MATH 1071 — Calculus for Business & Economics

Lecture Notes. This is NOT meant to be a substitute for coming to lecture or consulting the textbook directly, but rather a helpful resource to review for exams/ review what was discussed in lecture. Each section ends with a Self-Check taken from the in-class activity — try the problem first, then click the button to see the answer.

Antiderivatives

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Goals Define the antiderivative and its notation; find several basic rules for antiderivatives.

Going Backwards

We have learned how to take the derivative of many functions. Now let's go backwards: given a derivative, what function did it come from?

Remark (Derivative Rules We Will Reverse) As a reminder, here are the derivatives we will be "undoing" in this unit: \[ \frac{d}{dx}\left(x^n\right) = nx^{n-1}, \qquad \frac{d}{dx}\left(e^x\right) = e^x, \qquad \frac{d}{dx}\left(\ln x\right) = \frac1x, \qquad \frac{d}{dx}(k) = 0 \ \ (k \text{ a constant}). \] Also, the derivative of a sum is the sum of the derivatives, and constants multiplied in front just come along for the ride: $\frac{d}{dx}\big(5x^2\big) = 5\cdot 2x = 10x$.

Example Find a function whose derivative is $3x^2+2$.

We know $\frac{d}{dx}(x^3) = 3x^2$ and $\frac{d}{dx}(2x) = 2$, so one answer is

\[ x^3 + 2x. \]

Is there more than one possible answer? Yes! $x^3+2x+1$ and $x^3+2x-5$ also have derivative $3x^2+2$, because the derivative of a constant is $0$. In general, every function of the form

\[ x^3 + 2x + C \qquad (C \text{ any constant}) \]

works.

Definition A function $F$ is an antiderivative of the function $f$ if \[ F'(x) = f(x). \]

For example, the derivative of $x^2$ is $2x$, so an antiderivative of $2x$ is $x^2$. Remember that "constants added go away" when we take the derivative:

\[ \frac{d}{dx}\left(x^2+1\right) = 2x, \qquad \frac{d}{dx}\left(x^2+3\right) = 2x, \qquad \frac{d}{dx}\left(x^2+56\right) = 2x. \]

So the general antiderivative of $2x$ is $x^2 + C$. Graphically, all of these antiderivatives are the same curve shifted up or down — shifting a graph vertically doesn't change its slope anywhere.

Three antiderivatives of $2x$: $x^2-2$, $x^2$, and $x^2+2$. They all have the same slope at every $x$.

Integral Notation

Definition (Indefinite Integral) We use integral notation to mean "find the (general) antiderivative": \[ \int f(x)\,dx = F(x) + C, \qquad\text{where } F'(x) = f(x). \] The symbol $\int$ is the integral sign, $f(x)$ is the integrand, the $dx$ tells us the variable is $x$, and $C$ is called the constant of integration. The result $F(x)+C$ is called the general antiderivative or indefinite integral.

Example \[ \int 2x\,dx = x^2 + C. \]

Basic Rules

Each rule below comes from reading a derivative rule backwards. For the power rule: to differentiate $x^n$ we multiply by the power and subtract 1; to antidifferentiate we do the opposite in reverse order: add 1 to the power, then divide by the new power.

General Rules for Antiderivatives \[ \begin{aligned} \int x^n\,dx &= \frac{x^{n+1}}{n+1} + C \qquad (n\neq -1) &&\text{(power rule)}\\ \int \frac1x\,dx &= \ln|x| + C\\ \int e^x\,dx &= e^x + C\\ \int k\,dx &= kx + C \qquad (k \text{ a constant})\\ \int \big(f(x)\pm g(x)\big)\,dx &= \int f(x)\,dx \pm \int g(x)\,dx &&\text{(sums/differences split up)}\\ \int c\,f(x)\,dx &= c\int f(x)\,dx &&\text{(constants pull out front)} \end{aligned} \]

Remark (Why $n\neq-1$? Why $|x|$?) If $n=-1$, the power rule would give $\frac{x^0}{0}$, dividing by zero. That's why $\frac1x = x^{-1}$ gets its own rule: $\ln|x|$. The absolute value is there because $\frac1x$ makes sense for negative $x$, but $\ln x$ does not; $\ln|x|$ works for all $x\neq 0$ and still has derivative $\frac1x$.

Remark (Check Your Work) You can always check an antiderivative by taking the derivative of your answer. If you get back the integrand, you're right.

Before You Integrate: Rewrite as Powers

Rewrite roots as fractional exponents: $\sqrt x = x^{1/2}$, $\sqrt[3]{x} = x^{1/3}$.
Rewrite $x$'s in the denominator as negative exponents: $\frac{6}{x^2} = 6x^{-2}$. (But leave $\frac1x$ alone — it becomes $\ln|x|$.)
Multiply out products and split up fractions. There is no product rule for antiderivatives: $\int (x+1)(x-1)\,dx$ is not $\int(x+1)\,dx\cdot\int(x-1)\,dx$.

Example Find the general antiderivatives below, then check your work.

\[ \int x^3\,dx = \frac{x^4}{4} + C. \qquad \text{Check: } \frac{d}{dx}\left(\tfrac14x^4 + C\right) = \tfrac44x^3 = x^3. \ ✓ \]
\[ \int \left(x^4 + 5x\right)dx = \frac{x^5}{5} + 5\cdot\frac{x^2}{2} + C = \frac15x^5 + \frac52x^2 + C. \] Check: $\frac{d}{dx}\left(\frac15x^5+\frac52x^2+C\right) = x^4 + 5x$. ✓
\[ \int\left(\frac1x + e^x\right)dx = \ln|x| + e^x + C. \qquad \text{Check: } \frac{d}{dx}\left(\ln|x|+e^x+C\right) = \frac1x + e^x. \ ✓ \]
Rewrite first: $\dfrac4x = 4\cdot\dfrac1x$ and $\dfrac{6}{x^2} = 6x^{-2}$. For the second term, add 1 to the power ($-2+1 = -1$) and divide by the new power: \[ \int\left(\frac4x + \frac{6}{x^2}\right)dx = \int\left(4\cdot\frac1x + 6x^{-2}\right)dx = 4\ln|x| + \frac{6x^{-1}}{-1} + C = 4\ln|x| - 6x^{-1} + C. \] Check: $\frac{d}{dx}\left(4\ln|x| - 6x^{-1}+C\right) = \frac4x + 6x^{-2}$. ✓
Multiply out first: \[ \int (x+1)(x-1)\,dx = \int\left(x^2-1\right)dx = \frac{x^3}{3} - x + C. \] Check: $\frac{d}{dx}\left(\frac{x^3}3 - x + C\right) = x^2 - 1 = (x+1)(x-1)$. ✓

Self-Check

From Activity 29: Antiderivatives/ Indefinite Integrals

(a) Compute the following antiderivatives:

\[ \text{(i)}\ \int \left(x^2+2x+1\right)dx \qquad\qquad \text{(ii)}\ \int\left(e^x - 7\sqrt x\right)dx \qquad\qquad \text{(iii)}\ \int\left(\sqrt[3]{x} - \frac2x\right)dx \]

(b) Knowing that $\frac{d}{dx}\left(e^{ax}\right) = ae^{ax}$ by the chain rule when $a$ is a constant, what might you guess is $\displaystyle\int e^{ax}\,dx$?

Approximating Areas

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Goals Use sums of rectangular areas to approximate the area under a curve.

The Idea

We want to find the area under the curve $y = f(x)$ (and above the $x$-axis) from $x = a$ to $x = b$. Curved regions are hard to measure, but rectangles are easy: area $=$ base $\times$ height. So we chop the region into thin vertical strips and replace each strip with a rectangle. Two questions come up:

How many rectangles? We call this number $n$.
How tall should each rectangle be? We use the height of the curve at one point of each strip — usually the left endpoint or the right endpoint of the strip.

Definition (Riemann Sum) Divide $[a,b]$ into $n$ equal pieces, each of width \[ \Delta x = \frac{b-a}{n}. \] A Riemann sum (or rectangular approximation) adds up the areas of the $n$ rectangles: $\sum (\text{base})(\text{height}) = \Delta x\cdot f(x_1) + \Delta x\cdot f(x_2) + \cdots + \Delta x\cdot f(x_n)$, where $x_1,\dots,x_n$ are the chosen points.

$R_n$: $n$ rectangles, heights from the right endpoints.
$L_n$: $n$ rectangles, heights from the left endpoints.

Computing $L_n$ or $R_n$

Find the width $\Delta x = \frac{b-a}{n}$.
List the endpoints: start at $a$ and keep adding $\Delta x$ until you reach $b$.
For $L_n$, use every endpoint except the last one ($b$). For $R_n$, use every endpoint except the first one ($a$).
Add up the function values at those points and multiply by $\Delta x$: $\ \Delta x\big(f(x_1)+\cdots+f(x_n)\big)$.

Example Estimate the area under $f(x) = x^2$ from $x = 1$ to $x=9$ using 4 rectangles and right endpoints ($R_4$).

1. Width: $\Delta x = \frac{9-1}{4} = 2$.

2. Endpoints: $1, 3, 5, 7, 9$. Right endpoints: $3, 5, 7, 9$.

3. Heights: $f(3) = 9,\ f(5) = 25,\ f(7) = 49,\ f(9) = 81$.

4. Add the rectangle areas:

\[ R_4 = 2(9) + 2(25) + 2(49) + 2(81) = 18 + 50 + 98 + 162 = 328. \]

$R_4$ for $f(x) = x^2$ on $[1,9]$. Each rectangle's top-right corner touches the curve. The numbers are the rectangle areas.

Notice the rectangles stick out above the curve, so 328 is an overestimate.

Example Estimate the area under $f(x) = e^x + 3x$ from $x=0$ to $x = 1.5$ with 3 rectangles and right endpoints ($R_3$).

Width: $\Delta x = \frac{1.5-0}{3} = 0.5$. Endpoints: $0,\ 0.5,\ 1,\ 1.5$, so the right endpoints are $0.5,\ 1,\ 1.5$.

\[ R_3 = \tfrac12 f\!\left(\tfrac12\right) + \tfrac12 f(1) + \tfrac12 f(1.5) = \tfrac12\Big(f(0.5) + f(1) + f(1.5)\Big). \]

$x$	$0.5$	$1$	$1.5$
$f(x) = e^x+3x$	$e^{0.5}+1.5\approx 3.1487$	$e+3\approx 5.7183$	$e^{1.5}+4.5\approx 8.9817$

\[ R_3 \approx \tfrac12(3.1487 + 5.7183 + 8.9817) = \tfrac12(17.8487) \approx 8.9243. \]

Example Estimate the area under $f(x) = x^3+2$ from $x = 3$ to $x = 11$ with 4 rectangles and left endpoints ($L_4$).

Width: $\Delta x = \frac{11-3}{4} = \frac84 = 2$. Endpoints: $3, 5, 7, 9, 11$. Left endpoints: $3, 5, 7, 9$ (we do not use 11).

\[ L_4 = 2f(3) + 2f(5) + 2f(7) + 2f(9) = 2\big(29 + 127 + 345 + 731\big) = 2(1232) = 2464. \]

$L_4$ for $f(x) = x^3+2$ on $[3,11]$. Each rectangle's top-left corner touches the curve.

This time the rectangles sit below the curve, so 2464 is an underestimate.

Remark

We focus on left and right endpoints, but you can also use midpoints or any other point in each strip.
If $f$ is increasing, left endpoints give an underestimate and right endpoints give an overestimate. If $f$ is decreasing, it's the other way around.
Using more rectangles (larger $n$) gives a better estimate. Using the Fundamental Theorem of Calculus later in this unit, we will see that the exact area under $x^2$ on $[1,9]$ is $\frac{728}{3}\approx 242.67$, so our $R_4 = 328$ was quite rough; for comparison $L_4 = 2(1+9+25+49) = 168$ is too small.

Self-Check

Practice in the style of the lecture examples (there is no in-class activity for this section)

(a) Estimate the area under $f(x) = x^2+1$ from $x = 0$ to $x = 4$ using 4 rectangles, first with left endpoints ($L_4$) and then with right endpoints ($R_4$).

(b) In part (a), which estimate is too big and which is too small? Why?

(c) Estimate the area under $f(x) = \sqrt{x}$ from $x = 0$ to $x = 6$ using 3 rectangles and right endpoints ($R_3$). Round to two decimal places.

\(x\)	0	1	2	3	4
\(f(x) = x^2+1\)	1	2	5	10	17

The Definite Integral

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Goals State the definition of a definite integral in terms of net area; see the connection between Riemann sums and the definite integral; find the value of definite integrals using geometry.

Notation

When approximating areas, we estimated the area under a curve with rectangles. Now we want the exact (signed) area.

Definition (Definite Integral) We write \[ \int_a^b f(x)\,dx \] (read "the definite integral from $x = a$ to $x=b$ of $f(x)\,dx$") to mean the signed area between the curve $y = f(x)$ and the $x$-axis from $x=a$ to $x=b$. The numbers $a$ and $b$ are the lower and upper limits (or bounds) of integration, and $dx$ means "with respect to $x$."

Notice this notation looks a lot like the indefinite integral $\int 2x\,dx$ (take the antiderivative). That is not an accident — the reason is the Fundamental Theorem of Calculus, coming up next. Keep the difference in mind: an indefinite integral $\int f(x)\,dx$ is a function (plus $C$), while a definite integral $\int_a^b f(x)\,dx$ is a number.

From Estimate to Exact

As we use more and more rectangles, the leftover slivers between the rectangles and the curve get thinner and thinner, and the Riemann sum gets closer to the true area. To get the exact area, we take the limit as the number of rectangles goes to infinity:

Definition \[ \int_a^b f(x)\,dx = \lim_{n\to\infty} R_n = \lim_{n\to\infty} L_n. \]

Signed Area

What happens if the function is negative? In a Riemann sum, each rectangle contributes $(\text{base})\cdot f(x_i)$. If $f(x_i)$ is negative, that term is negative. So the definite integral counts area above the $x$-axis as positive and area below the $x$-axis as negative. This is called signed area or net area:

\[ \int_a^b f(x)\,dx = (\text{area above the } x\text{-axis}) - (\text{area below the } x\text{-axis}). \]

Using Geometry

Soon we will use antiderivatives to find these areas. For now, when the graph is made of straight lines, we can find exact values using area formulas from geometry:

\[ \text{Rectangle: } A = bh, \qquad \text{Triangle: } A = \tfrac12 bh. \]

Example What is $\displaystyle\int_0^3 x\,dx$?

This is the area under the line $y=x$ from $x=0$ to $x=3$. That region is a triangle with base 3 and height 3 (since $y = 3$ when $x=3$):

\[ \int_0^3 x\,dx = \tfrac12\cdot 3\cdot 3 = \tfrac92 = 4.5. \]

Example Let $f(x) = 10-2x$, graphed below. Find each of the following using geometry.

$f(x) = 10-2x$. It crosses the $x$-axis at $x = 5$: above the axis to the left, below to the right.

$\displaystyle\int_0^5 f(x)\,dx$: a triangle with base 5 and height $f(0) = 10$, above the axis: \[ \int_0^5 f(x)\,dx = \tfrac12(5)(10) = 25. \]
$\displaystyle\int_2^5 f(x)\,dx$: a triangle with base $5-2 = 3$ and height $f(2) = 10 - 4 = 6$: \[ \int_2^5 f(x)\,dx = \tfrac12(3)(6) = 9. \]
$\displaystyle\int_0^8 f(x)\,dx$: the big triangle above from 0 to 5, minus a triangle below from 5 to 8 with base 3 and height $|f(8)| = |10-16| = 6$: \[ \int_0^8 f(x)\,dx = 25 - \tfrac12(3)(6) = 25 - 9 = 16. \]
$\displaystyle\int_0^{10} f(x)\,dx$: the triangle above (area 25) and the triangle below from 5 to 10 (base 5, height $|f(10)| = 10$, area 25) cancel: \[ \int_0^{10} f(x)\,dx = \tfrac12(5)(10) - \tfrac12(5)(10) = 0. \] The area above equals the area below.

Example The graph of $f$ below is made of line segments. Find $\int_0^3 f(x)\,dx$, $\int_1^6 f(x)\,dx$, and $\int_0^8 f(x)\,dx$.

The graph of $f$. Shaded pieces are labeled with their (unsigned) areas; the pieces from 0 to 1 and from 6 to 8 are below the axis.

Break the region into simple shapes:

$0$ to $1$: triangle below the axis, base 1, height 6: area $\frac12(1)(6) = 3$, counts as $-3$.
$1$ to $2$: triangle above, base 1, height 6: $+3$.
$2$ to $4$: rectangle above, base 2, height 6: $+12$. (From 2 to 3 alone it is $1\cdot 6 = 6$.)
$4$ to $6$: triangle above, base 2, height 6: $+6$.
$6$ to $8$: triangle below, base 2, height 6: $-6$.

\[ \begin{aligned} \int_0^3 f(x)\,dx &= -3 + 3 + 6 = 6,\\ \int_1^6 f(x)\,dx &= 3 + 12 + 6 = 21,\\ \int_0^8 f(x)\,dx &= -3 + 3 + 12 + 6 - 6 = 12. \end{aligned} \]

Properties of the Definite Integral

Since a definite integral is an area, areas of neighboring pieces add up:

Properties \[ \begin{aligned} \int_a^b f(x)\,dx + \int_b^c f(x)\,dx &= \int_a^c f(x)\,dx &&\text{(adjacent intervals add)}\\ \int_a^a f(x)\,dx &= 0 &&\text{(no width, no area)}\\ \int_a^b \big(f(x)\pm g(x)\big)\,dx &= \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx\\ \int_a^b c\,f(x)\,dx &= c\int_a^b f(x)\,dx \end{aligned} \]

Example Suppose $\int_1^3 f(x)\,dx = 4$ and $\int_3^5 f(x)\,dx = 6$. Find $\int_1^5 f(x)\,dx$.

The area from 1 to 5 is the area from 1 to 3 plus the area from 3 to 5.

\[ \int_1^5 f(x)\,dx = \int_1^3 f(x)\,dx + \int_3^5 f(x)\,dx = 4 + 6 = 10. \]

Example Suppose $\int_2^8 f(x)\,dx = 3$ and $\int_5^8 f(x)\,dx = 1$. Find $\int_2^5 f(x)\,dx$.

The interval $[2,8]$ is split at 5 into $[2,5]$ and $[5,8]$:

\[ \int_2^5 f(x)\,dx + \int_5^8 f(x)\,dx = \int_2^8 f(x)\,dx \quad\Longrightarrow\quad \int_2^5 f(x)\,dx = 3 - 1 = 2. \]

Self-Check

Practice in the style of the lecture examples (there is no in-class activity for this section)

The graph of $f$ below is made of line segments.

The graph of $f$: through $(0,4)$, $(2,4)$, $(4,0)$, $(6,-2)$, $(8,-2)$.

(a) Use geometry to find $\displaystyle\int_0^2 f(x)\,dx$, $\ \displaystyle\int_0^4 f(x)\,dx$, $\ \displaystyle\int_4^8 f(x)\,dx$, and $\ \displaystyle\int_0^8 f(x)\,dx$.

(b) A different function $g$ has $\int_1^4 g(x)\,dx = 7$ and $\int_1^9 g(x)\,dx = 2$. Find $\int_4^9 g(x)\,dx$.

The Fundamental Theorem of Calculus

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Goals Use the Fundamental Theorem of Calculus to evaluate definite integrals.

Connecting Antiderivatives and Area

We have seen how to find antiderivatives and how to find area under a curve. Now let's connect these ideas.

Example (Speed and Distance) A person's speed is graphed over time. Find the distance traveled.

Constant speed $v(t) = 5$ mph for $t=0$ to $t=3$ hours. Distance $=$ rate $\times$ time $= 5\cdot 3 = 15$ miles. On the graph, this is the area of the $5\times 3$ rectangle under $v(t)$.
Speed $6$ mph for the first hour, $5$ mph for the second hour, and $4$ mph for the third. Distance $= 6\cdot1 + 5\cdot 1 + 4\cdot 1 = 15$ miles — again the area under the graph (three rectangles).
Speed $v(t) = 5-t$ from $t=0$ to $t=5$. The speed changes constantly, but distance is still the area under the graph, here a triangle: \[ \text{distance} = \tfrac12(5)(5) = 12.5 \text{ miles}. \]

Alternate idea. Velocity is the derivative of position: $v(t) = p'(t)$. So position is an antiderivative of velocity. In (a), $p'(t) = 5$ gives $p(t) = 5t + C$, and the distance traveled is

\[ p(3) - p(0) = (15 + C) - (0 + C) = 15. \]

In (c), $p'(t) = 5-t$ gives $p(t) = 5t - \frac{t^2}{2} + C$, and

\[ p(5) - p(0) = \left(25 - \tfrac{25}{2} + C\right) - (0 + C) = 12.5. \]

Same answers! The area under the velocity curve equals the change in an antiderivative. Notice also that the $C$'s cancel.

Fundamental Theorem of Calculus (Part 2) If $F$ is any antiderivative of $f$ (that is, $F' = f$), then \[ \int_a^b f(x)\,dx = F(b) - F(a). \] In words: the signed area under $f$ from $x=a$ to $x = b$ is the antiderivative evaluated at the top minus the antiderivative evaluated at the bottom.

Notation (Evaluation Bar) \[ F(x)\Big|_a^b = F(b) - F(a). \]

Remark (No $+C$ Needed) For a definite integral we can skip the $+C$: whatever constant we pick, it shows up in both $F(b)$ and $F(a)$ and cancels when we subtract.

Evaluating $\int_a^b f(x)\,dx$

Rewrite the integrand as powers if needed (roots $\to$ fractional exponents, multiply out products, split fractions).
Find an antiderivative $F(x)$ and write $F(x)\Big|_a^b$.
Plug in the top bound, plug in the bottom bound, and subtract: $F(b) - F(a)$. Use parentheses around each — the minus sign applies to all of $F(a)$.

Example \[ \int_0^1 x\,dx = \frac{x^2}{2}\Big|_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac12. \]

This matches geometry: the region is a triangle with base 1 and height 1.

Example \[ \begin{aligned} \int_2^3 \left(x^2+1\right)dx &= \left(\frac{x^3}{3} + x\right)\Big|_2^3 = \left(\frac{3^3}{3} + 3\right) - \left(\frac{2^3}{3} + 2\right)\\ &= (9 + 3) - \left(\frac83 + 2\right) = 12 - \frac{14}{3} = \frac{22}{3}. \end{aligned} \]

(On an exam it is fine to stop after the line where the numbers are plugged in.)

Example \[ \begin{aligned} \int_{-2}^2 \left(t^2-4\right)dt &= \left(\frac{t^3}{3} - 4t\right)\Big|_{-2}^2 = \left(\frac{2^3}{3} - 4\cdot 2\right) - \left(\frac{(-2)^3}{3} - 4(-2)\right)\\ &= \left(\frac83 - 8\right) - \left(-\frac83 + 8\right) = \frac83 - 8 + \frac83 - 8 = \frac{16}{3} - 16 = -\frac{32}{3}. \end{aligned} \]

Be careful distributing the minus sign over the second parentheses. The answer is negative because $t^2 - 4 \le 0$ on $[-2,2]$: the whole region is below the axis.

Example Evaluate $\displaystyle\int_1^9 \frac{x-1}{\sqrt x}\,dx$.

First split the fraction and rewrite as powers: $\dfrac{x-1}{\sqrt x} = \dfrac{x}{x^{1/2}} - \dfrac{1}{x^{1/2}} = x^{1/2} - x^{-1/2}$. The new powers are $\frac12 + 1 = \frac32$ and $-\frac12 + 1 = \frac12$.

\[ \begin{aligned} \int_1^9 \left(x^{1/2} - x^{-1/2}\right)dx &= \left(\frac{x^{3/2}}{3/2} - \frac{x^{1/2}}{1/2}\right)\Big|_1^9 = \left(\frac23x^{3/2} - 2x^{1/2}\right)\Big|_1^9\\ &= \left(\frac23\cdot 9^{3/2} - 2\cdot 9^{1/2}\right) - \left(\frac23\cdot 1^{3/2} - 2\cdot1^{1/2}\right)\\ &= \left(\frac23(27) - 2\cdot 3\right) - \left(\frac23 - 2\right) = 12 - \frac23 + 2 = \frac{40}{3}. \end{aligned} \]

(Recall $9^{3/2} = \left(\sqrt9\right)^3 = 27$.)

More Practice from Lecture

Example

\[ \int_0^3 \left(x^2 + \sqrt x\right)dx = \left(\frac13x^3 + \frac23x^{3/2}\right)\Big|_0^3 = \left(\frac13(3)^3 + \frac23(3)^{3/2}\right) - 0 = 9 + \frac23\cdot3^{3/2} = 9 + 2\sqrt3. \] (Since $3^{3/2} = 3\sqrt3$, $\frac23\cdot 3\sqrt3 = 2\sqrt3$. Numerically $\approx 12.46$.)
\[ \int_1^5 \left(e^x + \frac1x\right)dx = \Big(e^x + \ln|x|\Big)\Big|_1^5 = \left(e^5 + \ln 5\right) - \left(e^1 + \ln 1\right) = e^5 + \ln 5 - e \approx 147.30, \] using $\ln 1 = 0$.
Multiply out first: $(x^2+1)(4x) = 4x^3 + 4x$. \[ \int_3^8 (x^2+1)(4x)\,dx = \int_3^8 \left(4x^3+4x\right)dx = \left(x^4 + 2x^2\right)\Big|_3^8 = \left(8^4 + 2\cdot 8^2\right) - \left(3^4 + 2\cdot 3^2\right) = 4224 - 99 = 4125. \]

Applications: Total Change from a Rate

The theorem says: integrating a rate of change $F'$ gives the total change in $F$,

\[ \int_a^b F'(t)\,dt = F(b) - F(a). \]

So if we know a marginal function (marginal revenue $R'$, marginal cost $C'$, ...), integrating it over an interval gives the total revenue, cost, etc. accumulated over that interval.

Example The marginal revenue for an item is $R'(t) = -t^2 + 36t$, where $t$ is in weeks. What is the revenue collected over the first four weeks?

Method 1 (find $R$). Antidifferentiate: $R(t) = -\frac{t^3}{3} + 18t^2 + C$. At time 0 nothing has been collected yet, so $R(0) = 0$, which forces $C = 0$. Then

\[ R(4) = -\frac{4^3}{3} + 18(4)^2 = -\frac{64}{3} + 288 = \frac{800}{3}\approx 266.67. \]

Method 2 (definite integral).

\[ \int_0^4 R'(t)\,dt = R(t)\Big|_0^4 = R(4) - R(0) = \frac{800}{3}\approx 266.67. \]

Example The marginal cost for making bagels is $C'(t) = 10\sqrt t$ dollars per day. What are the costs for the second week?

The first week is days 0 to 7, so the second week is from $t = 7$ to $t = 14$. Rewrite $10\sqrt t = 10t^{1/2}$, whose antiderivative is $10\cdot\frac{t^{3/2}}{3/2} = \frac{20}{3}t^{3/2}$:

\[ \text{Cost} = \int_7^{14} 10\sqrt t\,dt = \frac{20}{3}t^{3/2}\Big|_7^{14} = \frac{20}{3}\left(14^{3/2} - 7^{3/2}\right)\approx \frac{20}{3}(52.383 - 18.520) \approx 225.75. \]

The costs for the second week are about $225.75.

Self-Check

Practice in the style of the lecture examples (there is no in-class activity for this section)

Evaluate:

\[ \text{(a)}\ \int_1^4 \left(3x^2 - 2x\right)dx \qquad\qquad \text{(b)}\ \int_1^e \left(\frac2x + 1\right)dx \qquad\qquad \text{(c)}\ \int_0^4 \sqrt x\,(x+1)\,dx \]

(d) The marginal cost of producing $x$ items is $C'(x) = 50 - 0.2x$ dollars per item. How much does the total cost increase when production goes from 100 items to 200 items?

Substitution

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Goals Use the method of substitution ($u$-substitution) to evaluate definite and indefinite integrals.

Warm-Up: The Chain Rule

Recall the chain rule: to differentiate a function of an "inside" function, take the derivative of the outside (leaving the inside alone), then multiply by the derivative of the inside.

$f(x)$	$f'(x)$
$(3x+4)^6$	$6(3x+4)^5\cdot 3 = 18(3x+4)^5$
$(8x^3+6x)^5$	$5(8x^3+6x)^4(24x^2+6) = 30(8x^3+6x)^4(4x^2+1)$
$\frac18 e^{4x^2+3}$	$\frac18 e^{4x^2+3}\cdot 8x = xe^{4x^2+3}$
$\ln(8x+3x^2+5)$	$\dfrac{8+6x}{8x+3x^2+5}$

What do we see in these derivatives? Each one looks like

\[ f(\text{inside})\cdot(\text{derivative of inside}), \]

where the derivative of the inside may be off by a constant multiple (like the $4x^2+1$ instead of $24x^2+6$, or the $x$ instead of $8x$).

The Method of Substitution

The method of substitution is used to take the antiderivative of functions where the chain rule was used for the derivative. We look for integrals of the form

\[ \int f(\text{inside})\cdot(\text{derivative of inside})\,dx, \]

where the derivative of the inside only needs to be there up to multiplication by a constant.

$u$-Substitution

Let the inside be $u$.
Find $du$: take the derivative, $\frac{du}{dx} = u'$, and write $du = u'\,dx$.
Solve for $dx$: $\ dx = \frac{du}{u'}$.
Plug back in and cancel all the $x$'s. (If some $x$'s won't cancel, try a different $u$.)
Take the antiderivative with respect to $u$.
Plug back in what $u$ was in terms of $x$.

Example Find $\displaystyle\int 18(3x+4)^5\,dx$.

1. Inside: $u = 3x+4$.

2. $\frac{du}{dx} = 3$, so $du = 3\,dx$.

3. $dx = \frac{du}{3}$.

4–6. Plug in, integrate, and plug back:

\[ \int 18u^5\cdot\frac{du}{3} = \int 6u^5\,du = u^6 + C = (3x+4)^6 + C. \]

This matches the warm-up: the derivative of $(3x+4)^6$ was $18(3x+4)^5$.

Example Find $\displaystyle\int 5(8x^3+6x)^4(24x^2+6)\,dx$.

Let $u = 8x^3 + 6x$. Then $du = (24x^2+6)\,dx$, so $dx = \dfrac{du}{24x^2+6}$.

\[ \int 5u^4\,(24x^2+6)\cdot\frac{du}{24x^2+6} = \int 5u^4\,du = u^5 + C = (8x^3+6x)^5 + C. \]

Example Find $\displaystyle\int xe^{4x^2+3}\,dx$.

Let $u = 4x^2+3$. Then $du = 8x\,dx$, so $dx = \dfrac{du}{8x}$.

\[ \int x\,e^u\cdot\frac{du}{8x} = \int \frac18 e^u\,du = \frac18e^u + C = \frac18e^{4x^2+3} + C. \]

Here the derivative of the inside ($8x$) was only present up to a constant (we had $x$); the leftover $\frac18$ just comes out front.

Example Find $\displaystyle\int\frac{3x+4}{8x+3x^2+5}\,dx$.

Let $u = 8x + 3x^2+5$ (the denominator). Then $du = (8+6x)\,dx = 2(3x+4)\,dx$, so $dx = \dfrac{du}{2(3x+4)}$.

\[ \int\frac{3x+4}{u}\cdot\frac{du}{2(3x+4)} = \frac12\int\frac1u\,du = \frac12\ln|u| + C = \frac12\ln\left|8x+3x^2+5\right| + C. \]

Substitution with Definite Integrals

Same basic idea, but we have to be careful about the bounds: the original bounds are $x$-values, not $u$-values. We can either

change the bounds into $u$-values (and then never change back to $x$), or
keep the bounds as $x$-values, writing "$x = $" on them all the way through, and change back to $x$ before plugging in.

Example (Change Bounds) Evaluate $\displaystyle\int_1^3 xe^{x^2+1}\,dx$.

Let $u = x^2+1$, so $du = 2x\,dx$ and $dx = \dfrac{du}{2x}$. Convert the bounds:

\[ x = 1 \ \Rightarrow\ u = 1^2+1 = 2, \qquad x = 3\ \Rightarrow\ u = 3^2 + 1 = 10. \] \[ \int_{u=2}^{u=10} x\,e^u\cdot\frac{du}{2x} = \int_2^{10}\frac12e^u\,du = \frac12e^u\Big|_2^{10} = \frac12e^{10} - \frac12e^2. \]

Example (Keep Bounds) Same integral, same $u = x^2+1$, $dx = \frac{du}{2x}$: \[ \int_{x=1}^{x=3}\frac12e^u\,du = \frac12e^u\Big|_{x=1}^{x=3} = \frac12e^{x^2+1}\Big|_{x=1}^{x=3} = \frac12e^{10} - \frac12e^2. \]

Which do you prefer? Both give the same answer.

Example Evaluate $\displaystyle\int_0^1\frac{x+2}{x^2+4x+1}\,dx$.

Let $u = x^2+4x+1$. Then $du = (2x+4)\,dx = 2(x+2)\,dx$, so $dx = \dfrac{du}{2(x+2)}$.

Changing bounds: $x=0\Rightarrow u = 1$ and $x = 1 \Rightarrow u = 1+4+1 = 6$.

\[ \int_1^6\frac{x+2}{u}\cdot\frac{du}{2(x+2)} = \int_1^6\frac12\cdot\frac1u\,du = \frac12\ln|u|\Big|_1^6 = \frac12\ln 6 - \frac12\ln 1 = \frac12\ln 6. \]

Keeping bounds:

\[ \int_{x=0}^{x=1}\frac12\cdot\frac1u\,du = \frac12\ln|u|\Big|_{x=0}^{x=1} = \frac12\ln\left|x^2+4x+1\right|\Big|_{x=0}^{x=1} = \frac12\ln 6 - \frac12\ln 1 = \frac12\ln 6. \]

Example Evaluate $\displaystyle\int_5^8\left(e^{8x} - \frac1x\right)dx$.

Split it into two integrals. Only the first needs substitution; the second we can do directly:

\[ \int_5^8\left(e^{8x}-\frac1x\right)dx = \int_5^8 e^{8x}\,dx - \int_5^8\frac1x\,dx. \]

First integral: $u = 8x$, $du = 8\,dx$, $dx = \frac{du}{8}$; bounds $x = 5\Rightarrow u = 40$, $x=8\Rightarrow u = 64$.

\[ \int_{40}^{64}e^u\,\frac{du}{8} = \frac18e^u\Big|_{40}^{64} = \frac18e^{64} - \frac18e^{40}. \]

Second integral: $\displaystyle\int_5^8\frac1x\,dx = \ln|x|\Big|_5^8 = \ln 8 - \ln 5$. Altogether,

\[ \int_5^8\left(e^{8x}-\frac1x\right)dx = \left(\frac18e^{64} - \frac18e^{40}\right) - \left(\ln 8 - \ln 5\right). \]

Self-Check

From the Final Exam Review (part a), plus practice in the style of the lecture examples (there is no in-class activity for this section)

(a) Evaluate $\displaystyle\int_3^8\frac{5e^x + \frac5x}{\left(e^x + \ln x\right)^3}\,dx$.

(b) Find $\displaystyle\int x^2\left(x^3+1\right)^4dx$.

(c) Evaluate $\displaystyle\int_0^1\frac{2x}{x^2+1}\,dx$.

Area Between Curves

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Goals Find the area between curves using integration; solve for bounds for more complicated regions.

Top Minus Bottom

We know how to find the area under a curve. Now we want to find the area between two curves.

Example Find the area between $y = x^2$ and $y = x$ (from $x = 0$ to $x = 1$, where they meet).

On $[0,1]$ the line $y=x$ is on top and the parabola $y = x^2$ is on the bottom.

The area under the top curve $y = x$ includes the shaded region plus the area under the bottom curve $y = x^2$. So subtract:

\[ \int_0^1 x\,dx - \int_0^1 x^2\,dx = \int_0^1\left(x - x^2\right)dx = \left(\frac{x^2}{2} - \frac{x^3}{3}\right)\Big|_0^1 = \frac12 - \frac13 = \frac16. \]

Area Between Curves If $f(x)\ge g(x)$ for all $x$ in $[a,b]$ (so $f$ is on top and $g$ is on the bottom), then the area between them from $x = a$ to $x = b$ is \[ \int_a^b\big(f(x) - g(x)\big)\,dx = \int_a^b(\text{top} - \text{bottom})\,dx. \]

Remark This formula still works when part of the region is below the $x$-axis: "top minus bottom" is always the (positive) height of the region. Unlike a plain definite integral, the area between curves is never negative.

Example Find the area between $y = x+1$ and $y = x^2-1$ from $x=0$ to $x = 1$.

On $[0,1]$, the line is on top (e.g. at $x = 0$: $1 > -1$).

\[ \begin{aligned} \int_0^1\Big((x+1) - (x^2-1)\Big)dx &= \int_0^1\left(x + 1 - x^2 + 1\right)dx = \int_0^1\left(x - x^2 + 2\right)dx\\ &= \left(\frac{x^2}{2} - \frac{x^3}{3} + 2x\right)\Big|_0^1 = \frac12 - \frac13 + 2 = \frac{13}{6}. \end{aligned} \]

Careful: distribute the minus sign to every term of the bottom function.

Finding the Bounds

When a problem asks for the area of the region bounded by two curves (with no interval given), the bounds are where the curves meet. To find them, set the two functions equal and solve.

Example Find the area of the region bounded between $y = x+1$ and $y = x^2-1$.

Points of intersection:

\[ x + 1 = x^2 - 1 \;\Longrightarrow\; x^2 - x - 2 = 0 \;\Longrightarrow\; (x+1)(x-2) = 0 \;\Longrightarrow\; x = -1,\ x = 2. \]

The region bounded by $y=x+1$ (top) and $y = x^2-1$ (bottom), between the intersection points $x=-1$ and $x = 2$.

\[ \begin{aligned} \int_{-1}^2\Big((x+1) - (x^2-1)\Big)dx &= \left(\frac{x^2}{2} - \frac{x^3}{3} + 2x\right)\Big|_{-1}^2\\ &= \left(\frac{2^2}{2} - \frac{2^3}{3} + 2(2)\right) - \left(\frac{1}{2} - \frac{(-1)^3}{3} + 2(-1)\right)\\ &= \left(2 - \frac83 + 4\right) - \left(\frac12 + \frac13 - 2\right) = \frac{10}{3} - \left(-\frac76\right) = \frac{20}{6} + \frac76 = \frac{27}{6} = \frac92. \end{aligned} \]

When the Curves Cross

If the curves cross inside the interval, the top and bottom curves switch. Then we must split the integral at each crossing point and use "top minus bottom" on each piece.

Area Between Curves That Cross

Set the functions equal to find where they cross.
Split $[a,b]$ at the crossing points that lie inside it.
On each piece, decide which function is on top (sketch, or plug in a test point).
Integrate top minus bottom on each piece, and add the results.

Example Find the area between $y = x^2-1$ and $y = x+5$ from $x=-5$ to $x=5$.

Where do they cross?

\[ x^2 - 1 = x + 5 \;\Longrightarrow\; x^2 - x - 6 = 0 \;\Longrightarrow\; (x+2)(x-3) = 0 \;\Longrightarrow\; x = -2,\ x = 3. \]

From $-5$ to $-2$ and from $3$ to $5$ the parabola is on top; from $-2$ to $3$ the line is on top.

Test points: at $x = -3$, $x^2-1 = 8 > 2 = x+5$; at $x = 0$, $x + 5 = 5 > -1$; at $x = 4$, $x^2 - 1 = 15 > 9$. So

\[ \text{Area} = \int_{-5}^{-2}\Big((x^2-1) - (x+5)\Big)dx + \int_{-2}^{3}\Big((x+5) - (x^2-1)\Big)dx + \int_3^5\Big((x^2-1)-(x+5)\Big)dx. \]

Each integrand is $\pm\left(x^2 - x - 6\right)$, and an antiderivative of $x^2 - x - 6$ is $F(x) = \frac{x^3}{3} - \frac{x^2}{2} - 6x$. We need

\[ F(-5) = -\frac{145}{6}, \qquad F(-2) = \frac{22}{3}, \qquad F(3) = -\frac{27}{2}, \qquad F(5) = -\frac56. \] \[ \begin{aligned} A_1 &= F(-2) - F(-5) = \frac{44}{6} + \frac{145}{6} = \frac{189}{6} = \frac{63}{2},\\ A_2 &= -\big(F(3) - F(-2)\big) = \frac{27}{2} + \frac{22}{3} = \frac{81}{6} + \frac{44}{6} = \frac{125}{6},\\ A_3 &= F(5) - F(3) = -\frac56 + \frac{27}{2} = -\frac56 + \frac{81}{6} = \frac{76}{6} = \frac{38}{3}. \end{aligned} \] \[ \text{Area} = \frac{189}{6} + \frac{125}{6} + \frac{76}{6} = \frac{390}{6} = 65. \]

Example Find the area between $y = -x$ and $y = 2-x^2$ on $[-2,2]$.

Crossings: $-x = 2 - x^2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x+1)(x-2) = 0$, so $x = -1$ and $x = 2$. Only $x=-1$ is strictly inside $[-2,2]$, so we split there.

On $[-2,-1]$ the line $y=-x$ is on top; on $[-1,2]$ the parabola $y = 2-x^2$ is on top.

On $[-2,-1]$ (test $x = -1.5$: $-x = 1.5$, $2-x^2 = -0.25$) the line is on top. On $[-1,2]$ (test $x=0$: $2 > 0$) the parabola is on top.

\[ \begin{aligned} \int_{-2}^{-1}\Big((-x) - (2-x^2)\Big)dx &= \left(-\frac{x^2}{2} - 2x + \frac{x^3}{3}\right)\Big|_{-2}^{-1} = \left(-\frac12 + 2 - \frac13\right) - \left(-2 + 4 - \frac83\right) = \frac76 + \frac23 = \frac{11}{6},\\ \int_{-1}^{2}\Big((2-x^2) - (-x)\Big)dx &= \left(2x - \frac{x^3}{3} + \frac{x^2}{2}\right)\Big|_{-1}^{2} = \left(4 - \frac83 + 2\right) - \left(-2 + \frac13 + \frac12\right) = \frac{10}{3} + \frac76 = \frac92. \end{aligned} \] \[ \text{Area} = \frac{11}{6} + \frac92 = \frac{11}{6} + \frac{27}{6} = \frac{38}{6} = \frac{19}{3}. \]

Self-Check

Practice in the style of the lecture examples (there is no in-class activity for this section)

(a) Find the area of the region bounded between $y = 2x$ and $y = x^2$.

(b) Find the area between $y = x$ and $y = x^3$ on $[-1,1]$. (Hint: where do they cross, and which is on top on each piece?)

\(x\)	\(0.5\)	\(1\)	\(1.5\)
\(f(x) = e^x+3x\)	\(e^{0.5}+1.5\approx 3.1487\)	\(e+3\approx 5.7183\)	\(e^{1.5}+4.5\approx 8.9817\)

\(f(x)\)	\(f'(x)\)
\((3x+4)^6\)	\(6(3x+4)^5\cdot 3 = 18(3x+4)^5\)
\((8x^3+6x)^5\)	\(5(8x^3+6x)^4(24x^2+6) = 30(8x^3+6x)^4(4x^2+1)\)
\(\frac18 e^{4x^2+3}\)	\(\frac18 e^{4x^2+3}\cdot 8x = xe^{4x^2+3}\)
\(\ln(8x+3x^2+5)\)	\(\dfrac{8+6x}{8x+3x^2+5}\)