An antiderivative of a function $f(x)$ is any function $F(x)$ such that $F'(x) = f(x)$. The process of finding antiderivatives is called integration or antidifferentiation.
The indefinite integral of $f(x)$ is the family of all antiderivatives:
$$\int f(x)\,dx = F(x) + C$$
where $F'(x) = f(x)$ and $C$ is the constant of integration (any real number). Because the derivative of a constant is 0, every antiderivative differs from another by only a constant.
| Rule | Formula |
|---|---|
| Power Rule ($n \neq -1$) | $\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ |
| Constant Multiple | $\displaystyle\int kf(x)\,dx = k\int f(x)\,dx$ |
| Sum/Difference | $\displaystyle\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$ |
| Natural Exponential | $\displaystyle\int e^x\,dx = e^x + C$ |
| Natural Log | $\displaystyle\int \frac{1}{x}\,dx = \ln|x| + C$ |
| Sine / Cosine | $\displaystyle\int \sin x\,dx = -\cos x + C$; $\displaystyle\int \cos x\,dx = \sin x + C$ |
| $\sec^2 x$ | $\displaystyle\int \sec^2 x\,dx = \tan x + C$ |
(a) $\displaystyle\int (3x^2 - 5x + 2)\,dx$
Apply power rule to each term:
$= \dfrac{3x^3}{3} - \dfrac{5x^2}{2} + 2x + C = \mathbf{x^3 - \dfrac{5}{2}x^2 + 2x + C}$
(b) $\displaystyle\int \left(4e^x + \frac{3}{x}\right)dx$
$= 4e^x + 3\ln|x| + C$
(c) $\displaystyle\int \sqrt{x}\,dx = \int x^{1/2}\,dx = \dfrac{x^{3/2}}{3/2} + C = \dfrac{2}{3}x^{3/2} + C$
Evaluate: $\displaystyle\int (6x^2 + \cos x - e^x)\,dx$
Given an initial condition, we can solve for $C$ to find the unique antiderivative satisfying that condition.
Find $f(x)$ if $f'(x) = 4x^3 - 2x$ and $f(1) = 5$.
Step 1 — Integrate: $f(x) = \int (4x^3 - 2x)\,dx = x^4 - x^2 + C$
Step 2 — Apply initial condition: $f(1) = 1 - 1 + C = 5 \Rightarrow C = 5$
Answer: $f(x) = x^4 - x^2 + 5$
The definite integral $\displaystyle\int_a^b f(x)\,dx$ represents the net area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$. Area above the $x$-axis is positive; area below is negative.
A Riemann sum approximates the definite integral by dividing $[a,b]$ into $n$ subintervals of width $\Delta x = \dfrac{b-a}{n}$ and summing rectangle areas:
$$\int_a^b f(x)\,dx \approx \sum_{i=1}^{n} f(x_i^*)\,\Delta x$$
where $x_i^*$ is a sample point in the $i$th subinterval (left endpoint, right endpoint, or midpoint).
Interactive: Riemann sum approximation — adjust $n$ to see rectangles approach the true area
Figure 3.1 — Left Riemann Sum Approximation for $f(x) = x^2$ on $[0, 2]$
Approximate $\displaystyle\int_0^2 x^2\,dx$ using $n = 4$ left-endpoint rectangles.
$\Delta x = (2-0)/4 = 0.5$. Left endpoints: $x = 0, 0.5, 1, 1.5$.
$$L_4 = \Delta x[f(0) + f(0.5) + f(1) + f(1.5)]$$ $$= 0.5[0 + 0.25 + 1 + 2.25] = 0.5(3.5) = \mathbf{1.75}$$
The exact value is $\dfrac{2^3}{3} = \dfrac{8}{3} \approx 2.667$. The left sum underestimates because $f(x) = x^2$ is increasing on $[0,2]$.
The FTC connects differentiation and integration — the two core operations of calculus. It has two parts.
If $f$ is continuous on $[a, b]$, then the function $g(x) = \displaystyle\int_a^x f(t)\,dt$ is differentiable on $(a, b)$ and:
$$g'(x) = \frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$
Integration and differentiation are inverse operations — differentiating an integral with respect to its upper limit returns the integrand.
If $F$ is an antiderivative of $f$ on $[a, b]$, then:
$$\int_a^b f(x)\,dx = F(b) - F(a) = \Big[F(x)\Big]_a^b$$
This is the primary tool for evaluating definite integrals exactly.
(a) $\displaystyle\int_1^3 (2x + 1)\,dx$
$F(x) = x^2 + x$
$\Big[x^2 + x\Big]_1^3 = (9 + 3) - (1 + 1) = 12 - 2 = \mathbf{10}$
(b) $\displaystyle\int_0^{\pi} \sin x\,dx$
$F(x) = -\cos x$
$\Big[-\cos x\Big]_0^{\pi} = (-\cos\pi) - (-\cos 0) = (1) - (-1) = \mathbf{2}$
(c) $\displaystyle\int_1^e \frac{1}{x}\,dx$
$\Big[\ln|x|\Big]_1^e = \ln e - \ln 1 = 1 - 0 = \mathbf{1}$
Definite integral as area under the curve — adjust $a$ and $b$ to explore
Figure 3.2 — Definite Integral as Net Area
AP Exam Tip: When using FTC1 with a chain rule upper limit, e.g., $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)$. Also remember: $\int_a^a f(x)\,dx = 0$ and $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$.
(a) Evaluate $\displaystyle\int_0^4 (3x^2 - 2x + 1)\,dx$. (b) Find $\dfrac{d}{dx}\displaystyle\int_2^{x^3} \cos(t)\,dt$.
U-substitution is the chain rule in reverse. It is the most important integration technique for the AP exam. Use it when the integrand contains a function and (essentially) its derivative.
(a) $\displaystyle\int 2x\cos(x^2)\,dx$
Let $u = x^2$, so $du = 2x\,dx$.
$\displaystyle\int \cos(u)\,du = \sin(u) + C = \mathbf{\sin(x^2) + C}$
(b) $\displaystyle\int \frac{3x^2}{x^3 + 1}\,dx$
Let $u = x^3 + 1$, so $du = 3x^2\,dx$.
$\displaystyle\int \frac{du}{u} = \ln|u| + C = \mathbf{\ln|x^3 + 1| + C}$
(c) $\displaystyle\int (5x+2)^7\,dx$
Let $u = 5x+2$, $du = 5\,dx$, so $dx = du/5$.
$\dfrac{1}{5}\displaystyle\int u^7\,du = \dfrac{1}{5}\cdot\dfrac{u^8}{8} + C = \mathbf{\dfrac{(5x+2)^8}{40} + C}$
Evaluate $\displaystyle\int_0^1 2x e^{x^2}\,dx$.
Let $u = x^2$, $du = 2x\,dx$. Change limits: when $x=0$, $u=0$; when $x=1$, $u=1$.
$$\int_0^1 e^u\,du = \Big[e^u\Big]_0^1 = e^1 - e^0 = e - 1 \approx \mathbf{1.718}$$
U-substitution: the shaded area under $f(x) = 2xe^{x^2}$ on $[0,1]$ equals $e-1$
Figure 3.3 — Area Under $f(x) = 2xe^{x^2}$ from 0 to 1
Evaluate using u-substitution: (a) $\displaystyle\int 6x^2(x^3-4)^5\,dx$ (b) $\displaystyle\int_0^{\pi/2} \sin^2(x)\cos(x)\,dx$
Evaluate: $\displaystyle\int (8x^3 - 6x + 4)\,dx$
Find $f(x)$ given $f'(x) = \cos x + 2x$ and $f(0) = 3$.
Evaluate $\displaystyle\int_1^4 \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)dx$.
Use u-substitution: $\displaystyle\int (3x^2+1)^4 \cdot 6x\,dx$
Find $\dfrac{d}{dx}\displaystyle\int_0^{x^2} \sqrt{1+t^3}\,dt$.
Evaluate $\displaystyle\int_0^{\ln 3} e^x\,dx$ and interpret as an area.
(AP Free Response Style) A particle moves with velocity $v(t) = t^2 - 4t + 3$ m/s. Find the total distance traveled on $[0, 3]$.
Evaluate: $\displaystyle\int_1^2 \frac{x}{x^2+1}\,dx$
A function $F$ such that $F'(x) = f(x)$. The general antiderivative is $F(x) + C$ where $C$ is the constant of integration.
$\displaystyle\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\,\Delta x$ — the signed area under $f$ from $a$ to $b$.
If $g(x) = \displaystyle\int_a^x f(t)\,dt$, then $g'(x) = f(x)$. Differentiation and integration are inverse operations.
Let $u = g(x)$, so $du = g'(x)\,dx$. Transforms $\int f(g(x))g'(x)\,dx$ into $\int f(u)\,du$.
$\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
$\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ where $F'(x) = f(x)$.
$\int e^x\,dx = e^x + C$ $\int \frac{1}{x}\,dx = \ln|x| + C$ $\int \cos x\,dx = \sin x + C$
$\int_a^b [f+g]\,dx = \int_a^b f\,dx + \int_a^b g\,dx$ $\int_a^b kf\,dx = k\int_a^b f\,dx$