1.7: Integrals Involving Exponential and Logarithmic Functions (2024)

Rule: Integrals of Exponential Functions

Exponential functions can be integrated using the following formulas.

\[ \begin{align} ∫e^x\,dx &= e^x+C \\[4pt] ∫a^x\,dx &=\dfrac{a^x}{\ln a}+C \end{align} \nonumber \]

Example \(\PageIndex{1}\): Finding an Antiderivative of an Exponential Function

Find the antiderivative of the exponential function \(e^{−x}\).

Solution

Use substitution, setting \(u=−x,\) and then \(du=−1\,dx\). Multiply the \(du\) equation by \(−1\), so you now have \(−du=\,dx\). Then,

\[∫e^{−x}\,dx=−∫e^u\,du=−e^u+C=−e^{−x}+C. \nonumber \]

Exercise \(\PageIndex{1}\)

Find the antiderivative of the function using substitution: \(x^2e^{−2x^3}\).

Hint

Let \(u\) equal the exponent on \(e\).

Answer

\(\displaystyle ∫x^2e^{−2x^3}\,dx=−\dfrac{1}{6}e^{−2x^3}+C\)

Example \(\PageIndex{2}\): Square Root of an Exponential Function

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

Solution

First rewrite the problem using a rational exponent:

\[∫e^x\sqrt{1+e^x}\,dx=∫e^x(1+e^x)^{1/2}\,dx.\nonumber \]

Using substitution, choose \(u=1+e^x\). Then, \(du=e^x\,dx\). We have

\[∫e^x(1+e^x)^{1/2}\,dx=∫u^{1/2}\,du.\nonumber \]

Then

\[∫u^{1/2}\,du=\dfrac{u^{3/2}}{3/2}+C=\dfrac{2}{3}u^{3/2}+C=\dfrac{2}{3}(1+e^x)^{3/2}+C\nonumber \]

Exercise \(\PageIndex{2}\)

Find the antiderivative of \(e^x(3e^x−2)^2\).

Hint

Let \(u=3e^x−2\).

Answer

\(\displaystyle ∫e^x(3e^x−2)^2\,dx=\dfrac{1}{9}(3e^x−2)^3+C\)

Example \(\PageIndex{3}\): Using Substitution with an Exponential Function

Use substitution to evaluate the indefinite integral \(\displaystyle ∫3x^2e^{2x^3}\,dx.\)

Solution

Here we choose to let \(u\) equal the expression in the exponent on \(e\). Let \(u=2x^3\) and \(du=6x^2\,dx\). Again, \(du\) is off by a constant multiplier; the original function contains a factor of \(3x^2,\) not \(6x^2\). Multiply both sides of the equation by \(\dfrac{1}{2}\) so that the integrand in \(u\) equals the integrand in \(x\). Thus,

\[∫3x^2e^{2x^3}\,dx=\frac{1}{2}∫e^u\,du. \nonumber \]

Integrate the expression in \(u\) and then substitute the original expression in \(x\) back into the \(u\)-integral:

\[\frac{1}{2}∫e^u\,du=\frac{1}{2}e^u+C=\frac{1}{2}e^{2x^3}+C. \nonumber \]

Exercise \(\PageIndex{3}\)

Evaluate the indefinite integral \(\displaystyle ∫2x^3e^{x^4}\,dx\).

Hint

Let \(u=x^4.\)

Answer

\(\displaystyle ∫2x^3e^{x^4}\,dx=\frac{1}{2}e^{x^4}+C\)

Example \(\PageIndex{4}\): Finding a Price–Demand Equation

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, \(p′(x),\) for \(x\) number of tubes per week, is given as

\[p'(x)=−0.015e^{−0.01x}. \nonumber \]

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

Solution

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

\[p(x)=∫−0.015e^{−0.01x}\,dx=−0.015∫e^{−0.01x}\,dx. \nonumber \]

Using substitution, let \(u=−0.01x\) and \(du=−0.01\,dx\). Then, divide both sides of the \(du\) equation by \(−0.01\). This gives

\[\dfrac{−0.015}{−0.01}∫e^u\,du=1.5∫e^u\,du=1.5e^u+C=1.5e^{−0.01}x+C. \nonumber \]

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

\[p(50)=1.5e^{−0.01(50)}+C=2.35. \nonumber \]

Now, just solve for \(C\):

\[C=2.35−1.5e^{−0.5}=2.35−0.91=1.44. \nonumber \]

Thus,

\[p(x)=1.5e^{−0.01x}+1.44. \nonumber \]

If the supermarket sells \(100\) tubes of toothpaste per week, the price would be

\[p(100)=1.5e−0.01(100)+1.44=1.5e−1+1.44≈1.99. \nonumber \]

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

Example \(\PageIndex{5}\): Evaluating a Definite Integral Involving an Exponential Function

Evaluate the definite integral \(\displaystyle ∫^2_1e^{1−x}\,dx.\)

Solution

Again, substitution is the method to use. Let \(u=1−x,\) so \(\,du=−1\,dx\) or \(−\,du=\,dx\). Then \(\displaystyle ∫e^{1−x}\,dx=−∫e^u\,du.\)

Next, change the limits of integration. Using the equation \(u=1−x\), we have:

\[\text{When }x = 1, \quad u=1−(1)=0, \nonumber \]

\[\text{and when }x = 2, \quad u=1−(2)=−1. \nonumber \]

The integral then becomes

\[\begin{align*} ∫^2_1e^{1−x}\,\,dx &= −∫^{−1}_0e^u\,\,du \\[4pt] &=∫^0_{−1}e^u\,\,du \\[4pt] &=e^u\bigg|^0_{−1}=e^0−(e^{−1}) \\[4pt] &=−e^{−1}+1. \end{align*}\]

See Figure \(\PageIndex{2}\).

Exercise \(\PageIndex{4}\)

Evaluate \(\displaystyle ∫^2_0e^{2x}\,dx.\)

Hint

Let \(u=2x.\)

Answer

\(\displaystyle \frac{1}{2}∫^4_0e^u\,du=\dfrac{1}{2}(e^4−1)\)

Example \(\PageIndex{6}\): Growth of Bacteria in a Culture

Suppose the rate of growth of bacteria 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?

Solution

We have

\[Q(t)=∫3^tdt=\dfrac{3^t}{\ln 3}+C. \nonumber \]

Then, at \(t=0\) we have \(Q(0)=10=\dfrac{1}{\ln 3}+C,\) so \(C≈9.090\) and we get

\[Q(t)=\dfrac{3^t}{\ln 3}+9.090. \nonumber \]

At time \(t=2\), we have

\[\begin{align*} Q(2) &=\dfrac{3^2}{\ln 3}+9.090 \\[4pt] &\approx 17.282. \end{align*}\]

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

Exercise \(\PageIndex{5}\)

From Example, suppose the bacteria grow at a rate of \(q(t)=2^t\). Assume the culture still starts with \(10,000\) bacteria. Find \(Q(t)\). How many bacteria are in the dish after \(3\) hours?

Hint

Use the procedure from Example \(\PageIndex{6}\) to solve the problem

Answer

\[\begin{align*} Q(t) &= \dfrac{2^t}{\ln 2} + 8.557. \\[4pt] Q(3) &\approx 20,099 \end{align*}\]

So there are \(20,099\) bacteria in the dish after \(3\) hours.

Example \(\PageIndex{7}\): Fruit Fly Population Growth

Suppose a population of fruit flies increases at a rate of \(g(t)=2e^{0.02t}\), in flies per day. If the initial population of fruit flies is \(100\) flies, how many flies are in the population after \(10\) days?

Solution

Let \(G(t)\) represent the number of flies in the population at time \(t\). Applying the net change theorem, we have

\[ \begin{align*} G(10)=G(0)+∫^{10}_02e^{0.02t}\,dt \\[4pt] &=100+\left[\dfrac{2}{0.02}e^{0.02t}\right]∣^{10}_0 \\[4pt] &=100+\left[100e^{0.02t}\right]∣^{10}_0 \\[4pt] &=100+100e^{0.2}−100 \\[4pt] &≈122. \end{align*}\]

There are \(122\) flies in the population after \(10\) days.

Exercise \(\PageIndex{6}\)

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

Hint

Use the process from Example \(\PageIndex{7}\) to solve the problem.

Answer

There are \(116\) flies.

Example \(\PageIndex{8}\): Evaluating a Definite Integral Using Substitution

Evaluate the definite integral using substitution: \[∫^2_1\dfrac{e^{1/x}}{x^2}\,dx.\nonumber \]

Solution

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

\[∫^2_1\dfrac{e^{1/x}}{x^2}\,\,dx=∫^2_1e^{x^{−1}}x^{−2}\,dx. \nonumber \]

Let \(u=x^{−1},\) the exponent on \(e\). Then

\[du=−x^{−2}\,dx \nonumber \]

\[−du=x^{−2}\,dx. \nonumber \]

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

\[−∫e^u\,du. \nonumber \]

Next, change the limits of integration:

\[u=(1)^{−1}=1 \nonumber \]

\[u=(2)^{−1}=\dfrac{1}{2}. \nonumber \]

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

\[−∫^{1/2}_1e^u\,du=∫^1_{1/2}e^u\,du=e^u\big|^1_{1/2}=e−e^{1/2}=e−\sqrt{e}.\nonumber \]

Exercise \(\PageIndex{7}\)

Evaluate the definite integral using substitution: \[∫^2_1\dfrac{1}{x^3}e^{4x^{−2}}\,dx.\nonumber \]

Hint

Let \(u=4x^{−2}.\)

Answer

\(\displaystyle ∫^2_1\dfrac{1}{x^3}e^{4x^{−2}}\,dx=\dfrac{1}{8}[e^4−e]\).

Rule: Integration Formulas Involving Logarithmic Functions

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

\[\begin{align*} ∫x^{−1}\,dx &=\ln |x|+C \\[4pt] ∫\ln x\,\,dx &= x\ln x−x+C =x (\ln x−1)+C \\[4pt] ∫\log_a x\,dx &=\dfrac{x}{\ln a}(\ln x−1)+C \end{align*}\]

Example \(\PageIndex{9}\): Finding an Antiderivative Involving \(\ln x\)

Find the antiderivative of the function \(\dfrac{3}{x−10}. \)

Solution

First factor the \(3\) outside the integral symbol. Then use the \(u^{−1}\) rule. Thus,

\[∫\dfrac{3}{x−10}\,dx=3∫\dfrac{1}{x−10}\,dx=3∫\dfrac{du}{u}=3\ln |u|+C=3\ln |x−10|+C,\quad x≠10. \nonumber \]

See Figure \(\PageIndex{3}\).

Exercise \(\PageIndex{8}\)

Find the antiderivative of \(\dfrac{1}{x+2}.\)

Hint

Follow the pattern from Example \(\PageIndex{9}\) to solve the problem.

Answer

\(\displaystyle \int \dfrac{1}{x+2}\,dx = \ln |x+2|+C\)

Example \(\PageIndex{10}\): Finding an Antiderivative of a Rational Function

Find the antiderivative of \(\dfrac{2x^3+3x}{x^4+3x^2}. \)

Solution

This can be rewritten as \(\displaystyle ∫(2x^3+3x)(x^4+3x^2)^{−1}\,dx.\) Use substitution.

Let \(u=x^4+3x^2\), then \(du=(4x^3+6x)\,dx.\) Alter \(du\) by factoring out the \(2\). Thus,

\[du=(4x^3+6x)\,dx=2(2x^3+3x)\,dx \nonumber \]

\[\dfrac{1}{2}\,du=(2x^3+3x)\,dx. \nonumber \]

Rewrite the integrand in \(u\):

\[∫(2x^3+3x)(x^4+3x^2)^{−1}\,dx=\dfrac{1}{2}∫u^{−1}\,du. \nonumber \]

Then we have

\[\dfrac{1}{2}∫u^{−1}\,du=\dfrac{1}{2}\ln |u|+C=\dfrac{1}{2}\ln ∣x^4+3x^2∣+C. \nonumber \]

Example \(\PageIndex{11}\): Finding an Antiderivative of a Logarithmic Function

Find the antiderivative of the log function \(\log_2 x.\)

Solution

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

\[∫\log_2 x\,dx=\dfrac{x}{\ln 2}(\ln x−1)+C.\nonumber \]

Exercise \(\PageIndex{9}\)

Find the antiderivative of \(\log_3 x\).

Hint

Follow Example \(\PageIndex{11}\) and refer to the rule on integration formulas involving logarithmic functions.

Answer

\(\displaystyle ∫\log_3 x\,dx=\dfrac{x}{\ln 3}(\ln x−1)+C\)

Example \(\PageIndex{12}\): Evaluating a Definite Integral

Evaluate the definite integral \[∫^{π/2}_0\dfrac{\sin x}{1+\cos x}\,dx.\nonumber \]

Solution

We need substitution to evaluate this problem. Let \(u=1+\cos x\) so \(du=−\sin x\,\,dx.\)

Rewrite the integral in terms of \(u\), changing the limits of integration as well. Thus,

\[ \begin{align*} u &= 1+\cos(0)=2 \\[4pt] u &=1+\cos \left(\dfrac{π}{2}\right)=1.\end{align*}\]

Then

\[ \begin{align*}∫^{π/2}_0\dfrac{\sin x}{1+\cos x} &=−∫^1_2 u^{−1}\,du \\[4pt] &=∫^2_1u^{−1}\,du \\[4pt] &=\ln |u|\,\bigg|^2_1 \\[4pt] &=[\ln 2−\ln 1]=\ln 2 \end{align*}\]