# 3.3: Antiderivatives of Formulas

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Now we can put the ideas of areas and antiderivatives together to get a way of evaluating definite integrals that is exact and often easy. To evaluate a definite integral $$\int\limits_a^b f(t)\, dt$$, we can find any antiderivative $$F(t)$$ of $$f(t)$$ and evaluate $$F(b) - F(a)$$. The problem of finding the exact value of a definite integral reduces to finding some (any) antiderivative $$F$$ of the integrand and then evaluating $$F(b) - F(a)$$. Even finding one antiderivative can be difficult, and we will stick to functions that have easy antiderivatives.

## Building Blocks

Antidifferentiation is going backwards through the derivative process. So the easiest antiderivative rules are simply backwards versions of the easiest derivative rules. Recall from Chapter 2:

##### Derivative Rules: Building Blocks

In what follows, $$f$$ and $$g$$ are differentiable functions of $$x$$.

#### Constant Multiple Rule

$\frac{d}{dx}\left( kf\right)=kf'\nonumber$

#### Sum and Difference Rule

$\frac{d}{dx}\left(f\pm g\right)=f' \pm g'\nonumber$

#### Power Rule

$\frac{d}{dx}\left(x^n\right)=nx^{n-1}\nonumber$

Special cases: $\frac{d}{dx}\left(k\right)=0 \quad \text{(Because $$k=kx^0$$.)}\nonumber$ $\frac{d}{dx}\left(x\right)=1 \quad \text{(Because $$x=x^1$$.)}\nonumber$

#### Exponential Functions

$\frac{d}{dx}\left(e^x\right)=e^x\nonumber$ $\frac{d}{dx}\left(a^x\right)=\ln(a)\,a^x\nonumber$

#### Natural Logarithm

$\frac{d}{dx}\left(\ln(x)\right)=\frac{1}{x}\nonumber$

Thinking about these basic rules was how we came up with the antiderivatives of $$2x$$ and $$e^x$$ before.

The corresponding rules for antiderivatives are next – each of the antiderivative rules is simply rewriting the derivative rule. All of these antiderivatives can be verified by differentiating.

There is one surprise – the antiderivative of $$\frac{1}{x}$$ is actually not simply $$\ln(x)$$, it's $$\ln|x|$$. This is a good thing – the antiderivative has a domain that matches the domain of $$\frac{1}{x}$$, which is bigger than the domain of $$\ln(x)$$, so we don’t have to worry about whether our $$x$$'s are positive or negative. But we must be careful to include those absolute values – otherwise, we could end up with domain problems.

##### Antiderivative Rules: Building Blocks

In what follows, $$f$$ and $$g$$ are differentiable functions of $$x$$, and $$k$$, $$n$$, and $$C$$ are constants.

#### Constant Multiple Rule

$\int k\cdot f(x)\, dx=k\cdot\int f(x)\, dx\nonumber$

#### Sum and Difference Rule

$\int \left(f(x)\pm g(x)\right)\, dx=\int f(x)\, dx \pm \int g(x)\, dx\nonumber$

#### Power Rule

$\int x^n \, dx = \frac{x^{n+1}}{n+1}+C, \text{ provided that } n\neq -1\nonumber$

Special case: $\int k\, dx =kx+C \quad \text{(Because $$k=kx^0$$.)}\nonumber$ (The other special case ($$n=-1$$) is covered next.)

#### Natural Logarithm

$\int x^{-1}\, dx =\int\frac{1}{x}\, dx = \ln|x|+C\nonumber$

#### Exponential Functions

$\int e^x\, dx=e^x +C \nonumber$ $\int a^x\, dx = \frac{a^x}{\ln(a)}+C \nonumber$

##### Example $$\PageIndex{1}$$

Find the antiderivative of $$y=3x^7-15\sqrt{x}+\frac{14}{x^2}$$.

Solution

\begin{align*} \int\left( 3x^7-15\sqrt{x}+\frac{14}{x^2} \right)\, dx & = \int\left( 3x^7-15x^{1/2}+14x^{-2} \right)\, dx \\ & = 3\frac{x^8}{8}-15\frac{x^{3/2}}{3/2}+14\frac{x^{-1}}{-1}+C \\ & = \frac{3}{8}x^8-10x^{3/2}-14x^{-1}+C \end{align*} \nonumber

##### Example $$\PageIndex{2}$$

Find $$\int\left(e^x+12-\frac{16}{x}\right)\, dx$$.

Solution

$\int\left(e^x+12-\frac{16}{x}\right)\, dx =e^x+12x-16\ln|x|+C\nonumber$

##### Example $$\PageIndex{3}$$

Find $$F(x)$$ so that $$F'(x)=e^x$$ and $$F(0)=10$$.

Solution

This time we are looking for a particular antiderivative; we need to find exactly the right constant. Let's start by finding the antiderivative: $\int e^x\, dx=e^x+C \nonumber$

So we know that $$F(x)=e^x+\text{(some constant)}$$, now we just need to find which one. To do that, we'll use the other piece of information (the initial condition): \begin{align*} F(x) & = e^x+C \\ F(0) & = e^0+C=1+C=10 \\ C & = 9 \end{align*} \nonumber

The particular constant we need is 9; thus, $$F(x)=e^x+9$$.

The reason we are looking at antiderivatives right now is so we can evaluate definite integrals exactly. Recall the Fundamental Theorem of Calculus:

##### Fundamental Theorem of Calculus

If $$F(x)$$ is a function where $$F'(x) = f(x)$$, then

$\int\limits_a^b f(x)\, dx = F(b)-F(a) \nonumber$

If we can find an antiderivative for the integrand, we can use that to evaluate the definite integral. The evaluation $$F(b) - F(a)$$ is represented as $$\left.F(x)\right]_a^b$$ or $$\left.F(x)\right|_a^b$$.

##### Example $$\PageIndex{4}$$

Evaluate $$\int\limits_1^3 x\, dx$$ in two ways:

1. By sketching the graph of $$y = x$$ and geometrically finding the area.
2. By finding an antiderivative of $$F(x)$$ of the integrand and evaluating $$F(3)-F(1)$$.

Solution

1. The graph of $$y = x$$ is shown below, and the shaded region corresponding to the integral has area 4.
2. One antiderivative of $$x$$ is $$F(x)=\frac{1}{2}x^2$$, and \begin{align*} \int\limits_1^3 x\, dx & = \left[\frac{1}{2}x^2\right]_1^3 \\ & = \left(\frac{1}{2}(3)^2\right) - \left(\frac{1}{2}(1)^2\right) \\ & = \frac{9}{2}-\frac{1}{2} \\ & = 4. \end{align*} \nonumber Note that this answer agrees with the answer we got geometrically.

If we had used another antiderivative of x, say $$F(x)=\frac{1}{2}x^2+7$$, then \begin{align*} \int\limits_1^3 x\, dx & = \left[\frac{1}{2}x^2+7\right]_1^3 \\ & = \left(\frac{1}{2}(3)^2+7\right) - \left(\frac{1}{2}(1)^2+7\right) \\ & = \frac{9}{2}+7-\frac{1}{2}-7 \\ & = 4. \end{align*} \nonumber

In general, whatever constant we choose gets subtracted away during the evaluation, so we might as well always choose the easiest one, where the constant is 0.

##### Example $$\PageIndex{5}$$

Find the area between the graph of $$y = 3x^2$$ and the horizontal axis for $$x$$ between 1 and 2.

Solution

This is $\int\limits_1^2 3x^2\, dx = \left.x^3\right|_1^2 = 2^3-1^3 = 7. \nonumber$

##### Example $$\PageIndex{6}$$

A robot has been programmed so that when it starts to move, its velocity after $$t$$ seconds will be $$3t^2$$ feet/second.

1. How far will the robot travel during its first 4 seconds of movement?
2. How far will the robot travel during its next 4 seconds of movement?

Solution

1. The distance during the first 4 seconds will be the area under the graph of velocity, from $$t = 0$$ to $$t = 4$$.

That area is the definite integral $$\int\limits_0^4 3t^2\, dt$$. An antiderivative of $$3t^2$$ is $$t^3$$, so $$\int\limits_0^4 3t^2\, dt =\left. t^3 \right]_0^4 =4^3-0^3 = 64$$ feet.

2. $$\int\limits_4^8 3t^2\, dt =\left. t^3 \right]_4^8=8^3-4^3 =512 - 64 = 448$$ feet.
##### Example $$\PageIndex{7}$$

Suppose that $$t$$ minutes after putting 1000 bacteria on a Petri plate the rate of growth of the population is $$6t$$ bacteria per minute.

1. How many new bacteria are added to the population during the first 7 minutes?
2. What is the total population after 7 minutes?

Solution

1. The number of new bacteria is the area under the rate of growth graph, and one antiderivative of $$6t$$ is $$3t^2$$.

So $\text{new bacteria}=\int\limits_0^7 6t\, dt= \left. 3t^2\right|_0^7=3(7)^2-3(0)^2=147\nonumber$

2. The new population = (old population) + (new bacteria) = 1000 + 147 = 1147 bacteria.
##### Example $$\PageIndex{8}$$

A company determines their marginal cost for production, in dollars per item, is $$MC(x)=\frac{4}{\sqrt{x}}+2$$ when producing $$x$$ thousand items. Find the cost of increasing production from 4 thousand items to 5 thousand items.

Solution

Remember that marginal cost is the rate of change of cost, and so the fundamental theorem tells us that $$\int\limits_a^b MC(x)\, dx = \int\limits_a^b C'(x)\, dx = C(b)-C(a)$$. In other words, the integral of marginal cost will give us a net change in cost. To find the cost of increasing production from 4 thousand items to 5 thousand items, we need to integrate $$\int\limits_4^5 MC(x)\, dx$$.

We can write the marginal cost as $$MC(x)=4x^{-1/2}+2$$. We can then use the basic rules to find an antiderivative: $C(x)=4\frac{x^{1/2}}{1/2}+2x=8\sqrt{x}+2x.\nonumber$

Using this, \begin{align*} \text{Net change in cost } & = \int\limits_4^5 \left(4x^{-1/2}+2\right)\, dx \\ & = \left[ 8\sqrt{x}+2x \right]_4^5 \\ & = \left( 8\sqrt{5}+2(5) \right)-\left( 8\sqrt{4}+2(4) \right) \\ & \approx 3.889 \end{align*} \nonumber

It will cost 3.889 thousand dollars to increase production from 4 thousand items to 5 thousand items. (The final answer would be better written as \$3889.)

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