# 5.6 Integrals involving exponential and logarithmic functions  (Page 3/4)

 Page 3 / 4

Suppose the rate of growth of the fly population is given by $g\left(t\right)={e}^{0.01t},$ and the initial fly population is 100 flies. How many flies are in the population after 15 days?

There are 116 flies.

## Evaluating a definite integral using substitution

Evaluate the definite integral using substitution: ${\int }_{1}^{2}\frac{{e}^{1\text{/}x}}{{x}^{2}}dx.$

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a power of x , then bring the x 2 in the denominator up to the numerator using a negative exponent. We have

${\int }_{1}^{2}\frac{{e}^{1\text{/}x}}{{x}^{2}}dx={\int }_{1}^{2}{e}^{{x}^{-1}}{x}^{-2}dx.$

Let $u={x}^{-1},$ the exponent on e . Then

$\begin{array}{cc}\hfill du& =\text{−}{x}^{-2}dx\hfill \\ \hfill -du& ={x}^{-2}dx.\hfill \end{array}$

Bringing the negative sign outside the integral sign, the problem now reads

$\text{−}\int {e}^{u}du.$

Next, change the limits of integration:

$\begin{array}{}\\ \\ u={\left(1\right)}^{-1}=1\hfill \\ u={\left(2\right)}^{-1}=\frac{1}{2}.\hfill \end{array}$

Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the limits. Thus,

$\begin{array}{}\\ \\ \\ \text{−}{\int }_{1}^{1\text{/}2}{e}^{u}du\hfill & ={\int }_{1\text{/}2}^{1}{e}^{u}du\hfill \\ & ={e}^{u}{|}_{1\text{/}2}^{1}\hfill \\ & =e-{e}^{1\text{/}2}\hfill \\ & =e-\sqrt{e}.\hfill \end{array}$

Evaluate the definite integral using substitution: ${\int }_{1}^{2}\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx.$

${\int }_{1}^{2}\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx=\frac{1}{8}\left[{e}^{4}-e\right]$

## Integrals involving logarithmic functions

Integrating functions of the form $f\left(x\right)={x}^{-1}$ result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ and $f\left(x\right)={\text{log}}_{a}x,$ are also included in the rule.

## Rule: integration formulas involving logarithmic functions

The following formulas can be used to evaluate integrals involving logarithmic functions.

$\begin{array}{ccc}\hfill \int {x}^{-1}dx& =\hfill & \text{ln}|x|+C\hfill \\ \hfill \int \text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx& =\hfill & x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x-x+C=x\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C\hfill \\ \hfill \int {\text{log}}_{a}x\phantom{\rule{0.2em}{0ex}}dx& =\hfill & \frac{x}{\text{ln}\phantom{\rule{0.1em}{0ex}}a}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C\hfill \end{array}$

## Finding an antiderivative involving $\text{ln}\phantom{\rule{0.2em}{0ex}}x$

Find the antiderivative of the function $\frac{3}{x-10}.$

First factor the 3 outside the integral symbol. Then use the u −1 rule. Thus,

$\begin{array}{ll}\int \frac{3}{x-10}dx\hfill & =3\int \frac{1}{x-10}dx\hfill \\ \\ \\ & =3\int \frac{du}{u}\hfill \\ & =3\phantom{\rule{0.1em}{0ex}}\text{ln}|u|+C\hfill \\ & =3\phantom{\rule{0.1em}{0ex}}\text{ln}|x-10|+C,x\ne 10.\hfill \end{array}$

Find the antiderivative of $\frac{1}{x+2}.$

$\text{ln}|x+2|+C$

## Finding an antiderivative of a rational function

Find the antiderivative of $\frac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.$

This can be rewritten as $\int \left(2{x}^{3}+3x\right){\left({x}^{4}+3{x}^{2}\right)}^{-1}dx.$ Use substitution. Let $u={x}^{4}+3{x}^{2},$ then $du=4{x}^{3}+6x.$ Alter du by factoring out the 2. Thus,

$\begin{array}{}\\ \hfill du& =\hfill & \left(4{x}^{3}+6x\right)dx\hfill \\ & =\hfill & 2\left(2{x}^{3}+3x\right)dx\hfill \\ \hfill \frac{1}{2}\phantom{\rule{0.2em}{0ex}}du& =\hfill & \left(2{x}^{3}+3x\right)dx.\hfill \end{array}$

Rewrite the integrand in u :

$\int \left(2{x}^{3}+3x\right){\left({x}^{4}+3{x}^{2}\right)}^{-1}dx=\frac{1}{2}\int {u}^{-1}du.$

Then we have

$\begin{array}{ll}\frac{1}{2}\int {u}^{-1}du\hfill & =\frac{1}{2}\text{ln}|u|+C\hfill \\ \\ & =\frac{1}{2}\text{ln}|{x}^{4}+3{x}^{2}|+C.\hfill \end{array}$

## Finding an antiderivative of a logarithmic function

Find the antiderivative of the log function ${\text{log}}_{2}x.$

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

$\int {\text{log}}_{2}xdx=\frac{x}{\text{ln}\phantom{\rule{0.1em}{0ex}}2}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C.$

Find the antiderivative of ${\text{log}}_{3}x.$

$\frac{x}{\text{ln}\phantom{\rule{0.1em}{0ex}}3}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C$

[link] is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.

## Evaluating a definite integral

Find the definite integral of ${\int }_{0}^{\pi \text{/}2}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1+\text{cos}\phantom{\rule{0.1em}{0ex}}x}dx.$

We need substitution to evaluate this problem. Let $u=1+\text{cos}\phantom{\rule{0.1em}{0ex}}x,,$ so $du=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx.$ Rewrite the integral in terms of u , changing the limits of integration as well. Thus,

$\begin{array}{c}u=1+\text{cos}\left(0\right)=2\hfill \\ u=1+\text{cos}\left(\frac{\pi }{2}\right)=1.\hfill \end{array}$

Then

$\begin{array}{cc}{\int }_{0}^{\pi \text{/}2}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1+\text{cos}\phantom{\rule{0.1em}{0ex}}x}\hfill & =\text{−}{\int }_{2}^{1}{u}^{-1}du\hfill \\ \\ \\ & ={\int }_{1}^{2}{u}^{-1}du\hfill \\ & ={\text{ln}|u||}_{1}^{2}\hfill \\ & =\left[\text{ln}\phantom{\rule{0.1em}{0ex}}2-\text{ln}\phantom{\rule{0.1em}{0ex}}1\right]\hfill \\ & =\text{ln}\phantom{\rule{0.1em}{0ex}}2.\hfill \end{array}$

## Key concepts

• Exponential and logarithmic functions arise in many real-world applications, especially those involving growth and decay.
• Substitution is often used to evaluate integrals involving exponential functions or logarithms.

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
diron
pweding paturo nsa calculus?
jimmy
how to use fundamental theorem to solve exponential
find the bounded area of the parabola y^2=4x and y=16x
what is absolute value means?
Chicken nuggets
Hugh
🐔
MM
🐔🦃 nuggets
MM
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
Ismael
find integration of loge x
find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3
how does this work
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
log tan (x/4+x/2)
Rohan
Rohan
y=(x^2 + 3x).(eipix)
Claudia
Ismael
A Function F(X)=Sinx+cosx is odd or even?
neither
David
Neither
Lovuyiso
f(x)=1/1+x^2 |=[-3,1]
apa itu?
fauzi
determine the area of the region enclosed by x²+y=1,2x-y+4=0
Hi
MP
Hi too
Vic
hello please anyone with calculus PDF should share
Which kind of pdf do you want bro?
Aftab
hi
Abdul
can I get calculus in pdf
Abdul
explain for me
Usman
How to use it to slove fraction
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
Jean
Im not in college but this will still help
nothing
how en where can u apply it
Migos
how can we scatch a parabola graph
Ok
Endalkachew
how can I solve differentiation?
with the help of different formulas and Rules. we use formulas according to given condition or according to questions
CALCULUS
For example any questions...
CALCULUS
v=(x,y) وu=(x,y ) ∂u/∂x* ∂x/∂u +∂v/∂x*∂x/∂v=1
log tan (x/4+x/2)
Rohan
what is the procedures in solving number 1? By OpenStax By Laurence Bailen By Brooke Delaney By OpenStax By Rohini Ajay By John Gabrieli By OpenStax By Jonathan Long By Mucho Mizinduko By Brooke Delaney