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Since $e^x$ is its own derivative, it's also its own antiderivative. This makes integrating exponential functions surprisingly straightforward—once you recognize the pattern.
Exponential integrals appear throughout science and engineering: computing total population growth, finding areas under decay curves, calculating accumulated interest, and solving differential equations. Mastering these integrals is essential preparation for applications in later chapters.
| Integral | Result | Notes |
|---|---|---|
| $\int e^x \, dx$ | $e^x + C$ | Basic formula |
| $\int e^{ax} \, dx$ | $\frac{1}{a} e^{ax} + C$ | Linear exponent |
| $\int e^{u} \, du$ | $e^{u} + C$ | After substitution |
| $\int a^x \, dx$ | $\frac{a^x}{\ln a} + C$ | General base ($a > 0$, $a \neq 1$) |
$$\boxed{\int e^x \, dx = e^x + C}$$
Why this works: Since $\frac{d}{dx}[e^x] = e^x$, the antiderivative of $e^x$ is $e^x$.
Verification: $\frac{d}{dx}[e^x + C] = e^x$ ✓
For $\int e^{ax} \, dx$ where $a$ is a constant:
$$\boxed{\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C}$$
Quick derivation: Let $u = ax$, so $du = a \, dx$, which means $dx = \frac{1}{a} du$. $$\int e^{ax} \, dx = \int e^u \cdot \frac{1}{a} \, du = \frac{1}{a} e^u + C = \frac{1}{a} e^{ax} + C$$
Memory aid: "Divide by the coefficient of $x$."
Examples:
For more complex exponents, use $u$-substitution:
$$\boxed{\int e^{u} \cdot u' \, dx = e^{u} + C}$$
Recognition pattern: Look for $e^{(\text{something})}$ multiplied by the derivative of that something.
Example: $\int 2x e^{x^2} \, dx$
Here $u = x^2$ and $u' = 2x$. The $2x$ is already present! $$\int 2x e^{x^2} \, dx = e^{x^2} + C$$
For $f(x) = a^x$ where $a > 0$, $a \neq 1$:
$$\boxed{\int a^x \, dx = \frac{a^x}{\ln a} + C}$$
Verification: $\frac{d}{dx}\left[\frac{a^x}{\ln a}\right] = \frac{a^x \ln a}{\ln a} = a^x$ ✓
Examples:
| Integrand | Antiderivative | Key Step |
|---|---|---|
| $e^x$ | $e^x + C$ | Direct |
| $e^{kx}$ | $\frac{1}{k}e^{kx} + C$ | Divide by coefficient |
| $f'(x) e^{f(x)}$ | $e^{f(x)} + C$ | Recognize derivative pattern |
| $a^x$ | $\frac{a^x}{\ln a} + C$ | Divide by $\ln a$ |
When evaluating definite integrals, remember to apply the Fundamental Theorem:
$$\int_a^b e^x \, dx = e^x \Big\vert _a^b = e^b - e^a$$
Example: Area under $y = e^x$ from $x = 0$ to $x = 1$: $$\int_0^1 e^x \, dx = e^1 - e^0 = e - 1 \approx 1.718$$
Find the area under $y = e^{-3x}$ from $x = 0$ to $x = 1$.
$$A = \int_0^1 e^{-3x} \, dx = -\frac{1}{3} e^{-3x} \Big\vert _0^1$$ $$= -\frac{1}{3}(e^{-3} - e^0) = -\frac{1}{3}(e^{-3} - 1) = \frac{1}{3}(1 - e^{-3})$$ $$\approx \frac{1}{3}(1 - 0.0498) \approx 0.317$$
Evaluate $\int x^2 e^{x^3} \, dx$.
Let $u = x^3$, so $du = 3x^2 \, dx$, which gives $x^2 \, dx = \frac{1}{3} du$.
$$\int x^2 e^{x^3} \, dx = \int e^u \cdot \frac{1}{3} \, du = \frac{1}{3} e^u + C = \frac{1}{3} e^{x^3} + C$$
Verification: $\frac{d}{dx}\left[\frac{1}{3} e^{x^3}\right] = \frac{1}{3} \cdot e^{x^3} \cdot 3x^2 = x^2 e^{x^3}$ ✓
Evaluate $\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx$.
Let $u = \sqrt{x} = x^{1/2}$, so $du = \frac{1}{2} x^{-1/2} dx = \frac{1}{2\sqrt{x}} dx$.
This means $\frac{dx}{\sqrt{x}} = 2 \, du$.
$$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx = \int e^u \cdot 2 \, du = 2e^u + C = 2e^{\sqrt{x}} + C$$
| Mistake | Correct Approach |
|---|---|
| $\int e^{2x} \, dx = e^{2x} + C$ | Need the $\frac{1}{2}$: $\int e^{2x} \, dx = \frac{1}{2}e^{2x} + C$ |
| $\int e^{x^2} \, dx = \frac{1}{2x}e^{x^2} + C$ | Can't integrate $e^{x^2}$ without the $2x$ factor present |
| $\int 2^x \, dx = 2^{x+1} + C$ | Wrong! Need: $\int 2^x \, dx = \frac{2^x}{\ln 2} + C$ |
| Forgetting the constant $C$ | Every indefinite integral needs $+ C$ |
The integral $\int e^{x^2} \, dx$ cannot be expressed in terms of elementary functions. It requires the "error function" erf$(x)$.
To integrate $e^{(\text{something})}$, you need the derivative of "something" as a factor in the integrand.
Evaluate: (a) $\int e^{4x} \, dx$ (b) $\int 5e^x \, dx$ (c) $\int e^{-x} \, dx$
Evaluate: $$\int_0^2 e^{-x} \, dx$$
Evaluate $\int (3x^2 + 1)e^{x^3 + x} \, dx$.
Evaluate $\int 3^{2x} \, dx$.
Evaluate $\int x e^{x^2} \, dx$.
Oil leaks from a tank at a rate of $r(t) = 100e^{-0.5t}$ liters per minute. How much oil leaks during the first 10 minutes?
Question 1: What is $\int e^{3x} \, dx$?
(A) $e^{3x} + C$ (B) $3e^{3x} + C$ (C) $\frac{1}{3}e^{3x} + C$ (D) $\frac{e^{3x+1}}{3x+1} + C$
(C) Using $\int e^{kx} dx = \frac{1}{k}e^{kx} + C$ with $k = 3$. Verify by differentiating: $\frac{d}{dx}[\frac{1}{3}e^{3x}] = \frac{1}{3} \cdot 3e^{3x} = e^{3x}$ ✓
Question 2: Which integral CANNOT be evaluated using elementary functions?
(A) $\int xe^{x^2} dx$ (B) $\int e^{x^2} dx$ (C) $\int e^{2x} dx$ (D) $\int x^2 e^x dx$
(B) The integral $\int e^{x^2} dx$ has no elementary antiderivative. Options (A), (C), and (D) can all be evaluated: (A) by substitution, (C) directly, and (D) by integration by parts.
Question 3: If $\int_0^a e^x dx = e - 1$, what is $a$?
(A) $0$ (B) $1$ (C) $e$ (D) $e - 1$
(B) We have $\int_0^a e^x dx = e^x \big\vert _0^a = e^a - 1$. Setting $e^a - 1 = e - 1$ gives $e^a = e$, so $a = 1$.
The "Reverse Copy" Rule:
Since $e^x$ copies itself when differentiated, it also copies itself when integrated:
For $e^{kx}$, the chain rule factor $k$ must be "undone":
The "Missing Piece" Pattern:
When you see $e^{f(x)}$ in an integral, ask: "Is $f'(x)$ also present?"
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Last updated: 2026-01-23