2.7 Integrals, Exponential Functions, and Logarithms

Natural Logarithms and Exponential Functions

The natural logarithm and the exponential function form the backbone of most integration techniques you'll encounter in Calculus II. The natural log gives us a way to integrate $\frac{1}{x}$, and $e^x$ is its own antiderivative. These two facts alone unlock a huge number of problems.

Definition of the Natural Logarithm

The natural logarithm is defined as an integral:

$\ln(x) = \int_1^x \frac{1}{t}\, dt$

This means $\ln(x)$ equals the area under the curve $\frac{1}{t}$ from $t = 1$ to $t = x$. That's not just a property; it's the definition you'll use to derive everything else.

The base of the natural logarithm is Euler's number $e \approx 2.71828$, defined as the unique number satisfying $\ln(e) = 1$. Equivalently, $e$ arises from the limit:

$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$

The natural exponential function $e^x$ and natural logarithm $\ln(x)$ are inverse functions, so they undo each other:

• $e^{\ln(x)} = x$ for $x > 0$
• $\ln(e^x) = x$ for all $x$

Differentiation and Antidifferentiation Rules

These four results come up constantly:

Notice the absolute value in $\ln|x| + C$. That's there because $\frac{1}{x}$ is defined for negative $x$ too, but $\ln(x)$ only accepts positive inputs. The absolute value extends the antiderivative to both sides of the domain.

The fact that $e^x$ is its own derivative (and its own antiderivative) is what makes it unique. No other function has this property.

Properties for Simplifying Integrals

Before integrating, you'll often need to rewrite expressions using these properties.

Logarithm properties:

• Product rule: $\ln(xy) = \ln(x) + \ln(y)$
• Quotient rule: $\ln\!\left(\frac{x}{y}\right) = \ln(x) - \ln(y)$
• Power rule: $\ln(x^n) = n\ln(x)$

Exponential properties:

• $e^{x+y} = e^x \cdot e^y$
• $e^{x-y} = \frac{e^x}{e^y}$
• $(e^x)^n = e^{nx}$

These aren't just algebra review. You'll actively use the power rule, for instance, to pull exponents out front before integrating, and the product/quotient rules to split complicated logarithmic expressions into simpler pieces.

Converting Between Bases

Most calculus work uses base $e$, so you need to convert other bases before differentiating or integrating.

Change of base formula for logarithms:

$\log_b(x) = \frac{\ln(x)}{\ln(b)}$

Since $\frac{1}{\ln(b)}$ is just a constant, this turns any base-$b$ logarithm into a constant times $\ln(x)$, which you already know how to handle.

Converting general exponentials:

$b^x = e^{x \ln(b)}$

This works because $b = e^{\ln(b)}$, so $b^x = (e^{\ln(b)})^x = e^{x\ln(b)}$. Once it's in the form $e^{kx}$, you can integrate using the standard rule $\int e^{kx}\, dx = \frac{1}{k}e^{kx} + C$.

Integration Techniques and Applications

Integration of Logarithmic and Exponential Expressions

Substitution is your first tool to try. A few common patterns:

• If you see $\frac{f'(x)}{f(x)}$, substitute $u = f(x)$. The integral becomes $\ln|u| + C = \ln|f(x)| + C$.
• If the exponent of $e$ is a function of $x$, try substituting $u$ equal to that exponent. For example, in $\int 2x\, e^{x^2}\, dx$, let $u = x^2$.

Integration by parts handles products where substitution doesn't simplify things. Here are two essential examples:

Example 1: $\int x\, e^x\, dx$

1. Let $u = x$ and $dv = e^x\, dx$
2. Then $du = dx$ and $v = e^x$
3. Apply the formula $\int u\, dv = uv - \int v\, du$:

$x\, e^x - \int e^x\, dx = x\, e^x - e^x + C$

Example 2: $\int \ln(x)\, dx$

1. Let $u = \ln(x)$ and $dv = dx$
2. Then $du = \frac{1}{x}\, dx$ and $v = x$
3. Apply the formula: $x\ln(x) - \int x \cdot \frac{1}{x}\, dx = x\ln(x) - x + C$

The trick with $\int \ln(x)\, dx$ is recognizing that you should set $dv = dx$. There's no obvious "second function," but integration by parts still works.

Behavior of Logarithmic and Exponential Functions

Understanding the graphs helps you check whether your answers make sense.

• $\ln(x)$: Domain $(0, \infty)$, range $(-\infty, \infty)$. Increasing for all $x > 0$, but slowly. Vertical asymptote at $x = 0$. Passes through $(1, 0)$.
• $e^x$: Domain $(-\infty, \infty)$, range $(0, \infty)$. Always increasing and always positive. Horizontal asymptote at $y = 0$ as $x \to -\infty$. Passes through $(0, 1)$.

Since $e^x$ grows faster than any polynomial and $\ln(x)$ grows slower than any polynomial, limits involving these functions often produce indeterminate forms like $\frac{\infty}{\infty}$ or $0 \cdot \infty$. L'Hôpital's rule resolves these by differentiating the numerator and denominator separately.

Applications of Logarithmic and Exponential Integrals

Exponential growth and decay models take the form $A(t) = A_0\, e^{kt}$, where $A_0$ is the initial quantity and $k$ is the growth ($k > 0$) or decay ($k < 0$) rate. Integrating this function over an interval gives the total accumulated quantity. For example, $\int_0^{10} A_0\, e^{kt}\, dt$ gives the total output over the first 10 time units.

These integrals appear in population growth, radioactive decay, continuously compounded interest, and Newton's law of cooling.

Fundamental Theorem of Calculus and Types of Integrals

Fundamental Theorem of Calculus

The FTC connects differentiation and integration, and both parts are directly relevant to working with $\ln(x)$ and $e^x$.

Part 1 (FTC1): If $F(x) = \int_a^x f(t)\, dt$, then $F'(x) = f(x)$. The derivative of a definite integral with respect to its upper limit equals the integrand evaluated at that limit. This is exactly how we know $\frac{d}{dx}[\ln(x)] = \frac{1}{x}$, since $\ln(x)$ is defined as $\int_1^x \frac{1}{t}\, dt$.

Part 2 (FTC2): If $F$ is any antiderivative of $f$, then $\int_a^b f(x)\, dx = F(b) - F(a)$. This is the evaluation theorem you use every time you compute a definite integral.

Types of Integrals

• Indefinite integrals represent the general antiderivative: $\int f(x)\, dx = F(x) + C$. The $+C$ accounts for the family of all antiderivatives.
• Definite integrals compute the signed area between the function and the $x$-axis over $[a, b]$. You evaluate them using FTC2: find an antiderivative, then compute $F(b) - F(a)$.