## Section1.5Indefinite Integrals

In this section we focus on the indefinite integral: its definition, the differences between the definite and indefinite integrals, some basic integral rules, and how to compute a definite integral.

### Subsection1.5.1Defining the Indefinite Integral

Recall the definition of the antiderivative from Section 1.1: A function $F$ is an antiderivative of $f$ on an interval $I$ if $F'(x)=f(x)$ for all $x$ in $I\text{.}$ Let us explore the antiderivative concretely by letting $f(x)=2x\text{.}$ Then we can readily determine that the antiderivative of $f$ is the function $F(x)=x^2\text{,}$ i.e. $F'(x)=f(x)\text{.}$ However, the function $F(x)+1 = x^2+1$ also has $f$ as its derivative:

\begin{equation*} \frac{d}{dx}\left(F(x)+1\right) = \frac{d}{dx} \left(x^2+1\right) = 2x\text{.} \end{equation*}

In fact, any function $F(x)+C=x^2+C$ for some real constant $C$ has $f$ as its derivative:

\begin{equation*} \frac{d}{dx}\left(F(x)+C\right) = \frac{d}{dx} \left(x^2+C\right) = 2x\text{.} \end{equation*}

It comes as no surprise to us that the graphs of the family of functions $F(x)+C$ are visually just vertical displacements of $F(x)\text{.}$ In the particular case when $F(x)=x^2\text{,}$ we can also see with the graphs of the family of functions $F(x)+C$ below that at any point $x$ the tangent lines are parallel, i.e. the tangent slopes are the same, i.e. the family of functions has the same derivative $f(x)=2x\text{.}$

Interactive Demonstration. Drag any point to investigate the tangent lines of the function family $F(x) = x^2 + C \text{.}$

This leads us to the following result:

###### Definition1.28.

{General Antiderivative} If a function $F$ is an antiderivative of $f$ on an interval $I\text{,}$ then the most general antiderivative of $f$ on an interval $I$ is

\begin{equation*} F(x)+C \end{equation*}

where $C$ is any real constant.

Let us now define the indefinite integral.

###### Definition1.29. The Indefinite Integral.

The set of all antiderivatives of a function $f(x)$ is the indefinite integral of $f(x)$ with respect to $x$ and denoted by

\begin{equation*} \int f(x)\, dx\text{,} \end{equation*}

where

 $\ds\int f(x)\, dx$ is read “the integral of $f$ w.r.t. $x$” , the symbol $\ds \int$ is called the integral sign, the function $f$ is referred to as the integrand of the integral, and the variable $x$ is called the variable of integration.

Note:

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

2. The above definition says that if a function $F$ is an antiderivative of $f\text{,}$ then

\begin{equation*} \int f(x)\, dx = F(x)+C \end{equation*}

for some real constant $C\text{.}$

3. Unlike the definite integral, the indefinite integral is a function.

### Subsection1.5.2Definite Integral versus Indefinite Integral

Due to the close relationship between an integral and an antiderivative, the integral sign is also used to mean “antiderivative”. You can tell which is intended by whether the limits of integration are included:

\begin{equation*} \int_1^2 x^2\,dx \end{equation*}

is an ordinary integral, also called a definite integral, because it has a definite value, namely

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

We use

\begin{equation*} \int x^2\,dx \end{equation*}

to denote the antiderivative of $\ds x^2\text{,}$ also called an indefinite integral. So this is evaluated as

\begin{equation*} \int x^2\,dx = {x^3\over 3}+C\text{,} \end{equation*}

which is clearly a function as opposed to the definite integral which is a value. It is customary to include the constant $C$ to indicate that there are really an infinite number of antiderivatives. We do not need this $C$ to compute definite integrals, but in other circumstances we will need to remember that the $C$ is there, so it is best to get into the habit of writing the $C\text{.}$

### Subsection1.5.3Computing Indefinite Integrals

We are finally ready to compute some indefinite integrals and introduce some basic integration rules from our knowledge of derivatives. We will first point out some common mistakes frequently observed in student work.

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 f(x)\, dx \neq \int f(x) \end{equation*}
2. Forgetting the $+C$ during the solution process and thereby not showing that the solution of an indefinite integration process is the set of all antiderivatives of the integrand. As an aside, constants of integration play a major role in the topic of Differential Equations. For example:

\begin{equation*} \int x^2 dx \neq \frac{x^3}{3} \end{equation*}

Caution: Note that we don't have properties to deal with products or quotients of functions, that is,

\begin{equation*} \int f(x)\cdot g(x)\,dx\neq \int f(x)\,dx\int g(x)\,dx\text{.} \end{equation*}
\begin{equation*} \int \frac{f(x)}{g(x)}\,dx\neq \frac{\int f(x)\,dx}{\int g(x)\,dx}\text{.} \end{equation*}

With derivatives, we had the product and quotient rules to deal with these cases. For integrals, we have no such rules, but we will learn a variety of different techniques to deal with these cases.

The following integral rules can be proved by taking the derivative of the functions on the right side.

###### Integral Rules.

 Constant Rule: $\ds\int k\,dx=kx+C\text{.}$ Constant Multiple Rule: $\ds\int kf(x)\,dx=k\int f(x)\,dx\text{,}$ where $k$ constant. Sum/Difference Rule: $\ds\int f(x)\pm g(x)\,dx=\int f(x)\,dx\pm\int g(x)\,dx\text{.}$ Power Rule: $\ds\int x^n\,dx=\frac{x^{n+1}}{n+1}+C, n\neq -1\text{.}$ Log Rule: $\ds\int \frac{1}{x}\,dx=\ln|x|+C, x\neq 0\text{.}$ Exponent Rule: $\ds\int a^{kx}=\frac{a^{kx}}{k\ln a}+C, x\neq 0\text{.}$ Trig Rules: $\ds\int \sin x\,dx=-\cos x+C\text{,}$ $\ds\int \cos x\,dx = \sin x + C\text{,}$ $\ds\int \sec^2 x\,dx = \tan x + C\text{,}$ $\ds\int \sec x \tan x \,dx = \sec x + C\text{,}$ $\ds\int \csc^2 x \,dx = -\cot x + C\text{,}$ $\ds\int \csc x \cot x \,dx = \csc x + C\text{,}$ $\ds\int \frac{dx}{1+x^2} = \arctan x + C\text{,}$ $\ds\int \frac{dx}{\sqrt{1-x^2}} = \arcsin x + C\text{.}$

###### Example1.30. Indefinite Integral.

If $f'(x)=x^4+2x-8\sin x$ then what is $f(x)\text{?}$

Solution

\begin{equation*} \begin{array}{rcl} \ds{f(x)=\int f'(x)\,dx}\amp =\amp \ds{\int \left(x^4+2x-8\sin x\right)\,dx}\\ \\ \amp =\amp \ds{ \int x^4 \,dx + 2\int x\,dx -8 \int \sin x\,dx }\\ \\ \amp =\amp \ds{\frac{x^5}{5}+x^2+8\cos x+C,} \end{array} \end{equation*}

where $C$ is a constant.

###### Example1.31. Indefinite Integral.

Find the indefinite integral $\ds\int 3x^2\,dx\text{.}$

Solution
\begin{equation*} \begin{array}{rcl} \ds{\int 3x^2\,dx} \amp = \amp \ds{3\int x^2\,dx}\\ \\ \amp =\amp \ds{3\frac{x^3}{3}+C}\\ \\ \amp =\amp x^3+C \end{array} \end{equation*}
###### Example1.32. Indefinite Integral.

Find the indefinite integral $\ds\int \frac{2}{\sqrt x}\,dx\text{.}$

Solution
\begin{equation*} \begin{array}{rcl} \ds{\int \frac{2}{\sqrt x}\,dx} \amp = \amp \ds{2\int x^{-\frac{1}{2}}\,dx}\\ \\ \amp =\amp \ds{2\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C}\\ \\ \amp =\amp 4\sqrt x+C \end{array} \end{equation*}

Note: There is an implicit (built-in) restriction in the above calculations, namely $x \geq 0\text{.}$ This comes from the fact that the integrand $\frac{2}{\sqrt{x}}$ is only defined for $x \geq 0\text{,}$ and, not surprisingly, the indefinite integral is also only defined over this interval.

###### Example1.33. Indefinite Integral.

Find the indefinite integral $\ds\int \left(\frac{1}{x}+e^{7x}+x^\pi+7\right)\,dx\text{.}$

Solution
\begin{equation*} \begin{array}{rcl} \ds{\int \left(\frac{1}{x}+e^{7x}+x^\pi+7\right)\,dx} \amp = \amp \ds{\int \frac{1}{x}\,dx+\int e^{7x}\,dx+\int x^\pi\,dx+\int 7\,dx}\\ \\ \amp =\amp \ds{\ln|x|+\frac{1}{7}e^{7x}+\frac{x^{\pi+1}}{\pi+1}+7x+C} \end{array} \end{equation*}

Note: Just like in the previous example, the rule for integration $\frac{1}{x}$ also comes with an implicit restriction, namely $x \neq 0\text{.}$ The function $\frac{1}{x}$ clearly has a well defined area, as long as you stick to values of $x \lt 0$ or $x > 0$ and is undefined for $x=0\text{,}$ but $\ln x$ is undefined for $x \leq 0\text{.}$ However, we know from differential calculus that $\frac{d}{dx}\ln |x| = \frac{1}{x}$ (try this derivative again to convince yourself by rewriting this function as a peicewise-defined function). Hence, the antiderivative of $\frac{1}{x}$ becomes $\ln|x|\text{.}$

###### Example1.34. Finding Cost Functions.

Suppose a publishing company has found that the marginal cost at a level of production of $x$ thousand magazines is given by

\begin{equation*} C'(x) = \frac{25}{\sqrt{x}} \end{equation*}

and that the fixed cost, i.e. the cost before the first book can be produced, is 36,000. Find the cost function $C(x)\text{.}$ Solution By the indefinite integral rules \begin{equation*} \int \frac{25}{\sqrt{x}} dx = \int 25 x^{-1/2} dx = 25(2x^{1/2})+k = 50x^{1/2}+k\text{,} \end{equation*} where $k$ represents the constant of integration to avoid confusion with the cost function. Notice that the production $x$ is always non-negative, and so we proceed with the integration with the implicit assumption that $x \geq 0$ is automatically satisfied. To find the value of $k\text{,}$ use the fact that $C(0)$ is $36,000\text{.}$ \begin{equation*} \begin{split} C(x) \amp = 50x^{1/2} + k \\ 36,000 \amp = 50 \cdot 0 + k \\ k \amp = 36,000 \end{split} \end{equation*} With this result, the cost function is $C(x)=50x^{1/2}+36,000\text{.}$ ###### Example1.35. Finding Revenue and Demand Functions. Suppose the marginal revenue from a product is given by \begin{equation*} 800e^{-0.2}+7.5\text{.} \end{equation*} 1. Find the revenue function for this product. 2. Find the demand function for this product. Solution 1. The marginal revenue is the derivative of the revenue function, so \begin{equation*} \begin{split} R'(q) \amp = 800e^{-0.2q}+7.5 \\ R(q) \amp = \int \left(800e^{-0.2q}+7.5\right)dq \\ \amp = 800 \frac{e^{-0.2q}}{-0.2} + 7.5q + C \\ \amp = -4000e^{-0.2q} + 7.5q + C \end{split} \end{equation*} If no items are sold, then there is no revenue. Hence, $q=0$ and $R=0\text{,}$ and so \begin{equation*} \begin{split} 0 \amp = -4000e^{-0.2(0)} + 7.5(0) + C \\ 0 \amp = -4000 + C \\ C \amp = 4000 \end{split} \end{equation*} Therefore, the revenue function is \begin{equation*} R(q) = -4000e^{-0.2q}+7.5q+4000\text{.} \end{equation*} 2. Recall that $R=qp\text{,}$ where $p$ is the demand function that represents the price $p$ as a function of $q\text{.}$ So \begin{equation*} \begin{split} -4000e^{-0.2q}+7.5q+4000 \amp = qp \\ \frac{-4000e^{-0.2q}+7.5q+4000 }{q} \amp = p \end{split} \end{equation*} Therefore, the demand function is \begin{equation*} p(q) = \frac{-4000e^{-0.2q}+7.5q+4000 }{q}\text{.} \end{equation*} ### Subsection1.5.4Differential Equations and Constants of Integration An equation involving derivatives where we want to solve for the original function is called a differential equation. For example, $f'(x)=2x$ is a differential equation with general solution $f(x)=x^2+C\text{.}$ Some solutions (i.e., specific values of $C$) are shown below. As seen with integral curves, we may have an infinite family of solutions satisfying the differential equation. However, if we were given a point (called an initial value) on the curve then we could determine $f(x)$ completely. Such a problem is known as an initial value problem. ###### Example1.36. Initial Value Problem. If $f'(x)=2x$ and $f(0)=2$ then determine $f(x)\text{.}$ Solution As previously stated, we have a solution of: \begin{equation*} f(x)=x^2+C\text{.} \end{equation*} But $f(0)=2$ implies: \begin{equation*} 2=0^2+C\to C=2\text{.} \end{equation*} Therefore, $f(x)=x^2+2$ is the solution to the initial value problem. ##### Exercises for Section 1.5. Find the following indefinite integrals. 1. $\ds\int 8\sqrt{x}\,dx$ Answer $\ds(16/3)x^{3/2}+C$ Solution An antiderivative of $f(x) = \sqrt{x}$ is $F(x) = \frac{2}{3} x^{3/2}\text{.}$ Therefore, \begin{equation*} \int 8\sqrt{x} \,dx = 8\int \sqrt{x}\,dx = 8\left(\frac{2}{3} x^{3/2} + C_1\right) = \frac{16}{3} x^{3/2} + C\text{,} \end{equation*} for some constant $C\text{.}$ 2. $\ds\int (3t^2+1) \,dt$ Answer $\ds t^3+t+C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int 3t^2+1\,dt \amp = 3\int t^2\,dt + \int 1 \,dt \\ \amp = 3 \left(\frac{1}{3}t^3+C_1\right) + \left(t+C_2\right)\\ \amp = t^3 + t + C. \end{split} \end{equation*} 3. $\ds\int \frac{4}{\sqrt{y}} \,dy$ Answer $\ds 8\sqrt{y}+C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int \frac{4}{\sqrt{y}}\,dy \amp = 4\int y^{-1/2}\,dy \\ \amp = 4\left(2\sqrt{y}+C_1\right)\\ \amp = 8\sqrt{y}+C \end{split} \end{equation*} 4. $\ds\int \frac{2}{z^2} \, dz$ Answer $-2/z+C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int \frac{2}{z^2}\,dy \amp = 2\int z^{-2}\,dz \\ \amp = 2\left(-z^{-1}+C_1\right)\\ \amp = -\frac{2}{z} + C \end{split} \end{equation*} 5. $\ds\int 7s^{-1}\,ds$ Answer $7 \ln s + C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int 7s^{-1}\,ds \amp = 7\int s^{-1}\,ds \\ \amp = 7\left(\ln |s| + C_1\right)\\ \amp = 7\ln|s|+C \end{split} \end{equation*} 6. $\ds\int (5x+1)^2\,dx$ Answer $\ds (5x+1)^3/15+C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int \left(5x+1\right)^2 \,dx \amp = \int \left(25x^2+10x+1\right)\,dx \\ \amp = 25\int x^2\,dx + 10\int x\,dx + \int 1\,dx \\ \amp = 25 \left(\frac{1}{3} x^3 + C_1\right) + 10\left(\frac{1}{2}x^2 + C_2\right) + x + C_3\\ \amp = \frac{25}{3} x^3 + 5x^2 + x + C \end{split} \end{equation*} 7. $\ds\int (t-6)^2 \,dt$ Answer $\ds (t-6)^3/3+C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int (t-6)^2\,dt \amp = \int \left(t^2-12t+36\right)\,dt\\ \amp = \int t^2\,dt -12\int t\,dt + 36\int\,dt \\ \amp = \left(\frac{1}{3}t^3+C_1\right) - 12\left(\frac{1}{2}t^2+C_2\right) + 36\left(t+C_3\right)\\ \amp = \frac{1}{3}t^3 - 6t^2 + 36t + C \end{split} \end{equation*} 8. $\ds\int z^{3/2}\,dz$ Answer $\ds 2z^{5/2}/5+C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int z^{3/2}\,dz \amp = \frac{2}{5} z^{5/2} + C \end{split} \end{equation*} 9. $\ds\int {2\over x\sqrt x} \, dx$ Answer $\ds -4/\sqrt{x}+C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int \frac{2}{x\sqrt{x}}\,dx \amp = 2\int x^{-3/2}\,dx \\ \amp = 2\left(-2x^{-1/2}+C_1\right)\\ \amp = -\frac{4}{\sqrt{x}}+ C \end{split} \end{equation*} 10. $\ds\int|2t-4| \, dt$ Answer $\ds 4t-t^2+C\text{,}$ $t\lt 2\text{;}$ $\ds t^2-4t+8+C\text{,}$ $t\ge 2$ Solution We first rewrite the integrand using the definition of the absolute value function: \begin{equation*} |2t-4| = \begin{cases}2t-4 \amp t \geq 2\\ 4-2t \amp t \lt 2 \end{cases} \end{equation*} We first consider the case where $t \geq 2\text{:}$ \begin{equation*} \begin{split} \int 2t-4\,dt \amp = 2\int t\,dt - 4\int \,dt \\ \amp = 2\left(\frac{1}{2}t^2+C_1\right) - 4\left(t+C_2\right)\\ \amp = t^2 -4t + C \end{split} \end{equation*} Similarly, when $t \lt 2\text{,}$ we have \begin{equation*} \begin{split} \int 4-2t\,dt \amp = -\int 2t-4\,dt \\ \amp = -t^2 + 4t + C \end{split} \end{equation*} All together, we have \begin{equation*} \int |2t-4|\,dt = \begin{cases}t^2 -4t + C \amp t \geq 2\\ 4t -t^2 + C \amp t \lt 2 \end{cases} \end{equation*} 11. $\ds\int \left(\cos(2x)+4\sin(x)\right)\,dx$ Answer $\ds (\sin x-4)\cdot \cos x + C$ Solution By linearity, \begin{equation*} \int \left(\cos(2x)+4\sin(x)\right)\,dx = \int \cos(2x) \,dx + 4 \int \sin(x)\,dx\text{.} \end{equation*} We know that $\diff{}{x}\cos(x) = -\sin(x)\text{.}$ Therefore, \begin{equation*} \int \sin(x)\,dx = -\cos(x) + C_1\text{.} \end{equation*} Now, by the chain rule, \begin{equation*} \diff{}{x} \sin(2x)= 2\cos(2x) \implies \frac{1}{2}\diff{}{x} \sin(2x)= \cos(2x)\text{,} \end{equation*} and so \begin{equation*} \int \cos(2x) \,dx = \frac{1}{2} \sin(2x) + C_2\text{.} \end{equation*} Together, we have \begin{equation*} \int \left(\cos(2x)+4\sin(x)\right)\,dx = \frac{1}{2}\sin(2x)-4\cos(x)+C\text{,} \end{equation*} for some constant $C\text{.}$ 12. $\ds\int \left(2y-\sec^2y\right)\,dy$ Answer $\ds y^2 - \tan y + C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int (2y - \sec^2 y) \,dy \amp = 2\int y\,dy - \int \sec^2 y\,dy\\ \amp = 2\left(\frac{1}{2}y^2+C_1\right) - \left(\tan y + C_2\right)\\ \amp = y^2 - \tan y + C \end{split} \end{equation*} 13. $\ds\int \left(\frac{\tan x}{\cos x} - \csc^2 x\right) \,dx$ Answer $\ds \cot x + \sec x + C$ Solution We use the indefinite integral rules: \begin{equation*} \begin{split} \int \left(\frac{\tan x}{\cos x} - \csc^2 x\right)\,dx \amp = \int \frac{\sin x}{\cos^2 x}\,dx - \int \csc^2x \,dx \\ \amp = \left(\sec x+ C_1\right) - \left(-\cot x + C_2\right)\\ \amp = \sec x + \cot x + C \end{split} \end{equation*} Find the following indefinite integrals. Hint Simplify the integrand first. 1. $\ds \int \frac{2s^3+s^2-s}{s} ds$ Answer $\ds \frac{s}{6}\left(4s^2+3s-6\right)+C$ Solution \begin{aligned}\displaystyle{\int\frac{2s^3+s^2-s}{s}\,ds} \amp = \int\left(2s^2+s-1\right)\,ds \\[1ex] \amp = \frac{2}{3}s^3 +\frac{1}{2}s^2-s + C \end{aligned} 2. $\ds \int \frac{3x^3-x^2+x-2}{x} dx$ Answer $\ds x^3 - \frac{1}{2}x^2 + x - 2\ln|x| + C$ Solution \begin{aligned}\displaystyle{\int \frac{3x^3-x^2+x-2}{x}\,dx} \amp = \int \left(3x^2-x+1-\frac{2}{x}\right)\,dx \\ \amp = x^3 - \frac{1}{2}x^2 + x - 2\ln|x| + C \end{aligned} 3. $\ds \int \frac{u^3 + \sqrt[3]{2u}}{2u^2} du$ Answer $\ds \frac{u^{8/3}-3\sqrt[3]{2}}{4u^{2/3}} + C$ Solution \begin{aligned}\int\frac{u^3+\sqrt[3]{2u}}{2u^2}\,du \amp =\int \left(\frac{1}{2}u + \frac{1}{2^{2/3}}u^{-5/3}\right)\,du \\[1ex] \amp = \frac{1}{4}u^2 + \frac{3}{2\cdot 2^{2/3}}u^{-2/3}+C \end{aligned} 4. $\ds \int \frac{\left(\sqrt{t}-2\right)^2}{4t^2} dt$ Answer $\ds \frac{8\sqrt{t}+t\ln t-4}{4t} + C$ Solution \begin{aligned}\int \frac{\left(\sqrt{t}-2\right)^2}{4t^2}\,dt \amp = \int \frac{t-4\sqrt{t}+4}{4t^2}\,dt \\ \amp = \int \left(\frac{1}{4}t^{-1} - t^{-3/2} + t^{-2}\right)\,dt\\ \amp = \frac{1}{4}\ln|t| +\frac{2}{\sqrt{t}}-\frac{1}{t} + C \end{aligned} Answer the initial value problems to find $f(t)\text{.}$ 1. $f'(t)=3t+2\text{,}$ where $f(0)=5$ Answer $f(t) = \frac{3}{2}t^2+2t+5$ Solution If $f'(t)=3t+2\text{,}$ then we must have \begin{equation*} f(t) = \int \left(3t+2\right)\,dt = \frac{3}{2}t^2 + 2t + C\text{.} \end{equation*} Therefore, if $f(0) = 5\text{,}$ then we must have $C=5\text{.}$ Hence, \begin{equation*} f(t) = \frac{3}{2}t^2+2t+5\text{.} \end{equation*} 2. $f'(t)=\frac{t+3}{t}\text{,}$ where $f(1)=-1$ Answer $f(t)= t + 3\ln t - 2$ Solution We first integrate: \begin{equation*} \int f'(t)\,dt = \int\frac{t+3}{t}\,dt = \int \left(1+\frac{3}{t}\right)\,dt = t -3\ln|t| + C\text{.} \end{equation*} Since we further know that $f(1)=-1\text{,}$ we can solve for the arbitrary constant $C\text{:}$ \begin{equation*} f(1) = 1- 3\ln(1) + C = 1+C = -1 \implies C = -2\text{.} \end{equation*} Therefore, the solution to the initial value problem is \begin{equation*} f(t) = t - 3\ln|t| -2\text{.} \end{equation*} 3. $f'(t)=\sin t\text{,}$ where $f(\pi) = 0$ Answer $f(t) = -\cos t-1$ Solution Since $f'(t) = \sin(t)\text{,}$ the function $f(t)$ must be of the form \begin{equation*} f(t) = \int \sin t\,dt = -\cos(t)+C\text{.} \end{equation*} To solve for the constant $C\text{,}$ we set \begin{equation*} f(\pi) = 0 \implies 1+C = 0 \implies C = -1\text{.} \end{equation*} All together, we find that \begin{equation*} f(t) = -\cos(t) -1\text{.} \end{equation*} 4. $f'(t) = e^{2t}-4\text{,}$ where $f(0) = 1$ Answer $f(t)=0.5e^{2t}-4t+0.5$ Solution We first integrate: \begin{equation*} \int e^{2t} - 4\,dt = \frac{1}{2}e^{2t} - 4t + C\text{.} \end{equation*} Therefore, if $f(0) = 1\text{,}$ then \begin{equation*} f(0) = \frac{1}{2} + C = 1 \implies C = \frac{1}{2}\text{.} \end{equation*} Therefore, the solution to the initial value problem is \begin{equation*} f(t) = \frac{1}{2} e^{2t} - 4t + \frac{1}{2}\text{.} \end{equation*} A retailer determines that the marginal revenue function associated with selling $q$ items is \begin{equation*} R'(q)=-0.008q + 16 \end{equation*} dollars per week per item. 1. Determine $R(q)\text{.}$ Answer $R(q) = 16q - 0.004q^2$ Solution We integrate the marginal revenue function: \begin{equation*} R(q) = \int \left(-0.008q + 16\right)\,dq = -0.004q^2+16q + C\text{.} \end{equation*} Since we must have that $R(0)=0$ (i.e. there is no revenue if no units are sold), it follows that $C=0\text{:}$ \begin{equation*} R(q) = -0.004q^2+16q\text{.} \end{equation*} 2. Determine the demand equation relating the unit price $p$ to the quantity $q$ demanded. Answer $p(q) = \frac{16q - 0.004q^2}{q}$ Solution The demand equation $p(q)$ must satisfy $R(q) = p \times q\text{.}$ Therefore, \begin{equation*} p(q) = \frac{R(q)}{q} = -0.004q + 16\text{.} \end{equation*} A supermarket determines that the marginal profit function associated with purchasing and selling $q$ heads of lettuce is \begin{equation*} P'(q)=-0.003q+15\text{,} \end{equation*} dollars per year per unit. Given that the fixed storage cost is1000 per year, what is the maximum possible daily profit?

$36,400 when $q=5000$ heads. Solution The marginal profit function is given by \begin{equation*} P'(q) = -0.003q + 15 \end{equation*} dollars per year per unit. Therefore, the profit function must be \begin{equation*} P(q) = \int P'(q)\,dq = \int \left(-0.003q+15\right)\,dq = -\frac{0.003}{2}q^2+15q + C\text{,} \end{equation*} for some arbitrary constant $C\text{.}$ There is a fixed cost of$1000 per year, which means that $P(0) = -1000\text{.}$ We can thus determine the value of the constant $C\text{:}$

\begin{equation*} P(0) = C = -1000\text{.} \end{equation*}

Thus, the yearly profit function is $P(q)= -0.0015q^2+15q-1000\text{.}$

The profit is maximized when

\begin{equation*} P'(q)=0 \implies -0.003q+15 =0 \implies q=5000\text{.} \end{equation*}

Since $P(q)$ is a concave downward parabola, this must be the global maximum. Therefore, the profit is maximized when $q=5000$ units with an associated profit of $P(5000) = \dollar36,500\text{.}$

The same supermarket determines that the marginal cost of storing $q$ bouquets of flowers is approximately

\begin{equation*} C'(q)=0.001q+50 \end{equation*}

dollars per month, with a fixed cost of $500 per month. Determine the total monthly cost of storing 500 bouquets a week. Answer $C(q)=0.0005q^2+50q+500\text{,}$ $C(500) = 25,625\text{.}$ Solution We first integrate the marginal cost function using an indefinite integral: \begin{equation*} C(q) = \int C'(q)\,dq = \int \left(0.001q+50\right)\,dq = 0.0005 q^2 + 50 q + C\text{.} \end{equation*} Since there is a fixed cost of$500 per month, then we must have $C(0) = 500\text{.}$ Therefore,

\begin{equation*} C(0) = C = 500\text{.} \end{equation*}

Therefore,

\begin{equation*} C(q) = 0.0005 q^2 + 50 q +500\text{.} \end{equation*}

The total monthly cost of storing 500 bouquest is thus \$25,625.

A microbiologist is in charge of counting bacteria in a Petri dish. Starting with a single bacterium, she determines that the organisms are increasing at a rate of

\begin{equation*} N'(t)=-2t^2+10t+100 \end{equation*}

for $t\in [0,5]$ hours.

1. Determine as a function of $t$ the number of organisms in the Petri dish.

$N(t)=\frac{-2}{3}t^3+5t^2+100t+1$ for $0\leq t\leq 5\text{.}$
Solution

We first integrate $N'(t)\text{:}$

\begin{equation*} \begin{split} N(t) \amp = \int N'(t)\,dt = \int \left(-2t^2 + 10t + 100\right)\,dt\\ \amp = -\frac{2}{3}t^3 + 5t^2 + 100t + C, \text{ for } t\in[0,5]. \end{split} \end{equation*}

Since $N(0)=1\text{,}$ we must have $C=1\text{.}$ Thus, the number of organisms in the Petri dish at any time in the 5 hour period is given by

\begin{equation*} N(t)=-\frac{2}{3}t^3 + 5t^2 + 100t + 1\text{.} \end{equation*}
2. Over the entire interval studied, what was the average number of organisms in the dish?