Integrals Involving Exponential and Logarithmic Functions: Learn It 1

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Integrals Involving Exponential and Logarithmic Functions: Learn It 1

Perform integrations on functions that include exponential terms
Solve integrals that feature 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, $y={e}^{x},$ is its own derivative and its own integral.

integrals of exponential functions

Exponential functions can be integrated using the following formulas.

\begin{array}{ccc} \int e^{x} d x & = & e^{x} + C \\ \int a^{x} d x & = & \frac{a^{x}}{ln a} + C \end{array}

$\begin{array}{ccc} {\displaystyle\int{e}^{x}dx} & {=} & {{e}^{x}+C} \\ {\displaystyle\int{a}^{x}dx} & {=} & {\dfrac{{a}^{x}}{\text{ln}a}+C}\end{array}$

The nature of the antiderivative of ${e}^{x}$ makes it fairly easy to identify what to choose as $u$ .

If only one $e$ exists, choose the exponent of $e$ as $u$ . If more than one $e$ exists, choose the more complicated function involving $e$ as $u$ .

Find the antiderivative of the exponential function $e^{-x}$ .

Use substitution, setting $u=\text{−}x,$ and then $du=-1dx.$

Multiply the du equation by $−1$ , so you now have $\text{−}du=dx.$

Then,

$\begin{array}{cc}{\displaystyle\int {e}^{\text{−}x}dx}\hfill & =\text{−}{\displaystyle\int {e}^{u}du}\hfill \\ & =\text{−}{e}^{u}+C\hfill \\ & =\text{−}{e}^{\text{−}x}+C.\hfill \end{array}$

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.

Find the antiderivative of the exponential function ${e}^{x}\sqrt{1+{e}^{x}}.$

First rewrite the problem using a rational exponent:

\int e^{x} \sqrt{1 + e^{x}} d x = \int e^{x} {(1 + e^{x})}^{1 / 2} d x .

$\displaystyle\int {e}^{x}\sqrt{1+{e}^{x}}dx=\displaystyle\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx.$

Using substitution, choose $u=1+{e}^{x}.u=1+{e}^{x}.$

Then, $du={e}^{x}dx.$ We have,

\int e^{x} {(1 + e^{x})}^{1 / 2} d x = \int u^{1 / 2} d u .

$\displaystyle\int {e}^{x}{(1+{e}^{x})}^{1\text{/}2}dx=\displaystyle\int {u}^{1\text{/}2}du.$

Then

\int u^{1 / 2} d u = \frac{u^{3 / 2}}{3 / 2} + C = \frac{2}{3} u^{3 / 2} + C = \frac{2}{3} {(1 + e^{x})}^{3 / 2} + C .

$\displaystyle\int {u}^{1\text{/}2}du=\frac{{u}^{3\text{/}2}}{3\text{/}2}+C=\frac{2}{3}{u}^{3\text{/}2}+C=\frac{2}{3}{(1+{e}^{x})}^{3\text{/}2}+C.$

A graph of the function f(x) = e^x * sqrt(1 + e^x), which is an increasing concave up curve, over [-3, 1]. It begins close to the x axis in quadrant two, crosses the y axis at (0, sqrt(2)), and continues to increase rapidly. — Figure 1. The graph shows an exponential function times the square root of an exponential function.

Use substitution to evaluate the indefinite integral $\displaystyle\int 3{x}^{2}{e}^{2{x}^{3}}dx.$

Here we choose to let $u$ equal the expression in the exponent on $e$ .

Let $u=2{x}^{3}$ and $du=6{x}^{2}dx..$ Again, du is off by a constant multiplier; the original function contains a factor of 3 $x^2$ , not 6 $x^2$ .

Multiply both sides of the equation by $\frac{1}{2}$ so that the integrand in $u$ equals the integrand in $x$ .

Thus,

\int 3 x^{2} e^{2 x^{3}} d x = \frac{1}{2} \int e^{u} d u .

$\displaystyle\int 3{x}^{2}{e}^{2{x}^{3}}dx=\frac{1}{2}\displaystyle\int {e}^{u}du.$

Integrate the expression in $u$ and then substitute the original expression in $x$ back into the $u$ integral:

\frac{1}{2} \int e^{u} d u = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{2 x^{3}} + C .

$\frac{1}{2}\displaystyle\int {e}^{u}du=\frac{1}{2}{e}^{u}+C=\frac{1}{2}{e}^{2{x}^{3}}+C.$

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.

Find the price–demand equation for a particular brand of toothpaste at a supermarket chain when the demand is $50$ tubes per week at $$2.35$ per tube, given that the marginal price—demand function, ${p}^{\prime }(x),$ for $x$ number of tubes per week, is given as

p' (x) = - 0.015 e^{- 0.01 x} .

$p\text{'}(x)=-0.015{e}^{-0.01x}.$

If the supermarket chain sells $100$ tubes per week, what price should it set?

To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative, then look at the particulars.

Thus,

\begin{matrix} p (x) & = \int - 0.015 e^{- 0.01 x} d x \\ = - 0.015 \int e^{- 0.01 x} d x . \end{matrix}

$\begin{array}{}\\ \\ p(x)\hfill & =\int -0.015{e}^{-0.01x}dx\hfill \\ & =-0.015\int {e}^{-0.01x}dx.\hfill \end{array}$

Using substitution, let $u=-0.01x$ and $du=-0.01dx.$ Then, divide both sides of the du equation by $−0.01$ .

This gives,

\begin{array}{cc} \frac{- 0.015}{- 0.01} \int e^{u} d u & = 1.5 \int e^{u} d u \\ = 1.5 e^{u} + C \\ = 1.5 e^{- 0.01 x} + C . \end{array}

$\begin{array}{cc}\frac{-0.015}{-0.01}\int {e}^{u}du\hfill & =1.5\int {e}^{u}du\hfill \\ \\ & =1.5{e}^{u}+C\hfill \\ & =1.5{e}^{-0.01x}+C.\hfill \end{array}$

The next step is to solve for $C$ . We know that when the price is $$2.35$ per tube, the demand is $50$ tubes per week. This means

\begin{matrix} p (50) & = 1.5 e^{- 0.01 (50)} + C \\ = 2.35. \end{matrix}

$\begin{array}{}\\ \\ p(50)\hfill & =1.5{e}^{-0.01(50)}+C\hfill \\ & =2.35.\hfill \end{array}$

Now, just solve for $C$ :

\begin{matrix} C & = 2.35 - 1.5 e^{- 0.5} \\ = 2.35 - 0.91 \\ = 1.44. \end{matrix}

$\begin{array}{}\\ C\hfill & =2.35-1.5{e}^{-0.5}\hfill \\ & =2.35-0.91\hfill \\ & =1.44.\hfill \end{array}$

Thus,

p (x) = 1.5 e^{- 0.01 x} + 1.44.

$p(x)=1.5{e}^{-0.01x}+1.44.$

If the supermarket sells $100$ tubes of toothpaste per week, the price would be,

p (100) = 1.5 e^{- 0.01 (100)} + 1.44 = 1.5 e^{- 1} + 1.44 \approx 1.99.

$p(100)=1.5{e}^{-0.01(100)}+1.44=1.5{e}^{-1}+1.44\approx 1.99.$

The supermarket should charge $$1.99$ per tube if it is selling $100$ tubes per week.

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.6.1” here (opens in new window).

Evaluate the definite integral ${\displaystyle\int }_{1}^{2}{e}^{1-x}dx.$

Again, substitution is the method to use. Let $u=1-x,$ so $du=-1dx$ or $\text{−}du=dx.$ Then $\displaystyle\int {e}^{1-x}dx=\text{−}\displaystyle\int {e}^{u}du.$ Next, change the limits of integration. Using the equation $u=1-x,$ we have

\begin{matrix} u = 1 - (1) = 0 \\ u = 1 - (2) = - 1. \end{matrix}

$\begin{array}{c}u=1-(1)=0\hfill \\ u=1-(2)=-1.\hfill \end{array}$

The integral then becomes

\begin{array}{cc} \int_{1}^{2} e^{1 - x} d x & = − \int_{0}^{- 1} e^{u} d u \\ = \int_{- 1}^{0} e^{u} d u \\ = {e^{u} |}_{- 1}^{0} \\ = e^{0} - (e^{- 1}) \\ = − e^{- 1} + 1. \end{array}

$\begin{array}{cc}{\displaystyle\int }_{1}^{2}{e}^{1-x}dx\hfill & =\text{−}{\displaystyle\int }_{0}^{-1}{e}^{u}du\hfill \\ & ={\displaystyle\int }_{-1}^{0}{e}^{u}du\hfill \\ & ={{e}^{u}|}_{-1}^{0}\hfill \\ & ={e}^{0}-({e}^{-1})\hfill \\ & =\text{−}{e}^{-1}+1.\hfill \end{array}$

A graph of the function f(x) = e^(1-x) over [0, 3]. It crosses the y axis at (0, e) as a decreasing concave up curve and symptotically approaches 0 as x goes to infinity. — Figure 2. The indicated area can be calculated by evaluating a definite integral using substitution.

Suppose the rate of in a Petri dish is given by $q(t)={3}^{t},$ where $t$ is given in hours and $q(t)$ is given in thousands of bacteria per hour. If a culture starts with $10,000$ bacteria, find a function $Q(t)$ that gives the number of bacteria in the Petri dish at any time $t$ . How many bacteria are in the dish after $2$ hours?

We have,

$Q(t)=\int {3}^{t}dt=\frac{{3}^{t}}{\text{ln}3}+C.$

Then, at $t=0$ we have

$Q(0)=10=\frac{1}{\text{ln}3}+C,$

so $C\approx 9.090$ and we get,

$Q(t)=\frac{{3}^{t}}{\text{ln}3}+9.090.$

Note: We are using $10$ in place of $10,000$ since $10 ,000$ bacteria are $10$ thousands of bacteria. We will multiple our final answer by a power of $1000$ at the end of our calculation to account for this.

At time $t=2,$ we have,

Q (2) = \frac{3^{2}}{ln 3} + 9.090 = 17.282

$Q(2)=\frac{{3}^{2}}{\text{ln}3}+9.090=17.282$

After $2$ hours, there are $17,282$ bacteria in the dish.

Evaluate the definite integral using substitution:

${\displaystyle\int }_{1}^{2}\dfrac{{e}^{1\text{/}x}}{{x}^{2}}dx.$

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 / x}}{x^{2}} d x = \int_{1}^{2} e^{x^{- 1}} x^{- 2} d x .

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

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

Then,

\begin{array}{cc} d u & = − x^{- 2} d x \\ - d u & = x^{- 2} d x . \end{array}

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

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

− \int e^{u} d u .

$\text{−}\displaystyle\int {e}^{u}du.$

Next, change the limits of integration:

\begin{matrix} u = {(1)}^{- 1} = 1 \\ u = {(2)}^{- 1} = \frac{1}{2} . \end{matrix}

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

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

Thus,

\begin{array}{cc} − \int_{1}^{1 / 2} e^{u} d u & = \int_{1 / 2}^{1} e^{u} d u \\ = e^{u} |_{1 / 2}^{1} \\ = e - e^{1 / 2} \\ = e - \sqrt{e} . \end{array}

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