# Indefinite Integrals (What is an antiderivative?)

What is an antiderivative? What is an indefinite integral? Maybe you have heard of them? Their reputation does often proceed them. Well, can you answer this question? What is the function, or what is a function that I need to have a given derivative? Read on to see how this idea leads to finding an area.

We discuss antidifferentiation by defining an antiderivative function and working out examples on finding antiderivatives. We also concentrate on the following problem: if a function is an antiderivative of a given continuous function, then any other antiderivative of must be the sum of the antiderivative and some constant. The Mean Value Theorem is used to show that all antiderivatives of a function have this form.

## Antiderivatives

We go over antidifferentiation by defining an antiderivative function and working out examples on finding antiderivatives. We also concentrate on the following problem: if $F$ is an antiderivative of the continuous function $f,$ then any other antiderivative of $f$ must have the form $F(x)+C$ where $C$ is some constant. Thus showing, if a function $F(x)$ is an antiderivative of the function $f(x)$ then so is $F(x)+C$ where $C$ is called an arbitrary constant. The mean value theorem will be used to show that all derivatives of $f(x)$ are of the form $F(x)+C$ and that there are no others.

Definition. A function $F$ is called an antiderivative of a given function $f$ on an interval $I$ if $F'(x)=f(x)$ for all $x$ in $I.$

Theorem. If $F$ is an antiderivative of the continuous function $f,$ then any other antiderivative $G$, of $f$ must have the form $G(x)=F(x)+C$ where $C$ is some constant.

Proof. We show that if $F$ is differentiable in $[a,b]$ and $F'(x)=0$ for all $x$ in $[a,b],$ then $F(x)=F(a)$ for all $x$ in $[a,b].$ By the mean-value theorem applied to $[a,x]$ for any $x$ such that $a<x\leq b,$ $$\frac{F(x)-F(a)}{x-a}=F'(c)=0$$ where $a<c<x.$ Thus $F(x)=F(a)=0,$ so $F(x)=F(a),$ and $F(x)$ is constant in $[a,b].$ So in fact, for $F'(x)=f(x)$ with all $x$ in $[a,b],$ if we suppose $G'(x)=f(x)$ also for $x$ in $[a,b]$ then $$\frac{d(G(x)-F(x))}{dx}=G'(x)-F'(x)=0.$$ Thus $G(x)-F(x)=G(a)-F(a).$ If we let $C=G(a)-F(a),$ then ${G(x)=F(x)+C.}$

The notation $$\int f(x) \, dx=F(x)+C$$ where $C$ is an arbitrary constant means that $F$ is an antiderivative of $f.$ The function $F$ is called the indefinite integral of $f$ and satisfies the condition that $F'(x)=f(x)$ for all $x$ in the domain of $f.$ It is important to remember that $F(x)+C$ represents a family of functions.

## Indefinite Integrals and Integral Notation

Example. Find the family of antiderivatives of the function $f(x)=\sin x$ and write an equation using the indefinite integral notation.

Solution. If $F(x)=-\cos x,$ then $F'(x)=\sin x,$ and so an antiderivative of sine is $-\cos x.$ Thus the general antiderivative is $G(x)=-\cos x+C.$ Therefore, $$\int \sin xdx=-\cos x+C$$ where $C$ is an arbitrary constant.

Example. Find the family of antiderivatives of the function $f(x)=x^n$, $n\geq 0,$ and write an equation using the indefinite integral notation.

Solution. Since $$\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=\frac{(n+1)x^n}{n+1}=x^n$$ the general antiderivative of $f$ is $$F(x)=\frac{x^{n+1}}{n+1}+C,$$ where $C$ is a constant, which is valid for $n\geq 0$ because $f(x)=x^n$ is defined on the interval $(-\infty ,+\infty ).$ Therefore, $$\int x^n \, dx=\frac{x^{n+1}}{n+1}+C$$ where $C$ is an arbitrary constant and $n\geq 0.$

Theorem. Suppose $f$ and $g$ are integrable functions and $a$, $b$, and $c$ are constants. Then

$(1) \quad \displaystyle \int c f(u)du=c\int f(u) \, du$

$(2) \quad \displaystyle \int [f(u)+g(u)] \, du=\int f(u) \, du+\int g(u) \, du$

$(3) \quad \displaystyle \int [f(u)-g(u)] \, du=\int f(u) \, du-\int g(u) \, du$

$(4) \quad \displaystyle \int [a f(u)du+b g(u)] \, du=a\int f(u)du+b \int g(u) \, du$

Theorem. Suppose $C$ is an arbitrary constant. Then \begin{equation*} \int x^n \, dx= \begin{cases} \dfrac{x^{n+1}}{n+1}+C & n\neq -1 \\[12px] \ln |x|+C & n=-1. \end{cases} \end{equation*}

Theorem. Suppose $u$ is a differentiable function of $x$ and $C$ is an arbitrary constant.

$(1) \quad \displaystyle \int 0 \, du=0+C$

$(2) \quad \displaystyle \int e^u \, du=e^u+C$

$(3) \quad \displaystyle \int \sin u \, du=-\cos u+C$

$(4) \quad \displaystyle \int \cos u \, du=\sin u+C$

$(5) \quad \displaystyle \int \sec ^2 u \, du=\tan u+C$

$(6) \quad \displaystyle \int \sec u \tan u \, du=\sec u+C$

$(7) \quad \displaystyle \int \csc u \cot u \, du=-\csc u+C$

$(8) \quad \displaystyle \int \csc ^2 u \, du=-\cot u+C$

$(9) \quad \displaystyle \int \frac{1}{\sqrt{1-u^2}} \, du=\sin ^{-1}u+C$

$(10) \quad \displaystyle \int \frac{1}{1+u^2} \, du=\tan ^{-1}u+C$

$(11) \quad \displaystyle \int \frac{1}{|u|\sqrt{u^2-1}} \, du=\sec ^{-1}u+C$

Example. Find the general antiderivative of the function $$f(x)=\frac{x^2}{x^2+1}.$$

Solution. Since $$\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}$$ we find \begin{align} \int \frac{x^2}{x^2+1} \, dx & =\int \left(1-\frac{1}{x^2+1}\right) \, dx \\ & =\int dx-\int \frac{1}{x^2+1} \, dx =x-\tan ^{-1}x+C. \end{align}
as desired.

Example. Find the general antiderivative of the function $$f(x)=\left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right).$$

Solution. We want to evaluate $\int \left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right)dx.$ Since $$\left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right)=1+\frac{1}{x}-\frac{4}{x^2}-\frac{4}{x^3}$$
we use the linearity rule and the power rule to find \begin{align} \int \left(1+\frac{1}{x}\right)\left(1-\frac{4}{x^2}\right)dx & =\int \left(1+x^{-1}-4x^{-2}-4x^{-3}\right) \, dx \\ & =x+\ln x+\frac{4}{x}+\frac{2}{x^2}+C. \end{align} as desired.

Example. Find $f(x)$ given $f”(x)=x+\sqrt{x},$ $f(1)=1,$ and $f'(1)=2.$

Solution. First we find $f’$ by $$f'(x)=\int \left(x+\sqrt{x}\right) \, dx=\frac{x^2}{2}+\frac{2 x^{3/2}}{3}+C$$ where $C$ is some constant which can be determined with $f'(1)=2.$ So $$f'(1)=\frac{(1)^2}{2}+\frac{2 (1)^{3/2}}{3}+C=\frac{7}{6}+C=2 \qquad \Longrightarrow \qquad C=\frac{5}{6}.$$ So in fact $$f'(x)=\frac{x^2}{2}+\frac{2 x^{3/2}}{3}+\frac{5}{6}.$$ Now to find $f$ we follow the same procedure $$f(x)=\int \left(\frac{x^2}{2}+\frac{2 x^{3/2}}{3}+\frac{5}{6}\right) \, dx=\frac{x^3}{6}+\frac{4 x^{5/2}}{15}+\frac{5 x}{6}+K$$ where $K$ is a constant which can be determined with $f(1)=1.$ So $$f(1)=\frac{(1)^3}{6}+\frac{4 (1)^{5/2}}{15}+\frac{5 (1)}{6}+K=\frac{19}{15}+K=1$$ which implies that $K=-4/15$. Therefore $$f(x)=\frac{x^3}{6}+\frac{4 x^{5/2}}{15}+\frac{5 x}{6}-\frac{4}{15}$$ as desired.

Does every function have an antiderivative?

## Finding Area

Theorem. (Area Function) If $f$ is a continuous function such that $f(x)\geq 0$ for all $x$ on the closed interval $[a,b],$ then the area bounded by the curve $y=f(x),$ the $x$-axis, and the vertical lines $x=a$ and $x=t,$ viewed as a function of $t,$ is an antiderivative of $f(t)$ on $[a,b].$

Example. Find the area under the parabola $y=x^2$ over the interval $[0,1].$

Solution. Since $f(x)=x^2$ is a continuous function with $f(x)\geq 0$ for all $x.$ The area function is given by $$A(t)=\int t^2 \, dt=\frac{1}{3}t^3+C$$ and we can determine $C$ using $A(0)=0$ and so $$A(0)=\frac{1}{3}(0)^3+C=0 \qquad \Longrightarrow \qquad C=0$$ which means $A(t)=\frac{1}{3}t^3.$ Therefore the area under the curve from $[0,1]$ is $A(1)=\frac{1}{3}(1)^3=\frac{1}{3}.$

## Applications of Integration

In the next example we find the demand function given the marginal revenue.

Example. A manufacturer estimates that the marginal revenue of a certain commodity is $R'(x)=240+0.1x$ when $x$ units are produced. Find the demand function $p(x).$

Solution. Since $$R(x)=\int R'(x)dx=\int (240+0.1x) \, dx=240 x+0.05x^2+C$$ and because $R(x)=x p(x),$ where $p(x)$ is the demand function, we must have $R(0)=0$ so that $240(0)+0.05(0)+C+0$ yielding $C=0$ and $$p(x)=\frac{R(x)}{x}=\frac{240x+0.05x^2}{x}=240+0.05x.$$ as desired.

Example. A ball is thrown upward with a speed of 48 ft/s from the edge of a cliff 432 feet above the ground. Find its height above the ground $t$ seconds later. When does it reach its maximum height? When does it hit the ground?

Solution. The motion is vertical and we choose the positive direction to be upward.
At time $t$ the distance above the ground is $s(t)$ and the velocity $v(t)$ is decreasing. Therefore the acceleration must be negative $$a(t)=\frac{dv}{dt}=-32.$$ Taking the antiderivative $$v(t)=\int a(t) \, dt=-32t+C.$$ To determine $C$ we use the given information of $v(0)=48.$ Thus $$v(0)=-32(0)+C=48$$ and so $C=48$ and $v(t)=-32t+48.$ It follows the ball reaches its maximum height at $v(t)=0$ which means $t=1.5s.$ Taking the antiderivative $$s(t)=v'(t)=\int (-32t+48) \, dt=-16 t^2+48 t+K$$ and using $s(0)=432$ we find that $s(0)=-16 (0)^2+48 (0)+K=432$ and so $K=432.$ Therefore the height function is $$s(t)=-16 t^2+48 t+432$$ and so the ball hits the ground when $s(t)=0$ meaning $$t=\frac{3+3\sqrt{13}}{2}\approx 6.9 s$$ using the quadratic formula.

Example. A company has found that the rate of change of its average cost for a product is $$\bar{C}'(x)=\frac{1}{4}-\frac{100}{x^2}$$ where $x$ is the number of units and cost is in dollars. The average cost of producing 20 units is 40,000 dollars. Find the average cost function for the product. Find the average cost of 100 units of the product.

Solution. To find $\bar{C}(x)$ we integrate, so
\bar{C}(x)=\int \left(\frac{1}{4}-\frac{100}{x^2}\right) \, dx=\frac{x}{4}+\frac{100}{x}+K to find the constant $K$ we use the given that the average cost of producing 20 units is 40,000. So we find the constant $K$ by $$\bar{C}(20)=\frac{20}{4}+\frac{100}{20}+K=10+K=40000$$ which means $K=40000-10=39990.$ (a) So the average cost function for the product is $$\bar{C}(x)=\frac{x}{4}+\frac{100}{x}+39990.$$ (b) The average cost of 100 units of the product is $$\bar{C}(10)=\frac{10}{4}+\frac{100}{10}+39990=\frac{80005}{2}=40002.5$$ dollars.

Example. An excellent film with a very small advertising budget must depend largely on world-of-mouth advertising. In this case, the rate at which weekly attendance might grow can be given by $$\frac{dA}{dt}=\frac{-100}{(t+10)^2}+\frac{2000}{(t+10)^3}$$
where $t$ is in the time in weeks since release and $A$ is attendance in millions. Find the function that describes weekly attendance at this film. Find the attendance at this film in the tenth week.

Solution. The function that describes weekly attendance at this film is found by integration \begin{align} A(t)& =\int \frac{dA}{dt} \, dt =\int \left(\frac{-100}{(t+10)^2}+\frac{2000}{(t+10)^3}\right) \, dt \\ & =\int \left(\frac{-100}{(t+10)^2}+\frac{2000}{(t+10)^3}\right) \, dt =\frac{100}{t+10}-\frac{1000}{(t+10)^2}+K \end{align} where $t$ is in the time in weeks since release and $A$ is attendance in millions. When $t=0$ the attendance was 0 and so we find $K$ by $$A(0)=\frac{100}{0+10}-\frac{1000}{(0+10)^2}+K=K=0.$$ Thus $$A(t)=\frac{100}{t+10}-\frac{1000}{(t+10)^2}$$ and so the attendance at this film in the tenth week is $$A(10)=\frac{100}{10+10}-\frac{1000}{(10+10)^2}=\frac{5}{2}=2.5$$ million people.

Example. Suppose the marginal cost for a product is $$\overline{M C}=60\sqrt{x+1}$$ and its fixed cost is 340.00. If the marginal revenue for the product is $\overline{M R}=80x,$ find the profit or loss from the production and sale of (a) 3 units (b) 8 units.

Solution. The marginal cost is $$\overline{M C}=60\sqrt{x+1}$$ and so the cost function is fond by integration $$C(x)=\int 60\sqrt{x+1}dx=40 (1+x)^{3/2}+K$$ where $K$ is a constant. Since $C(0)=340$ we find $K=300$ and so the cost function is $$C(x)=40 (1+x)^{3/2}+30.$$ The marginal revenue is $\overline{MR}=80x$ and so the revenue function is found by integration $$R(x)=\int 80 xdx=40 x^2.$$ Thus the profit function for this product is $$P(x)=R(x)-C(x)=40 x^2-40 (1+x)^{3/2}-300.$$ (a) The loss from the sale of 3 units is $$P(3)=40 (3)^2-40 (1+(3))^{3/2}-300=-260$$ dollars. (b) The profit from the sale of 8 units is $$P(8)=40 (8)^2-40 (1+(8))^{3/2}-300=1180$$ dollars.

## Constant-Difference Theorem

The following theorem says that two functions with equal derivatives on an open interval differ by a constant on that interval.

Theorem. (Constant-Difference Theorem) Let $f$ and $g$ be functions that are continuous on $[a,b]$ and differentiable on $(a,b).$ If $f'(x)=g'(x)$ for all $x$ in $(a,b),$ then $f-g$ is constant on $(a,b);$ that is, $f(x)=g(x)+k$ where $k$ is a constant.

Proof. Let $F(x)=f(x)-g(x).$ Then $F'(x)=f'(x)-g'(x)=0$ for all $x$ in $(a,b).$ Thus by the Zero Derivative Theorem, $F(x)=k$ for some constant $k$ and so $f(x)=g(x)+k$ as desired.

Example. Let $g(x)=\sqrt{x^2+5}.$ Find a function $f$ with $f'(x)=g'(x)$ and $f(2)=1.$

Solution. Let $f(x)=\sqrt{x^2+5}+k$ where $k$ is some constant to be determined. Then $f'(x)=g'(x)$ and to determine $k$ we use $f(2)=1$ to obtain $$f(2)=\sqrt{2^2+5}+k=3+k=1.$$ Therefore, $f(x)=\sqrt{x^2+5}-2$ is the function we desire.

Example. Show that $$f(x)=\frac{x+4}{5-x} \quad \text{and}\quad g(x)=\frac{-9}{x-5}$$ differ by a constant. Are the conditions of the constant difference theorem satisfied? Does $f'(x)=g'(x).$

Solution. We simplify $$f'(x)=g'(x)=\frac{9}{(x-5)^2}$$ which is valid on any interval not containing $x=5.$ Thus on any interval not containing $x=5,$ the constant difference theorem applies. In fact, when $x\neq 5$, we have $$f(x)-g(x)=\frac{x+4}{5-x}-\frac{-9}{x-5}=-1$$ as desired.

Example. Let $f(x)=(x-2)^3$ and $g(x)=\left(x^2+12\right)(x-6).$ Use $f$ and $g$ to demonstrate the Constant-Difference Theorem.

Solution. The functions $f$ and $g$ are polynomial functions so they are continuous and differentiable for all real numbers. Also, $$f'(x)=3(x-2)^2=3x^2-12x+4=(2x)(x-6)+\left(x^2+12\right)=g'(x)$$ for all real numbers. Therefore, we have ${f(x)=g(x)+k}$ for some real number $k.$

The following theorem is a partial converse to the statement that the derivative of a constant is 0.

Theorem. (Zero-Derivative Theorem) Let $f$ be a function that is continuous on $[a,b]$ and differentiable on $(a,b).$ If $f'(c)=0$ for all $c$ in $(a,b)$ then $f$ is constant on $[a,b].$

Proof. If $x_1$ and $x_2$ are different points in $[a,b]$ then by the Mean Value Theorem there exists a $c$ in $\left(x_1,x_2\right)$ such that $$f'(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}.$$ By hypothesis $f'(c)=0$ and so $f\left(x_2\right)-f\left(x_1\right)=0.$ Since $x_1$ and $x_2$ were chosen arbitrarily, $f$ is a constant function on $[a,b].$

Example. Consider $$f(x)= \begin{cases} 1 & x\geq 0 \\ -1 & x<0. \end{cases}$$ Notice that $f'(x)=0$ for all $x$ in the domain, but $f$ is not a constant. Does this example contradict the Zero-Derivative Theorem?

Solution. No it does not, rather it shows that the assumptions of the zero-derivative theorem are necessary.

## Exercises on Indefinite Integrals

Exercise. Find an antiderivative for the following sets of functions $f(x)$, $g(x)$, and $h(x).$ Write out an equation using an indefinite integral and each of these functions.

$(1) \quad \displaystyle f(x)=2x,$ $g(x)=x^2,$ and $h(x)=x^2-2x+1$

$(2) \quad \displaystyle f(x)=-3x^{-4},$ $g(x)=x^{-4},$ and $h(x)=x^{-4}+2x+3$

$(3) \quad \displaystyle f(x)=\frac{-2}{x^3},$ $g(x)=\frac{1}{2 x^3}$ and $h(x)=x^3-\frac{1}{x^3}$

$(4) \quad \displaystyle f(x)=\frac{2}{3}x^{-1/3},$ $g(x)=\frac{1}{3}x^{-2/3}$ and $h(x)=\frac{-1}{3}x^{-4/3}$

$(5) \quad \displaystyle f(x)=\frac{1}{3x},$ $g(x)=\frac{2}{5x},$ and $h(x)=1+\frac{4}{3x}-\frac{1}{x^2}$

$(6) \quad \displaystyle f(x)=\sec ^2x,$ $g(x)=\frac{2}{3}\sec ^2\frac{x}{3}$ and $h(x)=-\sec ^2\frac{3x}{2}$

$(7) \quad \displaystyle f(x)=e^{-2x}$ $g(x)=e^{4x/3},$ and $h(x)=e^{-x/5}.$

$(8) \quad \displaystyle f(x)=3^x,$ $g(x)=2^{-x},$ and $h(x)=\left(\frac{5}{3}\right)^x.$

$(9) \quad \displaystyle f(x)=x-\left(\frac{1}{2}\right)^x$ $g(x)=x^2+2^x,$ and $h(x)=\pi ^x-x^{-1}.$

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.