# Reading: Antiderivative Formulas

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 [latex] \int_{a}^{b}f(t)dt [/latex], we can find any antiderivative*F*of

*f*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 the derivative rules### Derivative Rules

In what follows,*f*and

*g*are differentiable functions of

*x*and

*and*

*k**are constants.*

*n***Constant Multiple Rule:**[latex] \frac{d}{dx} (kf) = kf\prime [/latex]**Sum (or Difference) Rule:**[latex] \frac{d}{dx}(f+g)=f\prime + g\prime [/latex]**or**[latex] \frac{d}{dx}(f-g)=f\prime - g\prime [/latex]**Power Rule:**[latex] \frac{d}{dx}(x^n) = nx^{n-1} [/latex] Special cases: [latex] \frac{d}{dx}(k) = 0 [/latex] (because*k*=*kx*^{0})**and**[latex] \frac{d}{dx}(x) = 1 [/latex] (because*x*=*x*^{1})**Exponential Functions:**[latex] \frac{d}{dx}(e^x) = e^x [/latex] [latex] \frac{d}{dx}(a^x) = \text{ln} a \times a^x [/latex]**Natural Logarithm:**[latex] \frac{d}{dx}(\text{ln}x) = \frac{1}{x} [/latex]

*x*and

*e*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 [latex] \frac{1}{x} [/latex] is actually not simply ln(

^{x}*x*), it’s ln|

*x*|. This is a good thing—the antiderivative has a domain that matches the domain of [latex] \frac{1}{x} [/latex], 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 you must be careful to include those absolute values—otherwise, you 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:**[latex] \int kf(x)dx = k \int f(x)dx [/latex]**Sum (or Difference) Rule:**[latex] \int f(x) \pm g(x)dx = \int f(x)dx \pm \int f(x)dx [/latex]**Power Rule:**[latex] \int x^n dx = \frac{x^{n+1}}{n + 1} + C [/latex] provided that*n*= −1 Special case: [latex] \int kdx = kx + C [/latex] (because*k*=*kx*^{0})**Exponential Functions:**[latex] \int e^x dx = e^x + C [/latex] [latex] \int a^x dx = \frac{a^x}{\text{ln}a} + C [/latex]**Natural Logarithm:**[latex] \int x^{-1}dx = \int\frac{1}{x}dx = \text{ln}|x| + C [/latex]

### Example

Find the antiderivative of [latex] 3x^7 - 15\sqrt x + \frac{14}{x^2} [/latex]#### Solution

[latex-display] \int(3x^7 - 15\sqrt x + \frac{14}{x^2})dx = \int(3x^7 - 15^{1/2} + 14x^{-2})dx = 3\frac{x^8}{8} - 15\frac{x^{3/2}}{3/2} + 14\frac{x^{-1}}{-1} + C [/latex-display] That’s a little hard to look at, so you might want to simplify a little: [latex-display] \int(3x^7 - 15\sqrt x + \frac{14}{x^2})dx = \frac{3x^8}{8} -10x^{3/2} - 14x^{-1} + C [/latex-display]### Example

[latex-display] \int(e^x + 12 - \frac{16}{x})dx [/latex-display]#### Solution

[latex-display] \int(e^x + 12 - \frac{16}{x})dx = e^x +12x - 16\mathrm{ln}|x| + C [/latex-display]### Example

Find*F*(

*x*) so that

*F*′(

*x*) =

*e*and

^{x}*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: [latex-display] \int e^xdx=e^x + C [/latex-display] So we know that*F*′(

*x*) =

*e*+ some constant; we just need to find which one. For that, we’ll use the other piece of information (the initial condition): [latex-display] F(x) = e^x + C [/latex-display] [latex-display] F(0) = e^0 + C = 1 + C = 10 [/latex-display] [latex-display] C = 9 [/latex-display] The particular constant we need is 9;

^{x}*F*(

*x*) =

*e*+ 9. The reason we are looking at antiderivatives right now is so we can evaluate definite integrals exactly. Recall the Fundamental Theorem of Calculus: [latex-display] \int_{a}^{b} F\prime(x)dx = F(b) -F(a) [/latex-display] 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 by the notation [latex] \left.F(x)\right]_a^b [/latex] or [latex] \left.F(x)\right|_a^b [/latex].

^{x}### Example

Evaluate [latex] \int_{1}^{3} xdx [/latex] in two ways:- By sketching the graph of
*y*=*x*and geometrically finding the area. - By finding an antiderivative of
*F*(*x*) of the integrand and evaluating*F*(*3*) –*F*(*1*).

#### Solution

- The graph of
*y*=*x*is shown in figure 1, and the shaded region has area 4. - One antiderivative of
*x*is [latex] F(x) = \frac{1}{2}x^2 [/latex] (check by differentiating ), and [latex] \int_{1}^{3}xdx=\frac{1}{2}x^2]_1^3 = [\frac{1}{2}(3)^2] - [\frac{1}{2}(1)^2] = \frac{9}{2} - \frac{1}{2} = 4 [/latex]. Note that this answer agrees with the answer we got geometrically.

*x*, say [latex] F(x) = \frac{1}{2}x^2 + 7 [/latex] (check by differentiating ), then [latex] F(x)|_{1}^{3} = F(3) - F(1) = [\frac{1}{2}(3^2)+7]-[\frac{1}{2}(1^2)+7]= \frac{23}{2} - \frac{15}{2} = 4 [/latex]. Whatever constant you choose, it gets subtracted away during the evaluation; we might as well always choose the easiest one, where the constant = 0.

### Example

Find the area between the graph of*y*= 3

*x*

^{2}and the horizontal axis for

*x*between 1 and 2.

#### Solution

This is [latex] \int_{1}^{2}3x^2dx=x^3]_{1}^{2} = (2^3) -(1^3) = 7 [/latex].### Example

A robot has been programmed so that when it starts to move, its velocity after*t*seconds will be 3

*t*

^{2}feet/second.

- How far will the robot travel during its first 4 seconds of movement?
- How far will the robot travel during its next 4 seconds of movement?

#### Solution

- The distance during the first 4 seconds will be the area under the graph (figure 2) of velocity, from
*t*= 0 to*t*= 4. That area is the definite integral [latex] \int_{0}^{4}3t^2dt [/latex]. An antiderivative of 3*t*^{2}is*t*^{3}, so [latex] \int_{0}^{4}3t^2dt =t^3 ]_{0}^{4} = 4^3 - 0^3 =64 [/latex] feet. - [latex] \int_{4}^{8}3t^2dt =t^3 ]_{4}^{8} = 8^3 - 4^3 = 512 - 64 = 448 [/latex] feet.

### Example

Suppose that*t*minutes after putting 1000 bacteria on a Petri plate the rate of growth of the population is 6

*t*bacteria per minute.

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

#### Solution

- The number of new bacteria is the area under the rate of growth graph (figure 3), and one antiderivative of 6
*t*is 3*t*^{2}. So new bacteria = [latex] \int_{0}^{7} 6t \mathrm{d}t = 3t^2|_{0}^{7} = 3(7)^2 - 3(0)^2 = 147 [/latex]. - The new population = {old population} + {new bacteria} = 1000 + 147 = 1147 bacteria.

## Licenses & Attributions

### CC licensed content, Shared previously

- Business Calculus.
**Provided by:**Washington State Colleges**Authored by:**Dale Hoffman and Shana Calaway.**Located at:**https://docs.google.com/file/d/0B1lkHWwO61QEM0gwOFhES2N5Tlk/edit.**License:**CC BY: Attribution.