## Section1.4The Definite Integral and FTC

### Subsection1.4.1Exploring an Example

We begin by exploring an example. Suppose that an object moves in a straight line so that its speed is $3t$ at time $t\text{.}$ How far does the object travel between time $t=a$ and time $t=b\text{?}$ We don't assume that we know where the object is at time $t=0$ or at any other time. It is certainly true that it is somewhere, so let's suppose that at $t=0$ the position is $k\text{.}$ Then we know that the position of the object at any time is $\ds 3t^2/2+k\text{.}$ This means that at time $t=a$ the position is $\ds 3a^2/2+k$ and at time $t=b$ the position is $\ds 3b^2/2+k\text{.}$ Therefore the change in position is

\begin{equation*} \frac{3b^2}{2}+k-\left(\frac{3a^2}{2}+k\right)=\frac{3b^2}{2}-\frac{3a^2}{2}\text{.} \end{equation*}

Notice that the $k$ drops out; this means that it doesn't matter that we don't know $k\text{,}$ it doesn't even matter if we use the wrong $k\text{,}$ we get the correct answer.

What about a second approach to this problem? We now want to approximate the change in position between time $a$ and time $b\text{.}$ We take the interval of time between $a$ and $b\text{,}$ divide it into $n$ subintervals, and approximate the distance traveled during each. The starting time of subinterval number $i$ is now $a+(i-1)(b-a)/n\text{,}$ which we abbreviate as $\ds t_{i-1}\text{,}$ so that $\ds t_0=a\text{,}$ $\ds t_1=a+(b-a)/n\text{,}$ and so on. The speed of the object is $f(t)=3t\text{,}$ and each subinterval is $(b-a)/n=\Delta t$ seconds long. The distance traveled during subinterval number $i$ is approximately $\ds f(t_{i-1})\Delta t\text{,}$ and the total change in distance is approximately

\begin{equation*} f(t_0)\Delta t+f(t_1)\Delta t+\cdots+f(t_{n-1})\Delta t\text{.} \end{equation*}

The exact change in position is the limit of this sum as $n$ goes to infinity. We abbreviate this sum using sigma notation:

\begin{equation*} \sum_{i=0}^{n-1} f(t_i)\Delta t = f(t_0)\Delta t+f(t_1)\Delta t+\cdots+f(t_{n-1})\Delta t \end{equation*}

\begin{equation*} \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t\text{.} \end{equation*}

Since this must be the same as the answer we have already obtained, we know that

\begin{equation*} \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t={3b^2\over 2}-{3a^2\over 2}\text{.} \end{equation*}

The significance of $\ds 3t^2/2\text{,}$ into which we substitute $t=b$ and $t=a\text{,}$ is of course that it is a function whose derivative is $f(t)\text{.}$ As we have discussed, by the time we know that we want to compute

\begin{equation*} \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\Delta t\text{,} \end{equation*}

it no longer matters what $f(t)$ stands for—it could be a speed, or the height of a curve, or something else entirely. We know that the limit can be computed by finding any function with derivative $f(t)\text{,}$ substituting $a$ and $b\text{,}$ and subtracting. We summarize this result in a theorem in Section 1.4.3 The Fundamental Theorem of Calculus, but first, we introduce the new notation definite integral and the terminology associated with it.

### Subsection1.4.2Defining the Definite Integral

Interactive Demonstration. Use the slider to investigate limit as $n \to \infty$ of the Riemann Sum. Note that the exact area under the curve is $4 \text{.}$

###### Definition1.17. The Definite Integral.

Let $f$ be a continuous function defined on the interval $[a,b]$ with partition $P=\{t_0,t_1,\dots,t_n\}$ of $[a,b]$ and subinterval width $\Delta t\text{.}$ If the limit

\begin{equation*} \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\,\Delta t \end{equation*}

exists, then the definite integral of $f$ over $[a,b]$ is defined by

\begin{equation*} \int_a^b f(t) dt = \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\,\Delta t \end{equation*}

where

 $\ds \int_a^b f(t)\,dt$ is read “the integral of $f$ from $a$ to $b$ w.r.t. $t$” , the symbol $\ds\int$ is called the integral sign, the value $a$ is the lower limit of integration, the value $b$ is the upper limit of integration, the function $f$ is referred to as the integrand of the integral, and the variable $t$ is called the variable of integration.

Note:

1. The process of finding the definite integral is called integration or integrating $f(x)\text{.}$

2. If the definite integral of $f$ exists over $[a,b]\text{,}$ then the function $f$ is integrable on $[a,b]\text{.}$

3. The definite integral is a number and not a function.

4. The value of the definite integral is independent of the variable of integration. Whether the integral is written as

\begin{equation*} \int_a^b f(x)\,dx \text{ or } \int_a^b f(t)\,dt \text{ or } \int_a^b f(u)\,du\text{,} \end{equation*}

it is still the limit of Riemann sums yielding the same value. Hence, the variable of integration is sometimes referred to as a dummy variable.

5. If $f$ is non-negative, then the definite integral represents the area of the region under the graph of $f$ on $[a,b]\text{;}$ otherwise, the definite integral represents the net area of the regions under the graph of $f$ on $[a,b]\text{,}$ which we will summarize in the geometric interpretation below.

We should ask ourselves, when is a function integrable? It turns out that no matter what choices are made in the Riemann sums associated with a continuous function, the Riemann sums always converge to the same limit. This is stated in the following theorem, which we will not prove.

Just like we emphasized the geometric interpretation of the derivative, we do not want to lose sight of the geometric interpretation of the definite integral.

###### Geometric Interpretation of the Definite Integral — Area.

Suppose $f$ is continuous on the interval $[a,b]\text{,}$ then

\begin{equation*} \int_a^b f(x) \, dx \end{equation*}

represents the area of the region under the graph of $f$ on $[a,b]$ if $f$ is non-negative.

###### Geometric Interpretation of the Definite Integral — Net Area.

Suppose $f$ is continuous on the interval $[a,b]\text{,}$ then

\begin{equation*} \int_a^b f(x) \,dx \end{equation*}

represents the net area of the region under the graph of $f$ on $[a,b]\text{.}$

Interactive Demonstration. Drag the points $a$ and $b$ by dragging the tick marks to investigate the geometric interpretation of the definite integral.

Lastly, we conclude this section by listing the properties of the definite integral.

###### Properties of Definite Integrals.

 Order of limits matters: $\ds \qquad\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx$ If interval is empty, integral is zero: $\ds\qquad\int_a^a f(x)\,dx=0$ Constant Multiple Rule: $\ds \qquad\int_a^b cf(x)\,dx=c\int_a^b f(x)\,dx$ Sum/Difference Rule: $\ds\qquad\int_a^b \left[f(x)\pm g(x)\right]\,dx=\int_a^b f(x)\,dx\pm\int_a^b g(x)\,dx$ Can split up interval $[a,b]=[a,c]\cup[c,b]\text{:}$ $\ds\qquad\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$ The variable does not matter: $\ds\qquad\int_a^b f(x)\,dx=\int_a^b f(t)\,dt$

The reason for the last property is that a definite integral is a number, not a function, so the variable is just a placeholder that won't appear in the final answer.

Some additional properties are comparison types of properties.

###### Comparison Properties of Definite Integrals.

 If $f(x)\geq 0$ for $x\in[a,b]\text{,}$ then $\ds \int_a^b f(x)\,dx\geq 0\text{.}$ If $f(x)\geq g(x)$ for $x\in[a,b]\text{,}$ then $\ds\int_a^b f(x)\,dx\geq \int_a^b g(x)\,dx\text{.}$ If $m\leq f(x)\leq M$ for $x\in[a,b]\text{,}$ then $\ds m(b-a)\leq \int_a^b f(x)\,dx\leq M(b-a)\text{.}$

###### Example1.19. Properties of Definite Integrals.

Suppose $\ds{\int_a^b f(x)~dx=7}$ and $\ds{\int_a^b g(x)~dx=3}\text{.}$ Find:

1. $\ds\int_a^b 2f(x)-3g(x)\,dx\text{.}$

2. $\ds\int_{b}^{a} 2g(x)\,dx\text{.}$

3. $\ds\int_a^a f(x)\cdot g(x)\,dx\text{.}$

4. $\ds\int_a^c f(x)~dx+\int_c^b f(x)\,dx\text{.}$

Solution
1. $\ds\int_a^b 2f(x)-3g(x)\,dx=\ds 2\int_a^b f(x)\,dx-3\int_a^b g(x)\,dx=2(7)-3(3)=5\text{.}$

2. $\ds\int_{b}^{a} 2g(x)\,dx=\ds -2\int_{a}^{b} g(x)\,dx=-2(3)=-6\text{.}$

3. $\ds\int_a^a f(x)\cdot g(x)\,dx=0\text{.}$

4. $\ds\int_a^c f(x)\,dx+\int_c^b f(x)\,dx=\ds\int_a^b f(x)\,dx=7\text{.}$

### Subsection1.4.3The Fundamental Theorem of Calculus

We are finally in a position to state the result from our exploration at the beginning of Section 1.4.1. What we have learned is that the integral of $f$ over the interval $[a,b]$ can be computed by finding a function, say $F(t)\text{,}$ with the property that $F'(t)= f(t)\text{,}$ and then computing $F(b)-F(a)\text{.}$ Recall that the function $F(t)$ is called an antiderivative of $f(t)\text{.}$ Let us now state the theorem:

Let's rewrite this slightly:

\begin{equation*} \int_a^x f(t)\,dt = F(x)-F(a) \end{equation*}

We've replaced the variable $x$ by $t$ and $b$ by $x\text{.}$ These are just different names for quantities, so the substitution doesn't change the meaning of the theorem statement. However, it does allow us to give a new interpretation of the theorem by thinking of the two sides of the equation as functions of $x\text{.}$ The expression

\begin{equation*} \int_a^x f(t)\,dt \end{equation*}

is a function: plug in a value for $x\text{,}$ get out some other value. The expression $F(x)-F(a)$ is of course also a function, and it has a nice property:

\begin{equation*} {d\over dx} (F(x)-F(a)) = F'(x) = f(x)\text{,} \end{equation*}

since $F(a)$ is a constant and has derivative zero. In other words, by shifting our point of view slightly, we see that the odd looking function

\begin{equation*} G(x)=\int_a^x f(t)\,dt \end{equation*}

has a derivative, and that in fact $G'(x)=f(x)\text{.}$

Note: This is really just a restatement of the Fundamental Theorem of Calculus, and indeed is often called the Fundamental Theorem of Calculus. To avoid confusion, some people call the two versions of the theorem “The Fundamental Theorem of Calculus, part I” and “The Fundamental Theorem of Calculus, part II”, although unfortunately there is no universal agreement as to which is part I and which part II. Since it really is the same theorem, differently stated, some people simply call them both “The Fundamental Theorem of Calculus.”

Interactive Demonstration. The graph of $f(x) = \cos(x)$ is shown below to the left. The net area $\int_a^x f(t)\,dt$ is plotted as a function of $x$ below to the right. Slide the point $x$ to investigate the FTC.

We have not really proved the Fundamental Theorem. In a nutshell, we gave the following argument to justify it: Suppose we want to know the value of

\begin{equation*} \int_a^b f(t)\,dt = \lim_{n\to\infty}\sum_{i=0}^{n-1} f(t_i)\,\Delta t\text{.} \end{equation*}

We can interpret the right hand side as the distance traveled by an object whose speed is given by $f(t)\text{.}$ We know another way to compute the answer to such a problem: find the position of the object by finding an antiderivative of $f(t)\text{,}$ then substitute $t=a$ and $t=b$ and subtract to find the distance traveled. This must be the answer to the original problem as well, even if $f(t)$ does not represent a speed.

What's wrong with this? In some sense, nothing. As a practical matter it is a very convincing argument, because our understanding of the relationship between speed and distance seems to be quite solid. From the point of view of mathematics, however, it is unsatisfactory to justify a purely mathematical relationship by appealing to our understanding of the physical universe, which could, however unlikely it is in this case, be wrong.

A complete proof is a bit too involved to include here, but we will indicate how it goes. First, if we can prove the second version of the Fundamental Theorem of Calculus, Theorem 1.21, then we can prove the first version from that:

Proof of Theorem 1.20.

We know from Theorem 1.21 that

\begin{equation*} G(x)=\int_a^x f(t)\,dt \end{equation*}

is an antiderivative of $f(x)\text{,}$ and therefore any antiderivative $F(x)$ of $f(x)$ is of the form $F(x)=G(x)+k\text{.}$ Then

\begin{equation*} \begin{split} F(b)-F(a) \amp=G(b)+k-(G(a)+k)\\ \amp = G(b)-G(a)\\ \amp = \int_a^b f(t)\,dt-\int_a^a f(t)\,dt \end{split} \end{equation*}

It is clear that $\ds \int_a^a f(t)\,dt=0\text{,}$ so this means that

\begin{equation*} F(b)-F(a)=\int_a^b f(t)\,dt\text{,} \end{equation*}

which is exactly what Theorem 1.20 says.

So the real job is to prove Theorem 1.21. We will sketch the proof, using some facts that we do not prove. First, the following identity is true of integrals:

\begin{equation*} \int_a^b f(t)\,dt = \int_a^c f(t)\,dt + \int_c^b f(t)\,dt \end{equation*}

This can be proved directly from the definition of the integral, that is, using the limits of sums. It is quite easy to see that it must be true by thinking of either of the two applications of integrals that we have seen. It turns out that the identity is true no matter what $c$ is, but it is easiest to think about the meaning when $a\le c\le b\text{.}$

First, if $f(t)$ represents a speed, then we know that the three integrals represent the distance traveled between time $a$ and time $b\text{;}$ the distance traveled between time $a$ and time $c\text{;}$ and the distance traveled between time $c$ and time $b\text{.}$ Clearly the sum of the latter two is equal to the first of these.

Second, if $f(t)$ represents the height of a curve, the three integrals represent the area under the curve between $a$ and $b\text{;}$ the area under the curve between $a$ and $c\text{;}$ and the area under the curve between $c$ and $b\text{.}$ Again it is clear from the geometry that the first is equal to the sum of the second and third.

Proof of Theorem 1.21.

We want to compute $G'(x)\text{,}$ so we start with the definition of the derivative in terms of a limit:

\begin{equation*} \begin{split} G'(x) =\amp \lim_{\Delta x\to0}{G(x+\Delta x)-G(x)\over\Delta x}\\ =\amp \lim_{\Delta x\to0}{1\over \Delta x}\left( \int_a^{x+\Delta x} f(t)\,dt - \int_a^x f(t)\,dt\right)\\ =\amp \lim_{\Delta x\to0}{1\over \Delta x}\left( \int_a^{x} f(t)\,dt + \int_x^{x+\Delta x} f(t)\,dt - \int_a^x f(t)\,dt\right)\\ =\amp \lim_{\Delta x\to0}{1\over \Delta x}\int_x^{x+\Delta x} f(t)\,dt \end{split} \end{equation*}

Now we need to know something about

\begin{equation*} \int_x^{x+\Delta x} f(t)\,dt \end{equation*}

when $\Delta x$ is small; in fact, it is very close to $\Delta x f(x)\text{,}$ but we will not prove this. Once again, it is easy to believe this is true by thinking of our two applications: The integral

\begin{equation*} \int_x^{x+\Delta x} f(t)\,dt \end{equation*}

can be interpreted as the distance traveled by an object over a very short interval of time. Over a sufficiently short period of time, the speed of the object will not change very much, so the distance traveled will be approximately the length of time multiplied by the speed at the beginning of the interval, namely, $\Delta x f(x)\text{.}$ Alternately, the integral may be interpreted as the area under the curve between $x$ and $x+\Delta x\text{.}$ When $\Delta x$ is very small, this will be very close to the area of the rectangle with base $\Delta x$ and height $f(x)\text{;}$ again this is $\Delta x f(x)\text{.}$ If we accept this, we may proceed:

\begin{equation*} \lim_{\Delta x\to0}{1\over \Delta x}\int_x^{x+\Delta x} f(t)\,dt =\lim_{\Delta x\to0}{\Delta x f(x)\over \Delta x}=f(x)\text{,} \end{equation*}

which is what we wanted to show.

It is still true that we are depending on an interpretation of the integral to justify the argument, but we have isolated this part of the argument into two facts that are not too hard to prove. Once the last reference to interpretation has been removed from the proofs of these facts, we will have a real proof of the Fundamental Theorem of Calculus.

Note: Now we know that to solve certain kinds of problems, those that lead to a sum of a certain form, we “merely” find an antiderivative and substitute two values and subtract. Unfortunately, finding antiderivatives can be quite difficult. While there are a small number of rules that allow us to compute the derivative of any common function, there are no such rules for antiderivatives. There are some techniques that frequently prove useful as you will see in Chapter 2 Techniques of Integration, but we will never be able to reduce the problem to a completely mechanical process.

### Subsection1.4.4Notation when Computing a Definite Integral

When we compute a definite integral, we first find an antiderivative of the integrand and then substitute. It is convenient to first display the antiderivative and then do the substitution; we need a notation indicating that the substitution is yet to be done. A typical solution would look like this:

\begin{equation*} \int_1^2 x^2\,dx=\left.{x^3\over 3}\right|_1^2 = {2^3\over3}-{1^3\over3}={7\over3}\text{.} \end{equation*}

The vertical line with subscript and superscript is used to indicate the operation “substitute and subtract” that is needed to finish the evaluation.

Common Mistakes:

1. Dropping the $dx$ at the end of the integral. This is required! Think of the integral as a set of parenthesis consisting of the integral sign and the $dx\text{.}$ Both are required so it is clear where the integrand ends and what variable you are integrating with respect to.

\begin{equation*} \int_a^b f(x)\,dx \neq \int_a^b f(x) \end{equation*}
2. Checking the antiderivative for the integrand using differentiation before following through on the lower and upper bounds of integration.

\begin{equation*} \int_a^b f(x)\,dx = F(x)\bigg\vert_a^b \text{ with } F'(x)=f(x) \end{equation*}
3. Dropping the lower and upper bounds of integration during the solution process. For example:

\begin{equation*} \int_1^2 x^2\,dx \neq \int x^2dx \neq \frac{x^3}{3}\bigg\vert_1^2 = \frac{2^3}{3}-\frac{1^3}{3}=\frac{7}{3} \end{equation*}
4. Forgetting to evaluate using the lower and upper bounds of integration that lead to the value of the definite integral and instead producing a function. See the note immediately after the definition of the definite integral. For example:

\begin{equation*} \int_1^2 x^2\, dx \neq \frac{x^3}{3} \end{equation*}
5. Switching the order in which the lower and upper bounds of integration are being dealt with. For example:

\begin{equation*} \int_1^2 x^2\,dx = \frac{x}{3}\bigg\vert_1^2 \neq \frac{1^3}{3}-\frac{2^3}{3} = -\frac{7}{3}\text{ , or } \end{equation*}
\begin{equation*} \int_1^2 x^2\,dx \neq \frac{x}{3}\bigg\vert_2^1 = \frac{1^3}{3}-\frac{2^3}{3} = -\frac{7}{3} \end{equation*}

### Subsection1.4.5Computing a Definite Integral

We seem to have found a pattern when dealing with power functions. When attempting to solve a previous question, we found the antiderivative of $x^2$ to be $x^3/3+c$ (as it was when solving the indefinite integral). Likewise, when we first began, we were trying to determine a position based on velocity, and $3t$ gave rise to $3t^2/2+k\text{.}$

As will be formalized later, we see that in these cases, the power is increased to $n+1\text{,}$ but we also divide through by this factor, $n+1\text{.}$ So the antiderivative of $x$ becomes $x^2/2\text{,}$ the antiderivative of $x^2$ becomes $x^3/3\text{,}$ and the antiderivative of $x^3$ will become $x^4/4\text{.}$

Now we will also try with negative and fractional values in the following example.

###### Example1.22. Fundamental Theorem of Calculus.

Evaluate $\ds\int_1^4 \left(x^3+\sqrt{x}+\frac{1}{x^2}\right)\,dx\text{.}$

Solution
\begin{equation*} \begin{array}{lcl} \ds\int_1^4 \left(x^3+\sqrt{x}+\frac{1}{x^2}\right)\,dx \amp = \amp \ds\left.\frac{x^4}{4}+\frac{2x^{3/2}}{3}-x^{-1}\right|_1^4\\ \\ \amp = \amp \ds\left(\frac{(4)^4}{4}+\frac{2(4)^{3/2}}{3}-4^{-1}\right) \\ \\ \amp \amp \ds-\left(\frac{(1)^4}{4}+\frac{2(1)^{3/2}}{3}-1^{-1}\right)\\ \\ \amp = \amp \ds\frac{415}{6} \end{array} \end{equation*}
Note: The above integral used parentheses around the integrand for clarity, but we can leave them out once we are familiar with the notation as shown in the next example.

We next evaluate a definite integral using three different techniques.

###### Example1.23. Three Different Techniques.

Evaluate $\ds{\int_0^2 x+1~dx}$ by

1. Using FTC I (the shortcut)

2. Using the definition of a definite integral (the limit sum definition)

3. Interpreting the problem in terms of areas (graphically)

Solution
1. The shortcut (FTC I) is the method of choice as it is the fastest. Integrating and using the ‘top minus bottom’ rule we have:

\begin{equation*} \begin{split} \int_0^2 x+1~dx\amp = \left.\frac{x^2}{2}+x\right|_0^2\\ =\amp \left[\frac{2^2}{2}+2\right]-\left[\frac{0^2}{2}+0\right]=4\end{split} \end{equation*}
2. We now use the definition of a definite integral. We divide the interval $[0,2]$ into $n$ subintervals of equal width $\Delta x\text{,}$ and from each interval choose a point $x_i^*\text{.}$ Using the formulas

we have

Then taking $x_i^*$'s as right endpoints for convenience (so that $x_i^*=x_i$), we have:

\begin{align*} \int_0^2 x+1~dx \amp = \ds\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\,\Delta x\\ \amp = \ds\lim_{n\to\infty}\sum_{i=1}^n f\left(\frac{2i}{n}\right) \frac{2}{n}\\ \amp = \ds\lim_{n\to\infty}\sum_{i=1}^n \left(\frac{2i}{n}+1\right) \frac{2}{n}\\ \amp = \ds\lim_{n\to\infty}\sum_{i=1}^n \left(\frac{4i}{n^2}+\frac{2}{n}\right)\\ \amp = \ds\lim_{n\to\infty}\left(\sum_{i=1}^n \frac{4i}{n^2}+\sum_{i=1}^n \frac{2}{n} \right)\\ \amp = \ds\lim_{n\to\infty}\left(\frac{4}{n^2}\sum_{i=1}^n i+\frac{2}{n}\sum_{i=1}^n 1 \right)\\ \amp = \ds\lim_{n\to\infty}\left(\frac{4}{n^2}\frac{n(n+1)}{2}+\frac{2}{n}n \right)\\ \amp = \ds\lim_{n\to\infty}\left(2+\frac{2}{n}+2 \right)\\ \amp = 4\text{.} \end{align*}
3. Finally, let's evaluate the net area under $x+1$ from $0$ to $2\text{.}$

Thus, the area is the sum of the areas of a rectangle and a triangle. Hence,

\begin{align*} \int_0^2 x+1~dx\amp = \mbox{Net Area}\\ \amp = \mbox{Area of rectangle} + \mbox{Area of triangle}\\ \amp = (2)(1)+\frac{1}{2}(2)(2)\\ \amp = 4\text{.} \end{align*}
###### Example1.24. FTC and Marginal Profit.

A small bakery determines that the marginal profit function for selling $q$ loaves of bread is

\begin{equation*} P'(q)=-0.0005q^2+2.5 \end{equation*}

dollars per day per loaf. If the fixed production cost is $10, 1. determine $P(50)\text{,}$ and 2. determine whether or not it is advisable to increase production from 50 loaves to 100 loaves. Solution 1. We first notice that \begin{equation*} \int_a^{50} P'(q)\,dq = P(50)-P(a)\text{.} \end{equation*} Since we know that $P(0)=-10$ (the fixed cost), we take $a=0$ and integrate: \begin{equation*} \begin{split} \int_0^{50} P'(q)\, dq \amp =\int_0^{50}\left(-0.0005q^2+2.5\right) \,dq \\[1ex] \amp =-0.0005\int_0^{50} q^2 dq+2.5\int_0^{50}\, dq \\[1ex] \amp = -0.0005\frac{q^3}{3}\bigg\vert_0^{50} + 2.5q\bigg\vert_0^{50} \\[1ex] \amp = -0.0005 \left(\frac{50^3}{3}-0\right)+2.5\left(50-0\right) = 104.167 \end{split} \end{equation*} Thus, \begin{equation*} P(50) = 104.167 - (-10) = 114.167\text{.} \end{equation*} Therefore, the daily profit the bakery realizes from selling 50 loaves of bread is approximately$114.

2. The change in the profit can be determined by integrating

\begin{equation*} \begin{split} \int_{50}^{100} P'(q)\, dq \amp = \int_{50}^{100}\left(-0.0005q^2+2.5\right) \,dq\\[1ex] \amp = -0.0005 \left(\frac{100^3}{3}-\frac{50^3}{3}\right)+2.5\left(100-50\right) \\ \amp= -20.8333 \end{split} \end{equation*}

This means that the bakery would lose about 21 per day if production were doubled. Therefore it is not advisable to increase production to 100 loaves. We next apply FTC to differentiate a function. ###### Example1.25. Using FTC. Differentiate the following function: \begin{equation*} g(x)=\int_{-2}^x \cos(1+5t)\sin t\,dt \end{equation*} Solution We simply apply the Fundamental Theorem of Calculus directly to get: \begin{equation*} g'(x)=\cos(1+5x)\sin x\text{.} \end{equation*} Using the Chain Rule we can derive a formula for some more complicated problems when the upper limit of integration is some function of $x$ rather than simply $x\text{.}$ We have \begin{equation*} \frac{d}{dx}\int_a^{v(x)}f(t)\,dt=f(v(x))\cdot v'(x)\text{.} \end{equation*} Now what if the upper limit is constant and the lower limit is a function of $x\text{?}$ Then we interchange the limits and add a minus sign to get \begin{equation*} \frac{d}{dx}\int_{u(x)}^af(t)\,dt=-\frac{d}{dx}\int_a^{u(x)} f(t)\,dt=-f(u(x))\cdot u'(x)\text{.} \end{equation*} Combining these two we can get a formula where both limits are functions of $x\text{.}$ We break up the integral as follows: \begin{equation*} \int_{u(x)}^{v(x)} f(t)\,dt=\int_{u(x)}^a f(t)\,dt+\int_a^{v(x)}f(t)\,dt \end{equation*} We just need to make sure $f(a)$ exists after we break up the integral. Then differentiating and using the above two formulas gives: ###### FTC and Chain Rule Formula:. \begin{equation*} \frac{d}{dx}\int_{{u(x)}}^{{v(x)}} f(t)\,dt=f({v(x)}){v'(x)}-f({u(x)}){u'(x)} \end{equation*} Many textbooks do not show this formula and instead to solve these types of problems will use FTC I along with the tricks we used to derive the formula above. Either method is perfectly fine. ###### Example1.26. FTC and Chain Rule I. Differentiate the following integral: \begin{equation*} \int_{10x}^{x^2} t^3\sin(1+t) \,dt \end{equation*} Solution We will use the formula above. We have $f(t)=t^3\sin(1+t)\text{,}$ $u(x)=10x$ and $v(x)=x^2\text{.}$ Then $u'(x)=10$ and $v'(x)=2x\text{.}$ Thus, \begin{align*} \frac{d}{dx}\int_{10x}^{x^2} t^3\sin(1+t) \,dt\amp = (x^2)^3\sin(1+(x^2))(2x)-(10x)^3\sin(1+(10x))(10)\\ \amp = 2x^7\sin(1+x^2)-10000x^3\sin(1+10x) \end{align*} ###### Example1.27. FTC and Chain Rule II. Differentiate the following integral with respect to $x\text{:}$ \begin{equation*} \int_{x^3}^{2x} 1+\cos t\,dt \end{equation*} Solution Using the formula we have: \begin{equation*} \frac{d}{dx}\int_{x^3}^{2x} 1+\cos t\,dt=(1+\cos(2x))(2)-(1+\cos(x^3))(3x^2)\text{.} \end{equation*} ##### Exercises for Section 1.4. Determine the area under the graph of each function on the given interval. 1. $\ds f(x)=3x+1$ on $[0,2]$ Answer 8 Solution The function $f(x)=3x+1$ is positive on the interval $[0,2]\text{.}$ Therefore, the area under the curve is given by \begin{equation*} \begin{split} A \amp = \int_0^2 f(x)\,dx= \int_0^2 \left(3x+1\right)\,dx\\ \amp = \left[\frac{3}{2}x^2+x\right]_0^2\\ \amp = \left(\frac{3}{2} (2^2) + 2\right)- \left(0 + 0\right)\\ \amp = 8. \end{split} \end{equation*} 2. $\ds f(t)=2t$ on $[1,e]$ Answer 6.39 Solution The function $f(t) = 2t$ is positive on the interval $[1,e]\text{,}$ and so the area under the curve can be calculated by the integral: \begin{equation*} \begin{split} A \amp = \int_1^e f(t)\,dt = \int_1^e 2t\,dt\\ \amp = t^2\bigg\vert_1^e\\ \amp = e^2 - 1^2 = e^2 -1 \end{split} \end{equation*} Therefore, $A \approx 6.39\text{.}$ 3. $\ds g(t)=1+\sqrt{t}$ on $[3,6]$ Answer 9.3339 Solution The function $g(t) = 1+\sqrt{t}$ is positive on the interval $[3,6]\text{.}$ Therefore, the area under the curve is calculated by \begin{equation*} \begin{split} A \amp = \int_3^6 g(t)\,dt = \int_3^6 \left(1+\sqrt{t}\right)\,dt \\[1ex] \amp = \left[t+\frac{2}{3}t^{3/2}\right]_3^6\\[1ex] \amp = \left(6+\frac{2}{3}(6)^{3/2}\right)-\left(3+\frac{2}{3}(3)^{3/2}\right). \end{split} \end{equation*} We find that $A \approx 9.3339\text{.}$ 4. $\ds h(x)=\sin(x)$ on $[0,\pi]$ Answer 2 Solution Again, we see that $h(x) = \sin(x)$ is positive on the interval $[0,\pi]\text{.}$ The area under the curve is then \begin{equation*} \begin{split} A \amp = \int_0^{\pi} h(x)\,dx = \int_0^{\pi} \sin(x)\,dx \\[1ex] \amp = -\cos(x)\bigg\vert_0^\pi = -\left[\left(-1\right) - 1\right] = 2. \end{split} \end{equation*} Evaluate the following definite integrals. 1. $\ds \int_1^4 t^2+3t\,dt$ Answer $87/2$ Solution \begin{aligned}\int_1^4 t^2+3t\,dt \amp = \left[\frac{1}{3}t^3+\frac{3}{2}t^2\right]_1^4\\ \amp = \left(\frac{1}{3}(4^3)+\frac{3}{2}(4^2)\right) - \left(\frac{1}{3}+\frac{3}{2}\right)\\ \amp = \frac{87}{2} \end{aligned} 2. $\ds \int_0^\pi \sin t\,dt$ Answer $2$ Solution \begin{aligned}\int_0^{\pi} \sin t\,dt \amp = -\cos t\bigg\vert_0^{\pi}\\ \amp = -\cos(\pi) + \cos(0)\\ \amp = 1+1 = 2 \end{aligned} 3. $\ds \int_1^{10} {1\over x}\,dx$ Answer $\ln(10)$ Solution \begin{aligned}\int_1^{10} \frac{1}{x}\,dx \amp = \ln|x|\bigg\vert_1^{10}\\ \amp = \ln(10)-\ln(1) \\ \amp =\ln(10) \end{aligned} 4. $\ds \int_0^5 e^x\,dx$ Answer $\ds e^5-1$ Solution \begin{aligned}\int_0^5 e^x \,dx \amp = e^x \bigg\vert_0^5\\ \amp = e^5 - e^0\\ \amp = e^5 - 1 \end{aligned} 5. $\ds \int_0^3 x^3\,dx$ Answer $\ds 3^4/4$ Solution \begin{aligned}\int_0^3 x^3\,dx \amp = \frac{1}{4}x^4\bigg\vert_0^3\\ \amp = \frac{1}{4}\left(3^4-0^4\right)\\ \amp = \frac{81}{4} \end{aligned} 6. $\ds \int_1^2 x^5\,dx$ Answer $\ds 2^6/6 -1/6$ Solution \begin{aligned}\int_1^2 x^5\,dx \amp = \frac{1}{6}x^6\bigg\vert_1^2\\ \amp = \frac{1}{6}\left(2^6-1^6\right)\\ \amp = \frac{63}{6} \end{aligned} 7. $\ds \int_1^2 7x^{5/2}\,dx$ Answer $\ds 16\sqrt{2}-2$ Solution \begin{aligned}\int_1^2 7x^{5/2}\,dx \amp = 7\left[\frac{2x^{7/2}}{7}\right]_1^2\\ \amp = 2\left(2^{7/2}-1\right)\\ \amp \approx 20.627 \end{aligned} 8. $\ds \int_0^2\left(2x^3+x^2-5\right)\,dx$ Answer $2/3$ Solution \begin{aligned}\int_0^2 \left(2x^3+x^2-5\right)\,dx \amp = \left[\frac{1}{2}x^4 +\frac{1}{3}x^3 - 5x\right]_0^2\\ \amp = \frac{1}{2} 2^4 + \frac{1}{3} 2^3 - 5(2) \\ \amp = \frac{2}{3} \end{aligned} 9. $\ds \int_0^2\left(\frac{1}{\sqrt{x}}-\sqrt{x}\right)\,dx$ Answer $\ds \frac{2\sqrt{2}}{3}$ Solution \begin{aligned}\int_0^2 \left(\frac{1}{\sqrt{x}}-\sqrt{x}\right)\,dx \amp = \left[2 \sqrt{x} - \frac{2}{3}x^(3/2)\right]_0^2\\ \amp = 2\sqrt{2} - \frac{2}{3} 2^{3/2}\\ \amp = \frac{2\sqrt{2}}{3} \end{aligned} 10. $\ds \int_2^5 \frac{1}{3x+2}\,dx$ Answer $\ds \frac{\log 17 - \log 8}{3}$ Solution We know that \begin{equation*} \diff{}{x} \ln(3x+2) = \frac{3}{2x+2}\text{,} \end{equation*} and so by the FTC: \begin{equation*} \begin{split} \int_2^5 \frac{1}{3x+2}\,dx \amp = \int_2^5 \frac{1}{3} \diff{}{x} \ln(3x+2)\,dx \\ \amp = \frac{1}{3} \ln(3x+2)\bigg\vert_2^5\\ \amp = \frac{1}{3} \left(\ln(17) - \ln(8)\right)\\ \amp \approx 0.2513 \end{split} \end{equation*} 11. $\ds \int_{-\pi/3}^0 \frac{\tan x}{\cos x}\,dx$ Answer $-1$ Solution We frist rewrite the integrand: \begin{equation*} \dfrac{\tan x}{\cos x} = \dfrac{\sin x}{\cos^2 x}\text{.} \end{equation*} Then we recall that \begin{equation*} \diff{}{x} \sec x = \frac{\sin x}{\cos^2 x}\text{.} \end{equation*} Therefore, by the FTC: \begin{equation*} \begin{split} \int_{-\pi/3}^0 \frac{\tan x}{\cos x}\,dx \amp = \int_{-\pi/3}^0 \diff{}{x} \sec x\,dx \\ \amp = \sec x\bigg\vert_{-\pi/3}^0\\ \amp = \sec (0) - \sec(-\pi/3) = -1 \end{split} \end{equation*} Find the derivative of the given function $G\text{.}$ 1. $\ds G(x)=\int_1^x t^2-3t\,dt$ Answer $\ds x^2-3x$ Solution We have \begin{equation*} G(x) = \int_1^x (t^2-3t)\,dt\text{.} \end{equation*} By the Fundamental Theorem of Calculus, \begin{equation*} G'(x) = \diff{}{x}\left(\int_1^x (t^2-3t)\,dt\right) = x^2 - 3x\text{.} \end{equation*} 2. $\ds G(x)=\int_1^{x^2} t^2-3t\,dt$ Answer $\ds 2x(x^4-3x^2)$ Solution We have \begin{equation*} G(x) = \int_1^{x^2} (t^2-3t)\,dt\text{.} \end{equation*} Let $F$ be the general antiderivative of the integrand, $t^2-3t\text{.}$ Then by the Chain Rule, \begin{equation*} \begin{split} G'(x) \amp = \diff{}{x} \left(F(x^2) - F(1)\right)\\[1ex] \amp =2xF'(x^2) - 0. \end{split} \end{equation*} Since $F'(t) = t^2-3t\text{,}$ we see that \begin{equation*} G'(x) = 2x(x^4-3x^2) = 2x^5-6x^3\text{.} \end{equation*} 3. $\ds G(x)=\int_1^x e^{t^2}\,dt$ Answer $\ds e^{x^2}$ Solution We have \begin{equation*} G(x) = \int_1^x e^{t^2}\,dt\text{.} \end{equation*} Therefore, by the Fundamental Theorem of Calculus, \begin{equation*} G'(x) = \diff{}{x}\int_1^x e^{t^2}\,dt = e^{x^2}. \end{equation*} 4. $\ds G(x)=\int_1^{x^2} e^{t^2}\,dt$ Answer $\ds 2xe^{x^4}$ Solution We have \begin{equation*} G(x) = \int_1^x e^{t^2}\,dt\text{.} \end{equation*} Let $F$ be the general antiderivative of the integrand, $e^{t^2}\text{.}$ Then by the Chain Rule, \begin{equation*} \begin{split} G'(x) \amp = \diff{}{x} \left(F(x^2) - F(1)\right)\\[1ex] \amp =2xF'(x^2) - 0. \end{split} \end{equation*} Since $F'(t) = e^{t^2}\text{,}$ we see that \begin{equation*} G'(x) = 2xe^{x^4}. \end{equation*} 5. $\ds G(x)=\int_1^x \tan(t^2)\,dt$ Answer $\ds \tan(x^2)$ Solution e have \begin{equation*} G(x) = \int_1^x \tan(t^2)\,dt\text{.} \end{equation*} Therefore, by the Fundamental Theorem of Calculus, \begin{equation*} G'(x) = \diff{}{x}\int_1^x\tan(t^2)\,dt = \tan(x^2)\text{.} \end{equation*} 6. $\ds G(x)=\int_{10x}^{x^2} \tan(t^2)\,dt$ Answer $\ds 2x\tan(x^4)-10\tan(100x^2)$ Solution We have \begin{equation*} G(x) = \int_{10x}^{x^2} \tan(t^2)\,dt\text{.} \end{equation*} Let $F$ be a general antiderivative of $\tan(t^2)\text{.}$ Then by the Chain Rule, \begin{equation*} \begin{split} G'(x) \amp = \diff{}{x} \left(F(x^2) - F(10x)\right)\\[1ex] \amp =2xF'(x^2) - 10F'(10x). \end{split} \end{equation*} We defined $F$ such that $F'(t)=\tan(t^2)\text{.}$ Thus, \begin{equation*} G'(x) = 2x\tan(x^4) - 10\tan(100x^2)\text{.} \end{equation*} Suppose $\ds\int_{1}^{4}f(x)\,dx=2$ and $\ds\int_{1}^{4}g(x)\,dx=7\text{.}$ Find: 1. $\ds\int_{1}^{4}(5f(x)+3g(x))\,dx$ Answer 31 Solution By linearity of the definite integral, \begin{equation*} \int_1^4 \left(5f(x) + 3g(x)\right)\,dx = 5\int_1^4 f(x)\,dx + 3\int_1^4 g(x)\,dx\text{.} \end{equation*} So from the given data, we have \begin{equation*} \begin{split} \int_1^4 \left(5f(x) + 3g(x)\right)\,dx \amp = 5\int_1^4 f(x)\,dx + 3\int_1^4 g(x)\,dx\\ \amp = 5(2) + 3(7)\\ \amp = 31. \end{split} \end{equation*} 2. $\ds\int_{1}^{4}(6-2f(x))\,dx$ Answer 41 Solution Similarly, \begin{equation*} \begin{split} \int_1^4 \left(6 -2f(x)\right)\,dx \amp = 6\int_1^4 \,dx -2\int_1^4 f(x)\,dx\\ \amp = 6(4-1) - 2(2)\\ \amp = 14. \end{split} \end{equation*} Suppose $\ds\int_{-2}^{5}f(x)\,dx=3$ and $\ds\int_{1}^{5}f(x)\,dx=-2\text{.}$ Find $\ds\int_{-2}^{1}f(x)\,dx\text{.}$ Answer 5 Solution We are given that \begin{equation*} \int_{-2}^5 f(x)\,dx = 3 \text{ and } \int_{1}^5 f(x)\,dx = -2\text{.} \end{equation*} Since the integrals are limits of Riemann sums, we can rewrite \begin{equation*} \int_{-2}^1 f(x)\,dx = \int_{-2}^5 f(x)\,dx - \int_{1}^5 f(x)\,dx\text{.} \end{equation*} Therefore, \begin{equation*} \int_{-2}^1 f(x)\,dx = 3-(-2) = 5\text{.} \end{equation*} Suppose $\ds\int_{-1}^{1}f(x)\,dx=10$ and $\ds\int_{0}^{1}f(x)\,dx=2\text{.}$ Find $\ds\int_{-1}^{0}2f(x)\,dx\text{.}$ Answer 16 Solution We are given that \begin{equation*} \int_{-1}^1 f(x)\,dx = 10 \text{ and } \int_{0}^1 f(x)\,dx = 2\text{.} \end{equation*} Since the integrals are limits of Riemann sums, we can rewrite \begin{equation*} \int_{-1}^0 f(x)\,dx = \int_{-1}^1 f(x)\,dx - \int_{0}^1 f(x)\,dx\text{.} \end{equation*} Furthermore, by linearity we have \begin{equation*} \int_{-1}^0 2f(x)\,dx = 2\int_{-1}^0 f(x)\,dx\text{.} \end{equation*} Therefore, \begin{equation*} \int_{-1}^0 2f(x)\,dx = 2\left(\int_{-1}^1 f(x)\,dx - \int_{0}^1 f(x)\,dx\right) = 2\left(10-2\right) = 16\text{.} \end{equation*} A manufacturer determines that the marginal cost function associated with the production of $q$ units is \begin{equation*} C'(q)=0.0006q^2-0.5q+25 \end{equation*} dollars per unit per day. Suppose the daily fixed cost is100.

1. Determine the total cost of producing 30 units.

$630.40 Solution The total cost of producing 30 units is \begin{equation*} \int_0^{30} C'(q)\,dq + 100 = \int_0^{30} 0.0006q^2-0.5q+25 \,dq +100 = 0.0002(30)^3-\frac{30^2}{4}+25(30) + 100 = 630.40 \end{equation*} dollars per day. 2. After 30 units have been made, determine the cost of producing an additional 20 units. Answer$119.60
Solution

After the first 30 units, the cost of producing an additional 20 units will be

\begin{equation*} \int_{30}^{50} C'(q)\,dq = \bigl(0.0002(50)^2-\frac{50^2}{4}+25(50)\bigr)- \bigl(0.0002(30)^3-\frac{30^2}{4}+25(30)\bigr) = 119.60\text{,} \end{equation*}

dollars per day.

A manufacturer determines that the marginal revenue function associated with the production of $q$ units is

\begin{equation*} R'(q)=-0.5q+100 \end{equation*}

dollars per unit per day.

1. Determine the total revenue from selling 100 units.

$7500 Solution The total revenue from selling 100 units is thus \begin{equation*} \int_0^{100} R'(q)\, dq = \left[-\frac{1}{4}q^2+100q\right]_0^{100} = 7500\text{,} \end{equation*} or,$7500 per day.
$2500 Solution After selling the first 100 units, the next 100 units will yield a revenue of \begin{equation*} \int_{100}^{200} R'(q)\,dq = \left[-\frac{1}{4}q^2+100q\right]_{100}^{200}=2500\text{.} \end{equation*} That is,$2500 per day.