Integrals Involving Exponential and Logarithmic Functions

Exponential and logarithmic functions are used to model population growth, cell growth, and financial growth, as well as depreciation, radioactive decay, and resource consumption, to name only a few applications. In this section, we explore integration involving exponential and logarithmic functions.

Integrals of Exponential Functions

The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function,

y = e^{x},

A common mistake when dealing with exponential expressions is treating the exponent on e the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent on e. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we’re using the right rules for the functions we’re integrating.

As mentioned at the beginning of this section, exponential functions are used in many real-life applications. The number e is often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Although the derivative represents a rate of change or a growth rate, the integral represents the total change or the total growth. Let’s look at an example in which integration of an exponential function solves a common business application.

A price–demand function tells us the relationship between the quantity of a product demanded and the price of the product. In general, price decreases as quantity demanded increases. The marginal price–demand function is the derivative of the price–demand function and it tells us how fast the price changes at a given level of production. These functions are used in business to determine the price–elasticity of demand, and to help companies determine whether changing production levels would be profitable.

Integrals Involving Logarithmic Functions

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 (x) = ln x

[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.

Key Concepts

Key Equations

In the following exercises, compute each indefinite integral.

\int e^{2 x} d x

\int e^{−3 x} d x

\frac{−1}{3} e^{−3 x} + C

\int 2^{x} d x

\int 3^{− x} d x

- \frac{3^{− x}}{ln 3} + C

\int \frac{1}{2 x} d x

\int \frac{2}{x} d x

ln (x^{2}) + C

\int \frac{1}{x^{2}} d x

\int \frac{1}{\sqrt{x}} d x

2 \sqrt{x} + C

In the following exercises, find each indefinite integral by using appropriate substitutions.

\int \frac{ln x}{x} d x

\int \frac{d x}{x {(ln x)}^{2}}

- \frac{1}{ln x} + C

\int \frac{d x}{x ln x} (x > 1)

\int \frac{d x}{x ln x ln (ln x)}

ln (ln (ln x)) + C

\int tan θ d θ

\int \frac{cos x - x sin x}{x cos x} d x

ln (x cos x) + C

\int \frac{ln (sin x)}{tan x} d x

\int ln (cos x) tan x d x

- \frac{1}{2} {(ln (cos (x)))}^{2} + C

\int x e^{− x^{2}} d x

\int x^{2} e^{− x^{3}} d x

\frac{− e^{− x^{3}}}{3} + C

\int e^{sin x} cos x d x

\int e^{tan x} {sec}^{2} x d x

e^{tan x} + C

\int e^{ln x} \frac{d x}{x}

\int \frac{e^{ln (1 - t)}}{1 - t} d t

t + C

In the following exercises, verify by differentiation that $\int ln x d x = x (ln x - 1) + C,$

then use appropriate changes of variables to compute the integral.

\int ln x d x

(H i n t : \int ln x d x = \frac{1}{2} \int x ln (x^{2}) d x)

\int x^{2} {ln}^{2} x d x

\frac{1}{9} x^{3} (ln (x^{3}) - 1) + C

\int \frac{ln x}{x^{2}} d x

(H i n t : Set u = \frac{1}{x} .)

\int \frac{ln x}{\sqrt{x}} d x

(H i n t : Set u = \sqrt{x} .)

2 \sqrt{x} (ln x - 2) + C

Write an integral to express the area under the graph of $y = \frac{1}{t}$

from $t = 1$

to e^x and evaluate the integral.

Write an integral to express the area under the graph of $y = e^{t}$

between $t = 0$

and $t = ln x,$

and evaluate the integral.

\int_{0}^{ln x} e^{t} d t = e^{t} {\|}_{0}^{ln x} = e^{ln x} - e^{0} = x - 1

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms.

\int tan (2 x) d x

\int \frac{sin (3 x) - cos (3 x)}{sin (3 x) + cos (3 x)} d x

- \frac{1}{3} ln (sin (3 x) + cos (3 x))

\int \frac{x sin (x^{2})}{cos (x^{2})} d x

\int x csc (x^{2}) d x

- \frac{1}{2} ln \| csc (x^{2}) + cot (x^{2}) \| + C

\int ln (cos x) tan x d x

\int ln (csc x) cot x d x

- \frac{1}{2} {(ln (csc x))}^{2} + C

\int \frac{e^{x} - e^{− x}}{e^{x} + e^{− x}} d x

In the following exercises, evaluate the definite integral.

\int_{1}^{2} \frac{1 + 2 x + x^{2}}{3 x + 3 x^{2} + x^{3}} d x

\frac{1}{3} ln (\frac{26}{7})

\int_{0}^{π / 4} tan x d x

\int_{0}^{π / 3} \frac{sin x - cos x}{sin x + cos x} d x

ln (\sqrt{3} - 1)

\int_{π / 6}^{π / 2} csc x d x

\int_{π / 4}^{π / 3} cot x d x

\frac{1}{2} ln \frac{3}{2}

In the following exercises, integrate using the indicated substitution.

\int \frac{x}{x - 100} d x; u = x - 100

\int \frac{y - 1}{y + 1} d y; u = y + 1

y - 2 ln \| y + 1 \| + C

\int \frac{1 - x^{2}}{3 x - x^{3}} d x; u = 3 x - x^{3}

\int \frac{sin x + cos x}{sin x - cos x} d x; u = sin x - cos x

ln \| sin x - cos x \| + C

\int e^{2 x} \sqrt{1 - e^{2 x}} d x; u = e^{2 x}

\int ln (x) \frac{\sqrt{1 - {(ln x)}^{2}}}{x} d x; u = ln x

- \frac{1}{3} {(1 - (ln x^{2}))}^{3 / 2} + C

In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate R₅₀ and solve for the exact area.

[T] $y = e^{x}$

over $[0, 1]$

[T] $y = e^{− x}$

over $[0, 1]$

Exact solution: $\frac{e - 1}{e}, R_{50} = 0.6258 .$

Since f is decreasing, the right endpoint estimate underestimates the area.

[T] $y = ln (x)$

over $[1, 2]$

[T] $y = \frac{x + 1}{x^{2} + 2 x + 6}$

over $[0, 1]$

Exact solution: $\frac{2 ln (3) - ln (6)}{2}, R_{50} = 0.2033 .$

Since f is increasing, the right endpoint estimate overestimates the area.

[T] $y = 2^{x}$

over $[−1, 0]$

[T] $y = − 2^{− x}$

over $[0, 1]$

Exact solution: $- \frac{1}{ln (4)}, R_{50} = −0.7164 .$

Since f is increasing, the right endpoint estimate overestimates the area (the actual area is a larger negative number).

In the following exercises, $f (x) \geq 0$

for $a \leq x \leq b .$

Find the area under the graph of $f (x)$

between the given values a and b by integrating.

f (x) = \frac{{log}_{10} (x)}{x}; a = 10, b = 100

f (x) = \frac{{log}_{2} (x)}{x}; a = 32, b = 64

\frac{11}{2} ln 2

f (x) = 2^{− x}; a = 1, b = 2

f (x) = 2^{− x}; a = 3, b = 4

\frac{1}{ln (65, 536)}

Find the area under the graph of the function $f (x) = x e^{− x^{2}}$

between $x = 0$

and $x = 5 .$

Compute the integral of $f (x) = x e^{− x^{2}}$

and find the smallest value of N such that the area under the graph $f (x) = x e^{− x^{2}}$

between $x = N$

and $x = N + 10$

is, at most, 0.01.

\int_{N}^{N + 1} x e^{− x^{2}} d x = \frac{1}{2} (e^{− N^{2}} - e^{− {(N + 1)}^{2}}) .

The quantity is less than 0.01 when $N = 2 .$

Find the limit, as N tends to infinity, of the area under the graph of $f (x) = x e^{− x^{2}}$

between $x = 0$

and $x = 5 .$

Show that $\int_{a}^{b} \frac{d t}{t} = \int_{1 / b}^{1 / a} \frac{d t}{t}$

when $0 < a \leq b .$

\int_{a}^{b} \frac{d x}{x} = ln (b) - ln (a) = ln (\frac{1}{a}) - ln (\frac{1}{b}) = \int_{1 / b}^{1 / a} \frac{d x}{x}

Suppose that $f (x) > 0$

for all x and that f and g are differentiable. Use the identity $f^{g} = e^{g ln f}$

and the chain rule to find the derivative of $f^{g} .$

Use the previous exercise to find the antiderivative of $h (x) = x^{x} (1 + ln x)$

and evaluate $\int_{2}^{3} x^{x} (1 + ln x) d x .$

Show that if $c > 0,$

then the integral of $1 / x$

from ac to bc $(0 < a < b)$

is the same as the integral of $1 / x$

from a to b.

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition $ln (x) = \int_{1}^{x} \frac{d t}{t},$

using properties of the definite integral and making no further assumptions.

Use the identity $ln (x) = \int_{1}^{x} \frac{d t}{t}$

to derive the identity $ln (\frac{1}{x}) = − ln x .$

We may assume that $x > 1, so \frac{1}{x} < 1 .$

Then, $\int_{1}^{1 / x} \frac{d t}{t} .$

Now make the substitution $u = \frac{1}{t},$

so $d u = - \frac{d t}{t^{2}}$

and $\frac{d u}{u} = - \frac{d t}{t},$

and change endpoints: $\int_{1}^{1 / x} \frac{d t}{t} = − \int_{1}^{x} \frac{d u}{u} = − ln x .$

Use a change of variable in the integral $\int_{1}^{x y} \frac{1}{t} d t$

to show that $ln x y = ln x + ln y for x, y > 0 .$

Use the identity $ln x = \int_{1}^{x} \frac{d t}{x}$

to show that $ln (x)$

is an increasing function of x on $[0, \infty),$

and use the previous exercises to show that the range of $ln (x)$

is $(− \infty, \infty) .$

Without any further assumptions, conclude that $ln (x)$

has an inverse function defined on $(− \infty, \infty) .$

Pretend, for the moment, that we do not know that $e^{x}$

is the inverse function of $ln (x),$

but keep in mind that $ln (x)$

has an inverse function defined on $(− \infty, \infty) .$

Call it E. Use the identity $ln x y = ln x + ln y$

to deduce that $E (a + b) = E (a) E (b)$

for any real numbers a, b.

Pretend, for the moment, that we do not know that $e^{x}$

is the inverse function of $ln x,$

but keep in mind that $ln x$

has an inverse function defined on $(− \infty, \infty) .$

Call it E. Show that $E' (t) = E (t) .$

x = E (ln (x)) .

Then, $1 = \frac{E' (ln x)}{x} or x = E' (ln x) .$

Since any number t can be written $t = ln x$

for some x, and for such t we have $x = E (t),$

it follows that for any $t, E' (t) = E (t) .$

The sine integral, defined as $S (x) = \int_{0}^{x} \frac{sin t}{t} d t$

is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x. Show that for $k \geq 1, \| S (2 π k) - S (2 π (k + 1)) \| \leq \frac{1}{k (2 k + 1) π} .$

(H i n t : sin (t + π) = − sin t)

[T] The normal distribution in probability is given by $p (x) = \frac{1}{σ \sqrt{2 π}} e^{− {(x - μ)}^{2} / 2 σ^{2}},$

where σ is the standard deviation and μ is the average. The standard normal distribution in probability, $p_{s},$

corresponds to $μ = 0 and σ = 1 .$

Compute the left endpoint estimates $R_{10} and R_{100}$

of $\int_{−1}^{1} \frac{1}{\sqrt{2 π}} e^{− x^{2 / 2}} d x .$

R_{10} = 0.6811, R_{100} = 0.6827

A graph of the function f(x) = .5 * ( sqrt(2)*e^(-.5x^2)) / sqrt(pi). It is a downward opening curve that is symmetric across the y axis, crossing at about (0, .4). It approaches 0 as x goes to positive and negative infinity. Between 1 and -1, ten rectangles are drawn for a right endpoint estimate of the area under the curve.

[T] Compute the right endpoint estimates $R_{50} and R_{100}$

of $\int_{−3}^{5} \frac{1}{2 \sqrt{2 π}} e^{− {(x - 1)}^{2} / 8} .$