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4.9 Antiderivatives

Not on Exam 3

antiderivative — reverse of derivative

  • → Given a function \( f(x) \), find \( F(x) \) such that \( F'(x) = f(x) \)

for example, \( F(x) = -\cos x \) is the antiderivative of \( f(x) = \sin x \) because \( F'(x) = -(-\sin x) = \sin x = f(x) \)

Antiderivative of \( x^n \)

an antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) if \( n \neq -1 \)

why? because

\[ \frac{d}{dx} \left( \frac{x^{n+1}}{n+1} \right) = \frac{1}{n+1} (n+1) x^n = x^n \]

for example, an antiderivative of \( x^2 \) is \( \frac{x^{2+1}}{2+1} = \frac{x^3}{3} = \frac{1}{3}x^3 \)

check:

\[ \frac{d}{dx} \left( \frac{1}{3}x^3 \right) = \frac{1}{3} \cdot 3x^2 = x^2 \]
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but notice \( \frac{1}{3}x^3 + 1 \) is also an antiderivative of \( x^2 \)

because \( \frac{d}{dx} (\frac{1}{3}x^3 + 1) = x^2 \)

in fact, so is \( \frac{1}{3}x^3 + 5 \), \( \frac{1}{3}x^3 + \pi \), \( \frac{1}{3}x^3 + e^{\pi} - 10 \), etc.

  • all are possible antiderivatives of \( x^2 \)
  • they differ only by a constant

to represent ALL of these, we write \( \frac{1}{3}x^3 + C \)

\( C \) is some constant (constant of integration)

so, the most general antiderivative of \( x^n \) is

\[ \frac{x^{n+1}}{n+1} + C \quad , \quad n \neq -1 \]
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Example: Find the antiderivative of \( f(x) = -\frac{10}{x^{12}} \)

(What function has this as the derivative?)

Rewrite:

\( f(x) = -10 x^{-12} \)

Constant multiple

(doesn't participate just like in differentiation)

\( x^n \) part

\( F(x) = -10 \left( \frac{x^{-12+1}}{-12+1} \right) + C \)

\( = -10 \left( \frac{x^{-11}}{-11} \right) + C \)

\( = \frac{10}{11} x^{-11} + C \)
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The indefinite integral is used to denote the process of finding the antiderivative.

\[ \int f(x) \, dx = F(x) + C \]

integral sign

(stretched out "S")

part of the indefinite integral symbol

So, from last example, we know:

\[ \int -\frac{10}{x^{12}} \, dx = \frac{10}{11} x^{-11} + C \]

Since finding antiderivatives is closely related to differentiation, many rules and properties from derivatives carry over.

For example, we can deal with terms separated by \( + \) or \( - \) individually.

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Example: Integration by Rewriting

\[ \int \left( \frac{2}{\sqrt{x}} + 2\sqrt{x} \right) dx \]

rewrite:

\[ \int (2x^{-1/2} + 2x^{1/2}) dx \]

+ or - separate them

deal with them one at a time

\[ = 2 \left( \frac{x^{-1/2+1}}{-1/2+1} \right) + 2 \left( \frac{x^{1/2+1}}{1/2+1} \right) + C \]

just need one + C

\[ = 2 \left( \frac{x^{1/2}}{1/2} \right) + 2 \left( \frac{x^{3/2}}{3/2} \right) + C \]
\[ = 4x^{1/2} + \frac{4}{3}x^{3/2} + C \]
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\[ \int x^n dx = \frac{x^{n+1}}{n+1} + C \quad \text{if } n \neq -1 \]

What happens if \( n = -1 \)?

\[ \int x^{-1} dx = \int \frac{1}{x} dx \]

what function has this as derivative? \( \rightarrow \ln x \)

so,

\[ \int x^{-1} dx = \int \frac{1}{x} dx = \ln |x| + C \]

need absolute value since \( x \) could be negative in \( \int \frac{1}{x} dx \)

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Trigonometric Integration Rules

With trig functions, think of the differentiation rules in reverse.

\[ \int \cos x \, dx = \sin x + C \quad \text{because } \frac{d}{dx}(\sin x + C) = \cos x \]
\[ \int \sin x \, dx = -\cos x + C \]
\[ \int \sec^2 x \, dx = \tan x + C \]

and so on...

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Exponential Integration

easy one:

\[ \int e^x \, dx = e^x + C \]

Example

\[ \int \left( \sin x + \sec x \tan x + \frac{2}{x} - \frac{10}{x^{1/3}} \right) dx \]

separated by +/- so one at a time

\[ = \int (\sin x + \sec x \tan x + 2x^{-1} - 10x^{-1/3}) \, dx \]\[ = -\cos x + \sec x + 2 \ln|x| - 10 \cdot \frac{x^{-1/3+1}}{-1/3+1} + C \]\[ = -\cos x + \sec x + 2 \ln|x| - 10 \cdot \frac{x^{2/3}}{2/3} + C \]

\[ = -\cos x + \sec x + 2 \ln|x| - 15x^{2/3} + C \]

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Finding the Constant of Integration

To find the value of "c" we need to know one point on the antiderivative.

Example

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

parabola with vertex at \( (0, c) \)

If we know one point that \( F(x) = x^2 + C \) passes through, then we know which one it is.

Graph showing a family of parabolas y = x^2 + 5, y = x^2 + 1, and y = x^2 - 3 on a coordinate plane.
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Example: Solving for C

\( f(x) = 2x \)

Find \( F(x) \) such that \( F'(x) = f(x) \) and \( F(1) = 3 \).

Note: \( (1, 3) \) is on \( F(x) \)

\[ F(x) = \int 2x \, dx = x^2 + C \]
Graph of parabolas with a specific point (1, 3) highlighted on one of the curves.

\( F(x) = x^2 + C \)

Since \( (1, 3) \) is on \( F(x) \), sub in \( x = 1 \), \( F(x) = 3 \) to find \( C \):

\( 3 = 1^2 + C \)

\( C = 2 \)

\[ F(x) = x^2 + 2 \]
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we handle higher-order derivatives the same way, but each order of derivative introduces one constant of integration

example

\( F''(x) = \sin x \) find \( F(x) \)

\[ F'(x) = \int \sin x \, dx = -\cos x + C \]

\[ F(x) = \int (-\cos x + C) \, dx \]

\[ = -\sin x + Cx + D \]

second constant of integration

\[ F(x) = -\sin x + Cx + D \]

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to find \( C \) and \( D \), we need one point on \( F(x) \) and one point on \( F'(x) \)

for example, if we know (for last example) \( F'(0) = 3 \), \( F(0) = 4 \)

from last page, \( F'(x) = -\cos x + C \)

  • \( (0, 3) \) on \( F' \)
  • plug in \( x = 0 \), \( F' = 3 \)

\[ 3 = -\cos(0) + C \]

\[ = -1 + C \]

\( C = 4 \)

from last page, \( F(x) = -\sin x + Cx + D \)

we now know \( C = 4 \)

\[ F(x) = -\sin x + 4x + D \]

  • \( (0, 4) \) on \( F \)
  • plug in \( x = 0 \), \( F = 4 \)

\[ 4 = -\sin(0) + 4(0) + D = D \]

\( D = 4 \)

so, \( F(x) = -\sin x + 4x + 4 \)

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MA 16100

Exam 3

Tue., Apr. 12, 2022

6:30 p.m. – 7:30 p.m.

Instructors Location
Isaac Chiu CL50 224
Rowan Desjardins CL50 224
Christopher Dewey CL50 224
Hanan Gadi WALC 1055
Mohit Pathak WALC 1055
Vittal Srinivasan PHYS 112
Ezekiel Yinka Kehinde PHYS 112