follows that $s(t) = t^3 + 40t$ is an antiderivative of $v\text{.}$ Therefore,

\begin{align*} D &= \int_1^5 3t^2 + 40 \, dt = s(5) - s(1)\\ &= (5^3 + 40 \cdot 5) - (1^3 + 40\cdot 1) = 284 \ \text{feet}\text{.} \end{align*}

Note the key lesson of Example4.59: to find the distance traveled, we need to compute the area under a curve, which is given by the definite integral. But to evaluate the integral, we can find an antiderivative, $s\text{,}$ of the velocity function, and then compute the total change in $s$ on the interval. In particular, we can evaluate the integral without computing the limit of a Riemann sum. Figure4.60The exact area of the region enclosed by $v(t) = 3t^2 + 40$ on $[1,5]\text{.}$

Mathematicians find it convenient to have a shorthand symbol for a function's antiderivative and so we introduce the following notation. For a continuous function $f\text{,}$ we often denote an antiderivative of $f$ by $F\text{.}$ That is, $F'(x) = f(x)$ for all relevant $x\text{.}$ Using a similar notation $V$ in place of $s$ (so that $V$ is an antiderivative of $v\text{,}$ that is $V'(x)=v(x)$) in Equation(4.3), we can write

\begin{equation} V(b) - V(a) = \int_a^b v(t) \, dt\text{.}\label{E-FTCV}\tag{4.4} \end{equation}

Now, to evaluate the definite integral $\int_a^b f(x) \, dx$ for an arbitrary continuous function $f\text{,}$ we could certainly think of $f$ as representing the velocity of some moving object, and $x$ as the variable that represents time. But Equations(4.3) and(4.4) hold for any continuous velocity function, even when $v$ is sometimes negative. So Equation(4.4) offers a shortcut route to evaluating any definite integral, provided that we can find an antiderivative of the integrand. The Fundamental Theorem of Calculus (FTC) summarizes these observations.

###### Fundamental Theorem of Calculus

If $f'(x)$ is a continuous function on $[a,b]\text{,}$ and $f$ is any antiderivative of $f'(x)\text{,}$ then $\int_a^b f'(x) \, dx = f(b) - f(a)\text{.}$

It is important to note that there is an alternative way of writing the fundamental theorem that is employed in many texts and examples using our convenient notation.

###### Fundamental Theorem of Calculus

If $f(x)$ is a continuous function on $[a,b]\text{,}$ and $F$ is any antiderivative of $f(x)\text{,}$ then $\int_a^b f(x) \, dx = F(b) - F(a)\text{.}$

These two statements of the fundamental theorem are equivalent. Specifically, many authors choose to denote a function by $f$ and the antiderivative of that function by $F\text{.}$

A common alternate notation for $F(b) - F(a)$ is

\begin{equation*} F(b) - F(a) = \left. F(x) \right|_a^b\text{,} \end{equation*}

where we read the righthand side as the function $F$ evaluated from $a$ to $b\text{.}$ In this notation, the FTC says that

\begin{equation*} \int_a^b f(x) \, dx = \left. F(x) \right|_a^b\text{.} \end{equation*}

The FTC opens the door to evaluating a wide range of integrals if we can find an antiderivative $F$ for the integrand $f\text{.}$ For instance since $\frac{d}{dx}[\frac{1}{3}x^3] = x^2\text{,}$ the FTC tells us that

\begin{align*} \int_0^1 x^2 \, dx =\mathstrut \amp \left. \frac{1}{3} \, x^3 \right|_0^1\\ =\mathstrut \amp \frac{1}{3} \, (1)^3 - \frac{1}{3} \, (0)^3\\ =\mathstrut \amp \frac{1}{3}\text{.} \end{align*}

But finding an antiderivative can be far from simple; it is often difficult or even impossible. While we can differentiate just about any function, even some relatively simple functions don't have an elementary antiderivative. A significant portion of integral calculus (which is the main focus of second semester college calculus) is devoted to the problem of finding antiderivatives.

###### Example4.61

Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly.

1. $\int_{-1}^4 (2-2x) \, dx$

2. $\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx$

3. $\int_0^1 e^x \, dx$

4. $\int_{-1}^{1} x^5 \, dx$

5. $\int_0^2 (3x^3 - 2x^2 - e^x) \, dx$

Hint
1. Find a function whose derivative is $2 - 2x\text{.}$

2. Which familiar function has derivative $\sin(x)\text{?}$

3. What is special about $e^x$ when it comes to differentiation?

4. Consider the derivative of $x^6\text{.}$

5. Find an antiderivative for each of the three individual terms in the integrand.

1. $\int_{-1}^4 (2-2x) \, dx = -5\text{.}$

2. $\int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = 1\text{.}$

3. $\int_0^1 e^x \, dx = e-1\text{.}$

4. $\int_{-1}^{1} x^5 \, dx = 0\text{.}$

5. $\int_0^2 (3x^3 - 2x^2 - e^x) \, dx = \frac{23}{3} - e^2\text{.}$

Solution
1. Because $\frac{d}{dx}[2x - x^2] = 2-2x\text{,}$ by the Fundamental Theorem of Calculus,

\begin{equation*} \int_{-1}^4 (2-2x) \, dx = \left. 2x - x^2 \right|_{-1}^4\text{,} \end{equation*}

and therefore

\begin{equation*} \int_{-1}^4 (2-2x) \, dx = (2 \cdot 4 - 4^2) - (2(-1) - (-1)^2) = -8 + 3 = -5\text{.} \end{equation*}
2. Since $\frac{d}{dx} [\cos(x)] = -\sin(x)\text{,}$ an antiderivative of $f(x) = \sin(x)$ is $F(x) = -\cos(x)\text{.}$ Therefore, by the FTC,

\begin{equation*} \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = \left. -\cos(x) \right|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\text{,} \end{equation*}

so

\begin{equation*} \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = -\cos(\frac{\pi}{2}) - (-\cos(0)) = -0 + 1 = 1\text{.} \end{equation*}
3. Since $e^x$ is its own derivative, it is also its own antiderivative. Hence,

\begin{equation*} \int_0^1 e^x \, dx = \left. e^x \right|_0^1 = e^1 - e^0 = e-1\text{.} \end{equation*}
4. Note that since $\frac{d}{dx} [x^6] = 6x^5\text{,}$ it follows that $\frac{d}{dx} [\frac{1}{6}x^6] = x^5\text{,}$ and thus

\begin{equation*} \int_{-1}^{1} x^5 \, dx = \left. \frac{1}{6}x^6 \right|_{-1}^1 = \frac{1}{6} (1)^6 - \frac{1}{6}(-1)^6 = 0\text{.} \end{equation*}
5. Using the sum and constant multiple rules for differentiation, we can see that similar results hold for antidifferentiation, and thus that $F(x) = \frac{3}{4} x^4 - \frac{2}{3} x^3 - e^x$ is an antiderivative of $f(x) = 3x^3 - 2x^2 - e^x\text{.}$ Now, by the FTC,

\begin{align*} \int_0^2 (3x^3 - 2x^2 - e^x) \, dx =\mathstrut \amp \left. \frac{3}{4} x^4 - \frac{2}{3} x^3 - e^x \right|_0^2\\ =\mathstrut \amp \frac{3}{4} (2)^4 - \frac{2}{3} (2)^3 - e^2 - (\frac{3}{4} (0)^4 - \frac{2}{3} (0)^3 - e^0)\\ =\mathstrut \amp 12 - \frac{16}{3} - e^2 - (0 - 0 - 1)\\ =\mathstrut \amp \frac{23}{3} - e^2\text{.} \end{align*}

### SubsectionThe Total Change Theorem

Let us review three interpretations of the definite integral.

• For a moving object with instantaneous velocity $v(t)\text{,}$ the object's change in position on the time interval $[a,b]$ is given by $\int_a^b v(t) \, dt\text{,}$ and whenever $v(t) \ge 0$ on $[a,b]\text{,}$ $\int_a^b v(t) \, dt$ tells us the total distance traveled by the object on $[a,b]\text{.}$

• For any continuous function $f\text{,}$ its definite integral $\int_a^b f(x) \, dx$ represents the net signed area bounded by $y = f(x)$ and the $x$-axis on $[a,b]\text{,}$ where regions that lie below the $x$-axis have a minus sign associated with their area.

• The value of a definite integral is linked to the average value of a function: for a continuous function $f$ on $[a,b]\text{,}$ its average value $f_{\operatorname{AVG} [a,b]}$ is given by

\begin{equation*} f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \int_a^b f(x) \, dx\text{.} \end{equation*}

The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand.

A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. Recall Equation(4.4), where we wrote the Fundamental Theorem of Calculus for a velocity function $v$ with antiderivative $V$ as

\begin{equation*} V(b) - V(a) = \int_a^b v(t) \, dt\text{.} \end{equation*}

If we instead replace $V$ with $s$ (which represents position) and replace $v$ with $s'$ (since velocity is the derivative of position), Equation(4.4) then reads as

\begin{equation} s(b) - s(a) = \int_a^b s'(t) \, dt\text{.}\label{E-FTCs}\tag{4.5} \end{equation}

In words, this version of the FTC tells us that the total change in an object's position function on a particular interval is given by the definite integral of the position function's derivative over that interval.

Of course, this result is not limited to only the setting of position and velocity. Writing the result in terms of a more general function $f\text{,}$ we have the Total Change Theorem.

###### Total Change Theorem

If $f$ is a continuously differentiable function on $[a,b]$ with derivative $f'\text{,}$ then $f(b) - f(a) = \int_a^b f'(x) \, dx\text{.}$ That is, the definite integral of the rate of change of a function on $[a,b]$ is the total change of the function itself on $[a,b]\text{.}$

The Total Change Theorem tells us more about the relationship between the graph of a function and that of its derivative. Recall that heights on the graph of the derivative function are equal to slopes on the graph of the function itself. If instead we know $f'$ and are seeking information about $f\text{,}$ we can say the following:

differences in heights on $f$ correspond to net signed areas bounded by $f'\text{.}$ Figure4.62The graphs of $f'(x) = 4 - 2x$ (at left) and an antiderivative $f(x) = 4x - x^2$ at right. Differences in heights on $f$ correspond to net signed areas bounded by $f'\text{.}$

To see why this is so, consider the difference $f(1) - f(0)\text{.}$ This value is 3, because $f(1) = 3$ and $f(0) = 0\text{,}$ but also because the net signed area bounded by $y = f'(x)$ on $[0,1]$ is 3. That is,

\begin{equation*} f(1) - f(0) = \int_0^1 f'(x) \, dx\text{.} \end{equation*}

In addition to this observation about area, the Total Change Theorem enables us to answer questions about a function whose rate of change we know.

###### Example4.63

Suppose that pollutants are leaking out of an underground storage tank at a rate of $r(t)$ gallons/day, where $t$ is measured in days. It is conjectured that $r(t)$ is given by the formula $r(t) = 0.0069t^3 -0.125t^2+11.079$ over a certain 12-day period. The graph of $y=r(t)$ is given in Figure4.64. What is the meaning of $\int_4^{10} r(t) \, dt$ and what is its value? What is the average rate at which pollutants are leaving the tank on the time interval $4 \le t \le 10\text{?}$ Figure4.64The rate $r(t)$ of pollution leaking from a tank, measured in gallons per day.
Solution

Since $r(t) \ge 0\text{,}$ the value of $\int_4^{10} r(t) \, dt$ is the area under the curve on the interval $[4,10]\text{.}$ A Riemann sum for this area will have rectangles with heights measured in gallons per day and widths measured in days, so the area of each rectangle will have units of

\begin{equation*} \frac{\text{gallons} }{\text{day} } \cdot \text{days} = \text{gallons}\text{.} \end{equation*}

Thus, the definite integral tells us the total number of gallons of pollutant that leak from the tank from day 4 to day 10. The Total Change Theorem tells us the same thing: if we let $R(t)$ denote the total number of gallons of pollutant that have leaked from the tank up to day $t\text{,}$ then $R'(t) = r(t)\text{,}$ and

\begin{equation*} \int_4^{10} r(t) \, dt = R(10) - R(4)\text{,} \end{equation*}

the number of gallons that have leaked from day 4 to day 10.

To compute the exact value of the integral, we use the Fundamental Theorem of Calculus. Antidifferentiating $r(t) = 0.0069t^3 -0.125t^2+11.079\text{,}$ we find that

\begin{align*} \int_4^{10} 0.0069t^3 -0.125t^2+11.079 \, dt =\mathstrut \amp \left. 0.0069 \cdot \frac{1}{4} \, t^4 - 0.125 \cdot \frac{1}{3} t^3 + 11.079t \right|_4^{10}\\ \approx\mathstrut \amp 44.282\text{.} \end{align*}

Thus, approximately 44.282 gallons of pollutant leaked over the six day time period.

To find the average rate at which pollutant leaked from the tank over $4 \le t \le 10\text{,}$ we compute the average value of $r$ on $[4,10]\text{.}$ Thus,

\begin{equation*} r_{\operatorname{AVG} [4,10]} = \frac{1}{10-4} \int_4^{10} r(t) \, dt \approx \frac{44.282}{6} = 7.380 \end{equation*}

gallons per day.

###### Example4.65

During a 40-minute workout, a person riding an exercise machine burns calories at a rate of $c$ calories per minute, where the function $y = c(t)$ is given in Figure4.66. On the interval $0 \le t \le 10\text{,}$ the formula for $c$ is $c(t) = -0.05t^2 + t + 10\text{,}$ while on $30 \le t \le 40\text{,}$ its formula is $c(t) = -0.05t^2 + 3t - 30\text{.}$ Figure4.66The rate $c(t)$ at which a person exercising burns calories, measured in calories per minute.
1. What is the exact total number of calories the person burns during the first 10 minutes of her workout?

2. Let $C(t)$ be an antiderivative of $c(t)\text{.}$ What is the meaning of $C(40) - C(0)$ in the context of the person exercising? Include units in your answer.

3. Determine the exact average rate at which the person burned calories during the 40-minute workout.

4. At what time(s), if any, is the instantaneous rate at which the person is burning calories equal to the average rate at which she burns calories, on the time interval $0 \le t \le 40\text{?}$

Hint
1. What are the units of the area of a rectangle found in a Riemann sum for the function $y= c(t)\text{?}$

2. Use the FTC.

3. Recall the formula for $c_{\operatorname{AVG} [0,40]}\text{.}$

4. Think carefully about which function tells you the instantaneous rate at which calories are burned.

1. The person burned approximately 133.33 calories in the first 10 minutes of the workout.

2. $C(40) - C(0) = \int_0^{40} C'(t) \, dt = \int_0^{40} c(t) \, dt$ is the total calories burned on $[0,40]\text{.}$

3. The exact average rate at which the person burned calories on $0 \le t \le 40$ is

\begin{equation*} c_{\operatorname{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt = \frac{1}{40} \cdot \frac{1700}{3} = \frac{1700}{120} \approx 14.17 \ \text{cal/min}\text{.} \end{equation*}
4. One time at which the instantaneous rate at which calories are burned equals the average rate on $[0,40]$ is $t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}$

Solution
1. Since the units on a rectangle in a Riemann sum are cal/min for the height and min for the width, the units of the area of such a rectangle are calories, and hence the units of area under the curve $y = c(t)$ are given in total calories. Hence, the total calories burned during the first 10 minutes of the workout is given by the definite integral $\int_0^{10} c(t) \, dt\text{.}$ We use the FTC and evaluate the integral, finding that

\begin{align*} \int_0^{10} (-0.05t^2 + t + 10) \, dt =\mathstrut \amp \left. \left( -\frac{0.05}{3} t^3 + \frac{1}{2} t^2 + 10t \right) \right|_0^{10}\\ =\mathstrut \amp \left( -\frac{0.05}{3} (10)^3 + \frac{1}{2} (10)^2 + 10(10) \right) - (0)\\ =\mathstrut \amp \frac{400}{3} \ \text{calories}\text{.} \end{align*}

Thus, the person burned approximately 133.33 calories in the first 10 minutes of the workout.

2. We observe first that by the Total Change Theorem, $C(40) - C(0) = \int_0^{40} C'(t) \, dt = \int_0^{40} c(t) \, dt\text{,}$ and therefore, as discussed in (a), the meaning of this value is the total calories burned on $[0,40]\text{.}$

3. The exact average rate at which the person burned calories on $0 \le t \le 40$ is given by

\begin{equation*} c_{\text{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt\text{.} \end{equation*}

To calculate $\int_0^{40} c(t) \, dt\text{,}$ we recognize that $c(t)$ is defined in piecewise fashion, and use the additive property of the definite integral, which tells us that

\begin{equation*} \int_0^{40} c(t) \, dt = \int_0^{10} c(t) \, dt + \int_{10}^{30} c(t) \, dt + \int_{30}^{40} c(t) \, dt\text{.} \end{equation*}

We know from our work in (a) that $\int_0^{10} c(t) \, dt = \frac{400}{3}\text{.}$ Since $c(t) = 15$ is constant on $10 \le t \le 20\text{,}$ it follows that $\int_{10}^{30} c(t) \, dt = 15 \cdot 30 = 300\text{.}$ And finally, it is straightforward to show using $c(t) = -0.05t^2 + 3t - 30$ (or symmetry) on $30 \le t \le 40$ that $\int_{30}^{40} c(t) \, dt = \frac{400}{3}\text{.}$ Hence,

\begin{align*} \int_0^{40} c(t) \, dt =\mathstrut \amp \int_0^{10} c(t) \, dt + \int_{10}^{30} c(t) \, dt + \int_{30}^{40} c(t) \, dt\\ =\mathstrut \amp \frac{400}{3} + 300 + \frac{400}{3}\\ =\mathstrut \amp \frac{1700}{3} \approx 566.67 \ \text{calories}\text{.} \end{align*}

Now, it follows that the exact average rate at which calories were burned on $[0,40]$ is

\begin{equation*} c_{\text{AVG} [0,40]} = \frac{1}{40-0} \int_0^{40} c(t) \, dt = \frac{1}{40} \cdot \frac{1700}{3} = \frac{1700}{120} \approx 14.17 \ \text{cal/min}\text{.} \end{equation*}
4. It makes sense intuitively that there must be at least one time at which the instantaneous rate at which calories are burned equals the average rate at which calories are burned, as it would be impossible for a continuous instantaneous rate of change to always be above its average value. Since we know from (c) that $c_{\text{AVG} [0,40]} = \frac{85}{6}\text{,}$ and $c(t)$ tells us the instantaneous rate at which calories are burned, it follows that we want to solve the equation

\begin{equation*} c(t) = \frac{85}{6}\text{.} \end{equation*}

From the graph, it appears that there are two such values of $t$ for which this equation is true, one in the first ten minutes, and one in the last ten. For instance, solving

\begin{equation*} -0.05t^2 + t + 10 = \frac{85}{6}\text{,} \end{equation*}

it follows that $t = \frac{5}{3}(6 \pm \sqrt{6}) \approx 14.082, 5.918\text{,}$ only the second of which lies in $0 \le t \le 10\text{.}$ So one time at which the instantaneous rate at which calories are burned equals the average rate on $[0,40]$ is $t = \frac{5}{3}(6 - \sqrt{6}) \approx 5.918\text{.}$ Similar reasoning leads to the second time that lies in $[30,40]\text{.}$

### SubsectionSummary

• We can find the exact value of a definite integral without taking the limit of a Riemann sum or using a familiar area formula by finding the antiderivative of the integrand, and hence applying the Fundamental Theorem of Calculus.

• The Fundamental Theorem of Calculus says that if $f$ is a continuous function on $[a,b]$ and $F$ is an antiderivative of $f\text{,}$ then

\begin{equation*} \int_a^b f(x) \, dx = F(b) - F(a)\text{.} \end{equation*}

Hence, if we can find an antiderivative for the integrand $f\text{,}$ evaluating the definite integral comes from simply computing the change in $F$ on $[a,b]\text{.}$

• A slightly different perspective on the FTC allows us to restate it as the Total Change Theorem, which says that

\begin{equation*} \int_a^b f'(x) \, dx = f(b) - f(a)\text{,} \end{equation*}

for any continuously differentiable function $f\text{.}$ This means that the definite integral of the instantaneous rate of change of a function $f$ on an interval $[a,b]$ is equal to the total change in the function $f$ on $[a,b]\text{.}$

### SubsectionExercises

When an aircraft attempts to climb as rapidly as possible, its climb rate (in feet per minute) decreases as altitude increases, because the air is less dense at higher altitudes. Given below is a table showing performance data for a certain single engine aircraft, giving its climb rate at various altitudes, where $c(h)$ denotes the climb rate of the airplane at an altitude $h\text{.}$

 $h$ (feet) $0$ $1000$ $2000$ $3000$ $4000$ $5000$ $6000$ $7000$ $8000$ $9000$ $10{,}000$ $c$ (ft/min) $925$ $875$ $830$ $780$ $730$ $685$ $635$ $585$ $535$ $490$ $440$

Let a new function called $m(h)$ measure the number of minutes required for a plane at altitude $h$ to climb the next foot of altitude.

1. Determine a similar table of values for $m(h)$ and explain how it is related to the table above. Be sure to explain the units.

2. Give a careful interpretation of a function whose derivative is $m(h)\text{.}$ Describe what the input is and what the output is. Also, explain in plain English what the function tells us.

3. Determine a definite integral whose value tells us exactly the number of minutes required for the airplane to ascend to 10,000 feet of altitude. Clearly explain why the value of this integral has the required meaning.

4. Use the Riemann sum $M_5$ to estimate the value of the integral you found in (c). Include units on your result.

In Chapter1, we showed that for an object moving along a straight line with position function $s(t)\text{,}$ the object's average velocity on the interval $[a,b]$ is given by

\begin{equation*} AV_{[a,b]} = \frac{s(b)-s(a)}{b-a}\text{.} \end{equation*}

More recently in Chapter4, we found that for an object moving along a straight line with velocity function $v(t)\text{,}$ the object's average value of its velocity function on $[a,b]$ is

\begin{equation*} \displaystyle v_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \int_a^b v(t) \, dt\text{.} \end{equation*}

Are the average velocity on the interval $[a,b]$ and the average value of the velocity function on $[a,b]$ the same thing? Why or why not? Explain.