Calculus I · Chapter 7
Chapter 7: Integration
Antiderivatives · Indefinite integrals · Substitution · Area & Riemann sums · The definite integral · The Fundamental Theorem of Calculus
Every operation in mathematics has an inverse, and the inverse of differentiation is integration. Given a rate, it recovers the total. The astonishing discovery, the Fundamental Theorem of Calculus, is that this reversal also computes the area under a curve. Two problems that look unrelated turn out to be the same problem.
Section 7.1
Antiderivatives
The Antiderivative Family
Definition: Antiderivative & Indefinite Integral
$F$ is an antiderivative of $f$ if $F'(x) = f(x)$. Since the derivative of a constant is zero, antiderivatives come in a whole family differing by a constant:
$$\int f(x)\,dx = F(x) + C.$$
The "$+C$" (the constant of integration) is not optional, it captures every member of the family at once.
The antiderivative family. Every curve here has the same derivative (same slope at each $x$), they are vertical shifts of one another, one for each value of $C$. Drag to slide $C$ and pick out one member. The "+C" is exactly this freedom.
Integration Rules
Basic Antiderivative Rules
| Function $f(x)$ | Antiderivative $\int f\,dx$ |
|---|---|
| $x^n$ ($n \neq -1$) | $\dfrac{x^{n+1}}{n+1} + C$ (reverse power rule) |
| $\dfrac{1}{x}$ | $\ln|x| + C$ |
| $e^{x}$ | $e^{x} + C$ |
| $e^{kx}$ | $\dfrac{1}{k}e^{kx} + C$ |
| $k$ (constant) | $kx + C$ |
⚠ Always Check by Differentiating
Integration has a built-in answer key: differentiate your result and you should get back $f(x)$. If you don't, you've made an error. This check catches almost every mistake.
📘 Example 7.1: Reverse Power Rule
Find $\displaystyle\int (6x^2 - 4x + 5)\,dx$.
Integrate term by term
$$\int 6x^2\,dx = 6\cdot\frac{x^3}{3} = 2x^3, \quad \int -4x\,dx = -2x^2, \quad \int 5\,dx = 5x.$$
$2x^3 - 2x^2 + 5x + C$
Section 7.2
Substitution
Substitution is the chain rule run backward. It handles integrals containing a function and its derivative, you spot an inner function $u$ whose derivative also appears.
The Substitution Method
- Choose $u = g(x)$ (usually the "inside" function).
- Compute $du = g'(x)\,dx$.
- Rewrite the integral entirely in terms of $u$, no $x$'s left.
- Integrate, then substitute $x$ back in.
📘 Example 7.2: Substitution
Find $\displaystyle\int 2x(x^2 + 1)^5\,dx$.
- Let $u = x^2 + 1$. Then $du = 2x\,dx$, and $2x\,dx$ is right there.
- Rewrite: $\displaystyle\int u^5\,du$.
- Integrate: $\dfrac{u^6}{6} + C$.
- Back-substitute: $\dfrac{(x^2+1)^6}{6} + C$.
📘 Example 7.3: Substitution with a Logarithm
Find $\displaystyle\int \frac{3x^2}{x^3 + 2}\,dx$.
- Let $u = x^3 + 2$, so $du = 3x^2\,dx$, the numerator.
- Rewrite: $\displaystyle\int \frac{1}{u}\,du = \ln|u| + C$.
- Back-substitute: $\ln|x^3 + 2| + C$.
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Section 7.3
Area & the Definite Integral
Riemann Sums, Area by Rectangles
To find the area under a curve, approximate it with rectangles, then let their number grow. The exact area is the limit of these Riemann sums.
Definition: The Definite Integral
$$\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i)\,\Delta x, \qquad \Delta x = \frac{b-a}{n}.$$
It is the exact (signed) area between $y = f(x)$ and the $x$-axis from $a$ to $b$. Area below the axis counts as negative.
Riemann sum approximation of the area under a curve. Slide $n$, as the rectangles multiply and thin out, the gap between the staircase and the curve vanishes, and the sum converges to the exact area (the definite integral). This limit is the integral.
sum ≈ , | exact = 9.000
Properties of the Definite Integral
- $\displaystyle\int_a^a f\,dx = 0$
- $\displaystyle\int_b^a f\,dx = -\int_a^b f\,dx$ (swapping limits flips the sign)
- $\displaystyle\int_a^b [f \pm g]\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$
- $\displaystyle\int_a^b k\,f\,dx = k\int_a^b f\,dx$
- $\displaystyle\int_a^c f\,dx + \int_c^b f\,dx = \int_a^b f\,dx$ (split the interval)
Section 7.4
The Fundamental Theorem of Calculus
Theorem: The Fundamental Theorem of Calculus
If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then
$$\int_a^b f(x)\,dx = F(b) - F(a).$$
This is the bridge between the two halves of calculus: to find an area (a limit of sums), you don't sum anything, you just evaluate an antiderivative at the two endpoints and subtract. Differentiation and integration are inverse operations.
The Fundamental Theorem in action. As the right endpoint $b$ sweeps right, the accumulated area (blue) grows, and its running total is tracked by the antiderivative $F$ (the height of the green curve below). The area from $a$ to $b$ equals $F(b) - F(a)$, no rectangles required.
area = ,
📘 Example 7.4: Evaluating a Definite Integral
Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$.
- Antiderivative: $F(x) = x^2 + x$.
- Evaluate at the limits: $F(3) = 9 + 3 = 12$, $F(1) = 1 + 1 = 2$.
- Subtract: $F(3) - F(1) = 12 - 2 = 10$.
📘 Example 7.5: Area Under a Parabola
Find the area under $f(x) = x^2$ from $x = 0$ to $x = 3$.
Apply the Fundamental Theorem
$$\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9.$$
This is the exact value the Riemann sum animation above converges to.
Area $= 9$ square units
📘 Example 7.6: Definite Integral with Substitution
Evaluate $\displaystyle\int_0^2 x\,e^{x^2}\,dx$.
- Let $u = x^2$, $du = 2x\,dx$, so $x\,dx = \tfrac{1}{2}du$.
- Change limits: $x = 0 \to u = 0$; $x = 2 \to u = 4$.
- $\displaystyle\frac{1}{2}\int_0^4 e^u\,du = \frac{1}{2}\big[e^u\big]_0^4 = \frac{1}{2}(e^4 - 1).$
The Course in One Sentence
The derivative breaks a total into its instantaneous rate; the integral reassembles a rate back into a total, and the same operation measures area. From the slope of a line in Chapter 1 to the Fundamental Theorem here, every idea has been about understanding change, and then undoing it. That is the whole of single-variable calculus.
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