Chapter 3: Integral Calculus – Master Integrals

Integrals are the inverse of derivatives, focusing on "accumulation"—they calculate total change, area under curves, volumes of solids, and more. This chapter connects indefinite integrals (antiderivatives) to definite integrals (accumulated change) and explores their wide-ranging applications.

3.1 Indefinite Integrals

An indefinite integral "reverses" differentiation: if $F'(x) = f(x)$, then $F(x)$ is an antiderivative of $f(x)$, and the indefinite integral of $f(x)$ is the set of all such antiderivatives.

3.1.1 Definition of Indefinite Integrals

The indefinite integral of a function $f(x)$, denoted $\int f(x) dx$, is defined as:

$$\int f(x) dx = F(x) + C$$

where $F'(x) = f(x)$ (so $F(x)$ is an antiderivative of $f(x)$) and $C$ is the constant of integration (since the derivative of a constant is 0, adding $C$ accounts for all possible antiderivatives).

Example: Find $\int 2x dx$

We know that $\frac{d}{dx}[x^2] = 2x$, so $F(x) = x^2$ is an antiderivative. Thus:

$$\int 2x dx = x^2 + C$$

3.1.2 Basic Integral Formulas

These formulas are derived directly from derivative rules (memorize them to speed up calculations):

Function $f(x)$	Indefinite Integral $\int f(x) dx$
$c$ (constant)	$cx + C$
$x^n$ ($n \neq -1$)	$\frac{x^{n+1}}{n+1} + C$ (Power Rule)
$\frac{1}{x}$ ($x > 0$)	$\ln
$e^x$	$e^x + C$
$a^x$ ($a > 0, a \neq 1$)	$\frac{a^x}{\ln a} + C$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$

3.1.3 Integration Methods

Unlike differentiation, integration often requires creativity. Here are three core techniques:

1. Substitution Method (Reverse of the Chain Rule)

Use substitution to simplify integrals of composite functions. For $\int f(g(x)) \cdot g'(x) dx$:

1. Let $u = g(x)$, so $du = g'(x) dx$
2. Rewrite the integral in terms of $u$: $\int f(u) du$
3. Integrate with respect to $u$, then substitute back $u = g(x)$

Example 1: Simple substitution
Find $\int x\sqrt{x^2 + 1} dx$

Step 1: Let $u = x^2 + 1$. Then $du = 2x dx$, so $\frac{1}{2} du = x dx$

Step 2: Rewrite the integral:

$$\int \sqrt{u} \cdot \frac{1}{2} du = \frac{1}{2} \int u^{1/2} du$$

Step 3: Integrate using the power rule:

$$\frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{3} u^{3/2} + C$$

Step 4: Substitute back $u = x^2 + 1$:

$$\int x\sqrt{x^2 + 1} dx = \frac{1}{3} (x^2 + 1)^{3/2} + C$$

Example 2: Trigonometric substitution
For integrals involving $\sqrt{a^2 - x^2}$, use $x = a\sin\theta$ (since $\sin^2\theta + \cos^2\theta = 1$ simplifies the square root).

Find $\int \sqrt{4 - x^2} dx$

Step 1: Let $x = 2\sin\theta$, so $dx = 2\cos\theta d\theta$. Then $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta$ (since $\cos\theta \geq 0$ for $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$)

Step 2: Rewrite the integral:

$$\int 2\cos\theta \cdot 2\cos\theta d\theta = 4 \int \cos^2\theta d\theta$$

Step 3: Use the identity $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$:

$$4 \int \frac{1 + \cos 2\theta}{2} d\theta = 2 \int (1 + \cos 2\theta) d\theta = 2\left(\theta + \frac{\sin 2\theta}{2}\right) + C = 2\theta + \sin 2\theta + C$$

Step 4: Substitute back. Since $x = 2\sin\theta$, $\theta = \arcsin\left(\frac{x}{2}\right)$, and $\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4 - x^2}}{2} = \frac{x\sqrt{4 - x^2}}{2}$

Final result:

$$\int \sqrt{4 - x^2} dx = 2\arcsin\left(\frac{x}{2}\right) + \frac{x\sqrt{4 - x^2}}{2} + C$$

2. Integration by Parts (Reverse of the Product Rule)

For integrals of the form $\int u dv$ (products of functions), use:

$$\int u dv = uv - \int v du$$

Choose $u$ and $dv$ such that $\int v du$ is simpler than the original integral. A common mnemonic: LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) – prioritize $u$ as the first type in this list.

Example: Find $\int x\sin x dx$

Step 1: Choose $u = x$ (Algebraic, higher in LIATE) and $dv = \sin x dx$

Step 2: Compute $du$ and $v$:

$$du = dx, \quad v = \int \sin x dx = -\cos x$$

Step 3: Apply integration by parts:

$$\int x\sin x dx = uv - \int v du = -x\cos x - \int (-\cos x) dx = -x\cos x + \int \cos x dx$$

Step 4: Integrate the remaining term:

$$= -x\cos x + \sin x + C$$

3. Integration of Rational Functions (Partial Fractions)

Rational functions ($\frac{P(x)}{Q(x)}$, where $P,Q$ are polynomials) can be simplified by decomposing them into "partial fractions"—simpler fractions that are easier to integrate.

Example: Integrate $\int \frac{2x + 3}{(x + 1)(x + 2)} dx$

Step 1: Decompose into partial fractions. Assume:

$$\frac{2x + 3}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$$

Step 2: Solve for $A$ and $B$. Multiply both sides by $(x + 1)(x + 2)$:

$$2x + 3 = A(x + 2) + B(x + 1)$$

Let $x = -2$: $2(-2) + 3 = B(-2 + 1) \rightarrow -1 = -B \rightarrow B = 1$

Let $x = -1$: $2(-1) + 3 = A(-1 + 2) \rightarrow 1 = A \rightarrow A = 1$

Thus:

$$\frac{2x + 3}{(x + 1)(x + 2)} = \frac{1}{x + 1} + \frac{1}{x + 2}$$

Step 3: Integrate term by term: $\int \left(\frac{1}{x + 1} + \frac{1}{x + 2}\right) dx = \ln|x + 1| + \ln|x + 2| + C = \ln|(x + 1)(x + 2)| + C$

3.2 Definite Integrals

Definite integrals calculate the "accumulated change" of a function over an interval $[a,b]$. They are defined using limits of sums and linked to indefinite integrals via the Fundamental Theorem of Calculus.

3.2.1 Definition: Riemann Sums

The definite integral of $f(x)$ from $a$ to $b$, denoted $\int_a^b f(x) dx$, is defined as the limit of Riemann sums:

$$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$$

• $\Delta x = \frac{b - a}{n}$: Width of each subinterval when $[a,b]$ is divided into $n$ equal parts
• $x_i^*$: A sample point in the $i$-th subinterval
• The sum approximates the area under the curve $y = f(x)$ from $a$ to $b$; the limit gives the exact area (if $f(x) \geq 0$)

Example: Approximate $\int_0^1 x^2 dx$ with a Riemann sum ($n = 4$)

Step 1: $\Delta x = \frac{1 - 0}{4} = 0.25$. Subintervals: $[0,0.25], [0.25,0.5], [0.5,0.75], [0.75,1]$

Step 2: Use right endpoints as sample points: $x_1^* = 0.25, x_2^* = 0.5, x_3^* = 0.75, x_4^* = 1$

Step 3: Compute the sum:

$$\begin{align*} \sum_{i=1}^4 f(x_i^*) \Delta x &= [(0.25)^2 + (0.5)^2 + (0.75)^2 + (1)^2] \cdot 0.25 \\ &= [0.0625 + 0.25 + 0.5625 + 1] \cdot 0.25 \\ &= 1.875 \cdot 0.25 = 0.46875 \end{align*}$$

The exact value (from later methods) is $\frac{1}{3} \approx 0.333$; as $n$ increases, the sum approaches this.

3.2.2 Fundamental Theorem of Calculus (FTC)

The FTC is the bridge between indefinite and definite integrals, making definite integrals computable without limits.

Part 1: If $f$ is continuous on $[a,b]$ and $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$. (The integral function is an antiderivative of $f$.)

Part 2: If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then:

$$\int_a^b f(x) dx = F(b) - F(a)$$

We write $F(b) - F(a)$ as $F(x) \big|_a^b$.

Example: Compute $\int_0^1 x^2 dx$ using FTC Part 2

Step 1: Find an antiderivative of $f(x) = x^2$. From the power rule: $F(x) = \frac{x^3}{3}$

Step 2: Apply FTC Part 2:

$$\int_0^1 x^2 dx = \frac{x^3}{3} \Big|_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$$

3.2.3 Calculation of Definite Integrals

• Use FTC Part 2: Find an antiderivative, evaluate at bounds, subtract
• For substitution in definite integrals, adjust the limits of integration to match the substitution variable (avoids substituting back)

Example: Compute $\int_0^2 xe^{x^2} dx$ with substitution

Step 1: Let $u = x^2$, so $du = 2x dx \rightarrow \frac{1}{2} du = x dx$

Step 2: Adjust limits: When $x = 0$, $u = 0^2 = 0$; when $x = 2$, $u = 2^2 = 4$

Step 3: Rewrite the integral with new limits:

$$\int_0^4 e^u \cdot \frac{1}{2} du = \frac{1}{2} \int_0^4 e^u du$$

Step 4: Integrate and evaluate:

$$\frac{1}{2} e^u \Big|_0^4 = \frac{1}{2} (e^4 - e^0) = \frac{1}{2} (e^4 - 1)$$

3.2.4 Improper Integrals

These involve infinite intervals (e.g., $\int_a^\infty f(x) dx$) or discontinuities in $[a,b]$. They are defined as limits:

• For infinite intervals: $\int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx$ (converges if the limit exists)

Example: Evaluate $\int_1^\infty \frac{1}{x^2} dx$

$$\begin{align*} \int_1^\infty \frac{1}{x^2} dx &= \lim_{b \to \infty} \int_1^b x^{-2} dx \\ &= \lim_{b \to \infty} \left(-\frac{1}{x}\right) \Big|_1^b \\ &= \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) = 0 + 1 = 1 \end{align*}$$

The integral converges to 1.

3.3 Applications of Definite Integrals

Definite integrals model accumulation, making them useful for calculating area, volume, work, and more.

3.3.1 Area Between Two Curves

For two curves $y = f(x)$ and $y = g(x)$ with $f(x) \geq g(x)$ on $[a,b]$, the area between them is:

$$\text{Area} = \int_a^b [f(x) - g(x)] dx$$

Example: Find the area between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$

Step 1: Identify which function is on top. For $0 \leq x \leq 1$, $x \geq x^2$ (e.g., $x = 0.5$: $0.5 \geq 0.25$)

Step 2: Set up the integral:

$$\text{Area} = \int_0^1 (x - x^2) dx$$

Step 3: Evaluate:

$$= \left(\frac{x^2}{2} - \frac{x^3}{3}\right) \Big|_0^1 = \left(\frac{1}{2} - \frac{1}{3}\right) - 0 = \frac{1}{6}$$

3.3.2 Volume of Solids of Revolution

When a region under a curve is rotated around an axis, the volume can be calculated with:

• Disk Method (rotating $y = f(x)$ around the x-axis, $a \leq x \leq b$):

  $$V = \pi \int_a^b [f(x)]^2 dx$$

• Washer Method (rotating the region between $y = f(x)$ and $y = g(x)$ around the x-axis, $f(x) \geq g(x) \geq 0$):

  $$V = \pi \int_a^b ([f(x)]^2 - [g(x)]^2) dx$$

Example: Volume of a cone (disk method)

A cone is formed by rotating $y = 2x$ (from $x = 0$ to $x = 3$) around the x-axis. Find its volume.

Step 1: Apply the disk method. $f(x) = 2x$, so $[f(x)]^2 = 4x^2$

Step 2: Set up the integral:

$$V = \pi \int_0^3 4x^2 dx = 4\pi \int_0^3 x^2 dx$$

Step 3: Evaluate:

$$= 4\pi \cdot \frac{x^3}{3} \Big|_0^3 = 4\pi \left(\frac{27}{3} - 0\right) = 4\pi \cdot 9 = 36\pi$$

3.3.3 Other Applications

• Arc Length: Length of a curve $y = f(x)$ from $a$ to $b$:

  $$L = \int_a^b \sqrt{1 + [f'(x)]^2} dx$$

• Work: Work done by a variable force $F(x)$ moving an object from $a$ to $b$:

  $$W = \int_a^b F(x) dx$$

• Average Value of a Function: The average value of $f(x)$ on $[a,b]$ is:

  $$f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x) dx$$

3.4 Practice Problems (with Answer Hints)

1. Find $\int \cos(3x) dx$ using substitution
   Hint: Let $u = 3x$, so $du = 3dx$

2. Evaluate $\int_0^\pi x\cos x dx$ using integration by parts

3. Find the area between $y = x^3$ and $y = x$ for $x \geq 0$
   Hint: Find where they intersect first ($x = 0$ and $x = 1$)

4. Use the disk method to find the volume of a sphere with radius $r$ (rotate $y = \sqrt{r^2 - x^2}$ around the x-axis from $-r$ to $r$)

Answer Hints

1. $\frac{1}{3} \sin(3x) + C$
2. $-2$ (use $u = x$, $dv = \cos x dx$; results in $x\sin x + \cos x$ evaluated from $0$ to $\pi$)
3. $\frac{1}{4}$ (area = $\int_0^1 (x - x^3) dx$)
4. $\frac{4}{3} \pi r^3$ (standard sphere volume formula)