Calculus

Integration

Integration is the inverse operation of differentiation. Given the derivative of a function, integration recovers the original function (up to an additive constant). It is a central operation in calculus with wide-ranging applications in geometry, physics, and probability.

Two principal forms exist:

• Indefinite integration (antidifferentiation): finds the general antiderivative of a function, producing a family of functions differing by a constant.
• Definite integration: computes a numerical value representing the signed area under a curve between two limits.

These are linked by the Fundamental Theorem of Calculus, which unifies differentiation and integration into a single coherent framework.

Notation

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

where $F'(x) = f(x)$ and $C$ is the constant of integration. The symbol $\int$ is an elongated $S$ (for "sum"), $f(x)$ is the integrand, and $dx$ indicates the variable of integration.

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1. Antiderivatives

Definition

A function $F$ is an antiderivative of $f$ on an interval $I$ if $F'(x) = f(x)$ for all $x \in I$. If $F$ is an antiderivative of $f$, then so is $F + C$ for any constant $C \in \mathbb{'\{'}R{'\}'}$.

Basic Rules

Power Rule

For $n \neq -1$:

$$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$$

Constant Multiple Rule

For $k \in \mathbb{'\{'}R{'\}'}$:

$$\int k f(x)\,dx = k \int f(x)\,dx$$

Sum and Difference Rule

$$\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$$

Reciprocal Rule (Logarithmic)

$$\int \frac{1}{x}\,dx = \ln|x| + C$$

Common Antiderivatives

| $f(x)$ | $\int f(x)\,dx$ | | -------------------------- | -------------------------------------------- | ------ | ---- | | $k$ | $kx + C$ | | | | $x^n$ | $\dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)$ | | | | $\dfrac{1}{x}$ | $\ln | x | + C$ | | $e^x$ | $e^x + C$ | | | | $a^x$ | $\dfrac{a^x}{\ln a} + C$ | | | | $\sin x$ | $-\cos x + C$ | | | | $\cos x$ | $\sin x + C$ | | | | $\sec^2 x$ | $\tan x + C$ | | | | $\csc^2 x$ | $-\cot x + C$ | | | | $\sec x \tan x$ | $\sec x + C$ | | | | $\csc x \cot x$ | $-\csc x + C$ | | | | $\tan x$ | $\ln | \sec x | + C$ | | $\cot x$ | $\ln | \sin x | + C$ | | $\frac{1}{\sqrt{1 - x^2}}$ | $\arcsin x + C$ | | | | $\frac{1}{1 + x^2}$ | $\arctan x + C$ | | |

Examples

Details

Expand

Find $\int (3x^4 - 5x^2 + 7)\,dx$

\begin`\{aligned}` \int (3x^4 - 5x^2 + 7)\,dx &= 3 \cdot \frac{x^5}{5} - 5 \cdot \frac{x^3}{3} + 7x + C \\[6pt] &= \frac{3x^5}{5} - \frac{5x^3}{3} + 7x + C \end`\{aligned}`

Find $\int \left(\frac{2}{x} + 3e^x - \frac{1}{\cos^2 x}\right)\,dx$

\begin`\{aligned}` \int \left(\frac{2}{x} + 3e^x - \sec^2 x\right)\,dx &= 2\ln|x| + 3e^x - \tan x + C \end`\{aligned}`

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2. Definite Integration

Definite Integral as a Limit of a Riemann Sum

For a function $f$ continuous on $[a, b]$, the definite integral is defined as:

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

where $\Delta x = \dfrac{b - a}{n}$ and $x_i^*$ is a sample point in the $i$-th subinterval.

Fundamental Theorem of Calculus -- Part 1

If $f$ is continuous on $[a, b]$ and $F$ is defined by:

$$F(x) = \int_a^x f(t)\,dt$$

then $F$ is differentiable on $(a, b)$ and:

$$F'(x) = f(x)$$

This establishes that differentiation and integration are inverse operations.

Fundamental Theorem of Calculus -- Part 2 (Evaluation Theorem)

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 typically written using bracket notation:

$$\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)$$

Area Under a Curve

If $f(x) \geq 0$ on $[a, b]$, then $\int_a^b f(x)\,dx$ gives the area between the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$.

If $f(x)$ changes sign on $[a, b]$, the integral gives the net (signed) area. The total area is computed by splitting at the zeros of $f$ and taking absolute values:

$$\mathrm{Total area} = \int_a^b |f(x)|\,dx$$

Integration as Area Under a Curve

Adjust the sliders to change the function and limits, and observe how the shaded area approximates the definite integral.

Examples

Details

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Evaluate $\int_1^3 (x^2 + 1)\,dx$

\begin`\{aligned}` \int_1^3 (x^2 + 1)\,dx &= \left[\frac{x^3}{3} + x\right]_1^3 \\[6pt] &= \left(\frac{27}{3} + 3\right) - \left(\frac{1}{3} + 1\right) \\[6pt] &= 12 - \frac{4}{3} = \frac{32}{3} \end`\{aligned}`

Find the area enclosed by $y = x^2 - 4$ and the $x$-axis.

The curve crosses the $x$-axis at $x = -2$ and $x = 2$.

\begin`\{aligned}` \mathrm{Area} &= \int_{-2}^2 |x^2 - 4|\,dx = \int_{-2}^2 (4 - x^2)\,dx \\[6pt] &= \left[4x - \frac{x^3}{3}\right]_{-2}^2 \\[6pt] &= \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{32}{3} \end`\{aligned}`

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3. Integration Techniques

3.1 Substitution ($u$-Substitution)

Substitution is the reverse of the chain rule. Given an integral containing a composite function, choose a substitution $u = g(x)$ such that $du = g'(x)\,dx$ transforms the integral into a simpler form.

General procedure:

1. Identify an inner function and set $u = g(x)$.
2. Compute $du = g'(x)\,dx$ and solve for $dx$.
3. Rewrite the entire integral in terms of $u$.
4. Evaluate the integral with respect to $u$.
5. Substitute back $x$ (for indefinite integrals) or change the limits (for definite integrals).

Changing limits for definite integrals:

$$\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$$

Examples

Details

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Find $\int 2x\sqrt{x^2 + 1}\,dx$

Let $u = x^2 + 1$, then $du = 2x\,dx$.

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

Evaluate $\int_0^2 \frac{x}{x^2 + 1}\,dx$

Let $u = x^2 + 1$, $du = 2x\,dx$. When $x = 0$, $u = 1$; when $x = 2$, $u = 5$.

$$\int_0^2 \frac{x}{x^2 + 1}\,dx = \frac{1}{2}\int_1^5 \frac{1}{u}\,du = \frac{1}{2}\Big[\ln|u|\Big]_1^5 = \frac{1}{2}\ln 5$$

3.2 Integration by Parts

Integration by parts is the reverse of the product rule. The formula is:

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

Choosing $u$ and $dv$: Use the LIATE priority rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to select $u$ as the function that appears first in the list.

Integration by parts may need to be applied repeatedly. For integrals of the form $\int e^{ax}\sin(bx)\,dx$ or $\int e^{ax}\cos(bx)\,dx$, apply integration by parts twice and solve the resulting equation algebraically.

Tabular (DI) method: For integrals of the form $\int f(x) g(x)\,dx$ where one factor differentiates to zero after finitely many steps, the tabular method provides an efficient alternative to repeated application of the formula.

Examples

Details

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Find $\int x e^x\,dx$

Let $u = x$, $dv = e^x\,dx$. Then $du = dx$ and $v = e^x$.

$$\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = (x - 1)e^x + C$$

Find $\int x^2 \sin x\,dx$

Apply integration by parts twice.

First: $u = x^2$, $dv = \sin x\,dx$, so $du = 2x\,dx$, $v = -\cos x$.

$$\int x^2 \sin x\,dx = -x^2 \cos x + 2\int x \cos x\,dx$$

Second: $u = x$, $dv = \cos x\,dx$, so $du = dx$, $v = \sin x$.

\begin`\{aligned}` \int x^2 \sin x\,dx &= -x^2 \cos x + 2\left(x\sin x - \int \sin x\,dx\right) \\[6pt] &= -x^2 \cos x + 2x\sin x + 2\cos x + C \end`\{aligned}`

3.3 Partial Fractions

When the integrand is a rational function $\dfrac{P(x)}{Q(x)}$ where $\deg P < \deg Q$ and $Q$ factors into linear or irreducible quadratic factors, partial fraction decomposition converts the integrand into a sum of simpler fractions.

Decomposition rules:

Factor in $Q(x)$	Term in decomposition
$(ax + b)$	$\dfrac{A}{ax + b}$
$(ax + b)^k$	$\dfrac{A_1}{ax + b} + \dfrac{A_2}{(ax+b)^2} + \cdots + \dfrac{A_k}{(ax+b)^k}$
$ax^2 + bx + c$ (irreducible)	$\dfrac{Ax + B}{ax^2 + bx + c}$

Examples

Details

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Find $\int \frac{5x + 1}{(x+1)(x-2)}\,dx$

Decompose: $\dfrac{5x + 1}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$

$$5x + 1 = A(x - 2) + B(x + 1)$$

Setting $x = -1$: $-5 + 1 = A(-3) \implies A = \frac{4}{3}$

Setting $x = 2$: $10 + 1 = B(3) \implies B = \frac{11}{3}$

$$\int \frac{5x + 1}{(x+1)(x-2)}\,dx = \frac{4}{3}\ln|x+1| + \frac{11}{3}\ln|x-2| + C$$

3.4 Trigonometric Integrals

Several standard techniques apply to integrals involving trigonometric functions.

Pythagorean identities for conversion:

$$\sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2}$$

These are used to reduce the power of trigonometric functions in integrands.

Products to sums:

$$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$$ $$\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$$ $$\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$$

Using substitution with trigonometric derivatives: Recall that $\dfrac{d}{dx}(\sin x) = \cos x$ and $\dfrac{d}{dx}(\cos x) = -\sin x$. Many trigonometric integrals yield to $u$-substitution when the derivative of one trigonometric factor is present.

Integrals of the form $\int \tan^n x\,dx$: Rewrite in terms of $\sec^2 x$ and use $\dfrac{d}{dx}(\tan x) = \sec^2 x$.

Examples

Details

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Find $\int \sin^2 x\,dx$

$$\int \sin^2 x\,dx = \int \frac{1 - \cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$$

Find $\int \sin^3 x\,dx$

\begin`\{aligned}` \int \sin^3 x\,dx &= \int \sin x (1 - \cos^2 x)\,dx \\[6pt] &= \int \sin x\,dx - \int \sin x \cos^2 x\,dx \\[6pt] &= -\cos x + \frac{\cos^3 x}{3} + C \end`\{aligned}`

(using $u = \cos x$, $du = -\sin x\,dx$ for the second integral)

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4. Applications of Integration

4.1 Area Between Two Curves

Given two continuous functions $f$ and $g$ with $f(x) \geq g(x)$ on $[a, b]$, the area between the curves is:

$$A = \int_a^b [f(x) - g(x)]\,dx$$

If the curves intersect, find the points of intersection and split the integral accordingly so that the integrand $|f(x) - g(x)|$ is always non-negative.

Horizontal strips: When integrating with respect to $y$, the formula becomes:

$$A = \int_c^d [f(y) - g(y)]\,dy$$

where the curves are expressed as $x = f(y)$ and $x = g(y)$.

Examples

Details

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Find the area between $y = x^2$ and $y = 2x$.

Intersection: $x^2 = 2x \implies x = 0$ or $x = 2$. On $[0, 2]$, $2x \geq x^2$.

$$A = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$$

4.2 Volumes of Revolution

When a region bounded by $y = f(x)$, the $x$-axis, and the lines $x = a$, $x = b$ is revolved about the $x$-axis, the volume of the solid of revolution is:

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

Revolution about the $y$-axis:

$$V = \pi \int_c^d [g(y)]^2\,dy$$

where $x = g(y)$.

Revolution about a horizontal line $y = k$:

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

Washer method (volume between two surfaces):

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

where $f(x) \geq g(x) \geq 0$ on $[a, b]$.

Examples

Details

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Find the volume generated by revolving $y = \sqrt{x}$ about the $x$-axis from $x = 0$ to $x = 4$.

\begin`\{aligned}` V &= \pi \int_0^4 (\sqrt{x})^2\,dx = \pi \int_0^4 x\,dx \\[6pt] &= \pi \left[\frac{x^2}{2}\right]_0^4 = 8\pi \end`\{aligned}`

4.3 Kinematics

Integration connects the kinematic quantities:

$$v(t) = \int a(t)\,dt + v_0$$ $$s(t) = \int v(t)\,dt + s_0$$

where $a(t)$ is acceleration, $v(t)$ is velocity, $s(t)$ is displacement, and $v_0$, $s_0$ are initial conditions.

Key relationships:

• The displacement over a time interval $[t_1, t_2]$ is $\displaystyle\int_{t_1}^{t_2} v(t)\,dt$.
• The total distance travelled is $\displaystyle\int_{t_1}^{t_2} |v(t)|\,dt$.
• The velocity is the derivative of displacement: $v = \dfrac{ds}{dt}$.
• The acceleration is the derivative of velocity: $a = \dfrac{dv}{dt}$.

Examples

Details

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A particle moves with velocity $v(t) = 6t^2 - 4t$ m/s for $t \geq 0$. Find the displacement and total distance in the first 3 seconds.

Displacement:

$$\int_0^3 (6t^2 - 4t)\,dt = \left[2t^3 - 2t^2\right]_0^3 = 54 - 18 = 36 \mathrm{ m}$$

Total distance: find when $v(t) = 0$: $6t^2 - 4t = 0 \implies t = 0$ or $t = \frac{2}{3}$.

\begin`\{aligned}` \mathrm{Distance} &= \int_0^{2/3} |6t^2 - 4t|\,dt + \int_{2/3}^3 |6t^2 - 4t|\,dt \\[6pt] &= \int_0^{2/3} (4t - 6t^2)\,dt + \int_{2/3}^3 (6t^2 - 4t)\,dt \\[6pt] &= \left[2t^2 - 2t^3\right]_0^{2/3} + \left[2t^3 - 2t^2\right]_{2/3}^3 \\[6pt] &= \frac{8}{27} + \left(36 - \left(-\frac{8}{27}\right)\right) = \frac{8}{27} + 36 + \frac{8}{27} = \frac{988}{27} \approx 36.6 \mathrm{ m} \end`\{aligned}`

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5. Properties of Definite Integrals

These properties follow from the definition of the definite integral and are essential for simplifying computations.

Linearity

$$\int_a^b [k f(x) + l g(x)]\,dx = k\int_a^b f(x)\,dx + l\int_a^b g(x)\,dx$$

for constants $k, l \in \mathbb{'\{'}R{'\}'}$.

Reversal of Limits

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

Interval of Zero Length

$$\int_a^a f(x)\,dx = 0$$

Additivity Over Intervals

$$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$

for any $c \in [a, b]$.

Comparison Properties

If $f(x) \geq 0$ for all $x \in [a, b]$, then $\displaystyle\int_a^b f(x)\,dx \geq 0$.

If $f(x) \geq g(x)$ for all $x \in [a, b]$, then $\displaystyle\int_a^b f(x)\,dx \geq \int_a^b g(x)\,dx$.

Bounds on the Integral

If $m \leq f(x) \leq M$ for all $x \in [a, b]$, then:

$$m(b - a) \leq \int_a^b f(x)\,dx \leq M(b - a)$$

Examples

Details

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Given $\int_1^3 f(x)\,dx = 5$ and $\int_1^3 g(x)\,dx = -2$, find $\int_3^1 [3f(x) - 2g(x)]\,dx$.

\begin`\{aligned}` \int_3^1 [3f(x) - 2g(x)]\,dx &= -\int_1^3 [3f(x) - 2g(x)]\,dx \\[6pt] &= -\left(3 \cdot 5 - 2 \cdot (-2)\right) = -(15 + 4) = -19 \end`\{aligned}`

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6. Improper Integrals (HL)

Improper integrals extend the concept of definite integration to cases where the interval of integration is unbounded or the integrand has a vertical asymptote within (or at an endpoint of) the interval of integration.

Type 1: Infinite Limits of Integration

$$\int_a^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_a^t f(x)\,dx$$ $$\int_{-\infty}^b f(x)\,dx = \lim_{t \to -\infty} \int_t^b f(x)\,dx$$ $$\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^{\infty} f(x)\,dx$$

Both limits must converge independently for the integral to converge.

Type 2: Integrands with Vertical Asymptotes

If $f$ has a vertical asymptote at $x = a$:

$$\int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx$$

If $f$ has a vertical asymptote at $x = b$:

$$\int_a^b f(x)\,dx = \lim_{t \to b^-} \int_a^t f(x)\,dx$$

If $f$ has a vertical asymptote at an interior point $x = c \in (a, b)$:

$$\int_a^b f(x)\,dx = \lim_{t \to c^-} \int_a^t f(x)\,dx + \lim_{t \to c^+} \int_t^b f(x)\,dx$$

Convergence and Divergence

An improper integral converges if the corresponding limit exists and is finite. It diverges otherwise.

$p$-test for convergence: The integral $\displaystyle\int_1^{\infty} \frac{1}{x^p}\,dx$ converges if and only if $p > 1$, in which case:

$$\int_1^{\infty} \frac{1}{x^p}\,dx = \frac{1}{p - 1}$$

Similarly, $\displaystyle\int_0^1 \frac{1}{x^p}\,dx$ converges if and only if $p < 1$.

Comparison test: If $0 \leq f(x) \leq g(x)$ for all $x \geq a$, and $\int_a^{\infty} g(x)\,dx$ converges, then $\int_a^{\infty} f(x)\,dx$ also converges. Conversely, if $\int_a^{\infty} f(x)\,dx$ diverges, then $\int_a^{\infty} g(x)\,dx$ also diverges.

Examples

Details

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Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx$ converges.

\begin`\{aligned}` \int_1^{\infty} \frac{1}{x^2}\,dx &= \lim_{t \to \infty} \int_1^t x^{-2}\,dx = \lim_{t \to \infty} \left[-\frac{1}{x}\right]_1^t \\[6pt] &= \lim_{t \to \infty} \left(-\frac{1}{t} + 1\right) = 1 \end`\{aligned}`

The integral converges to $1$. (Consistent with the $p$-test: $p = 2 > 1$.)

Determine whether $\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$ converges.

\begin`\{aligned}` \int_0^1 \frac{1}{\sqrt{x}}\,dx &= \lim_{t \to 0^+} \int_t^1 x^{-1/2}\,dx = \lim_{t \to 0^+} \Big[2\sqrt{x}\Big]_t^1 \\[6pt] &= \lim_{t \to 0^+} (2 - 2\sqrt{t}) = 2 \end`\{aligned}`

The integral converges to $2$. (Consistent with the $p$-test: $p = \frac{1}{2} < 1$.)

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7. Practice Problems

Problem 1

Details

Find $\int \frac{x^3 + 2x}{\sqrt{x^2 + 1}}\,dx$

Let $u = x^2 + 1$, $du = 2x\,dx$. Note that $x^3 + 2x = x(x^2 + 1) + x = xu + \frac{u - 1}{2} + \frac{1}{2}$.

Alternatively, split: $\dfrac{x^3 + 2x}{\sqrt{x^2 + 1}} = \dfrac{x(x^2 + 1)}{\sqrt{x^2 + 1}} + \dfrac{x}{\sqrt{x^2 + 1}} = x\sqrt{x^2 + 1} + \dfrac{x}{\sqrt{x^2 + 1}}$.

\begin`\{aligned}` \int x\sqrt{x^2 + 1}\,dx &= \frac{1}{3}(x^2 + 1)^{3/2} + C_1 \\[6pt] \int \frac{x}{\sqrt{x^2 + 1}}\,dx &= \sqrt{x^2 + 1} + C_2 \end`\{aligned}`$$\int \frac{x^3 + 2x}{\sqrt{x^2 + 1}}\,dx = \frac{1}{3}(x^2 + 1)^{3/2} + \sqrt{x^2 + 1} + C$$

Problem 2

Details

Evaluate $\int_0^{\pi/2} x\sin x\,dx$

Integration by parts: $u = x$, $dv = \sin x\,dx$, so $du = dx$, $v = -\cos x$.

\begin`\{aligned}` \int_0^{\pi/2} x\sin x\,dx &= \Big[-x\cos x\Big]_0^{\pi/2} + \int_0^{\pi/2} \cos x\,dx \\[6pt] &= (0 - 0) + \Big[\sin x\Big]_0^{\pi/2} = 1 \end`\{aligned}`

Problem 3

Details

Find the area enclosed by $y = x^3$ and $y = x$.

Intersection points: $x^3 = x \implies x(x^2 - 1) = 0 \implies x = -1, 0, 1$.

On $[-1, 0]$: $x^3 \geq x$. On $[0, 1]$: $x \geq x^3$.

\begin`\{aligned}` A &= \int_{-1}^0 (x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx \\[6pt] &= \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0 + \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 \\[6pt] &= \left(0 - \frac{1}{4} + \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \end`\{aligned}`

Problem 4

Details

Find the volume of revolution when the region bounded by $y = \ln x$, $x = e$, and the $x$-axis is rotated about the $x$-axis.

The curve $y = \ln x$ meets the $x$-axis at $x = 1$.

\begin`\{aligned}` V &= \pi \int_1^e (\ln x)^2\,dx \end`\{aligned}`

Integration by parts with $u = (\ln x)^2$, $dv = dx$: $du = \dfrac{2\ln x}{x}\,dx$, $v = x$.

\begin`\{aligned}` \int (\ln x)^2\,dx &= x(\ln x)^2 - 2\int \ln x\,dx \end`\{aligned}`

Apply parts again for $\int \ln x\,dx$: $u = \ln x$, $dv = dx$, so $\int \ln x\,dx = x\ln x - x$.

\begin`\{aligned}` \int (\ln x)^2\,dx &= x(\ln x)^2 - 2(x\ln x - x) + C \\[6pt] &= x(\ln x)^2 - 2x\ln x + 2x + C \end`\{aligned}`

Evaluating from $1$ to $e$:

\begin`\{aligned}` \Big[x(\ln x)^2 - 2x\ln x + 2x\Big]_1^e &= (e - 2e + 2e) - (0 - 0 + 2) = e - 2 \end`\{aligned}`$$V = \pi(e - 2)$$

Problem 5 (HL)

Use partial fractions to find $\int \frac{3x + 7}{(x + 2)(x - 1)}\,dx$.

$$\frac{3x + 7}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$$$$3x + 7 = A(x - 1) + B(x + 2)$$

$x = 1$: $10 = 3B \implies B = \dfrac{10}{3}$

$x = -2$: $1 = -3A \implies A = -\dfrac{1}{3}$

$$\int \frac{3x + 7}{(x+2)(x-1)}\,dx = -\frac{1}{3}\ln|x+2| + \frac{10}{3}\ln|x-1| + C$$

Problem 6

A particle moves with acceleration $a(t) = 6t - 2$ m/s$^2$. Initially, $v(0) = 4$ m/s and $s(0) = 0$ m. Find $s(t)$ and the displacement after 3 seconds.

$$v(t) = \int (6t - 2)\,dt + 4 = 3t^2 - 2t + 4$$$$s(t) = \int (3t^2 - 2t + 4)\,dt + 0 = t^3 - t^2 + 4t$$

Displacement at $t = 3$:

$$s(3) = 27 - 9 + 12 = 30 \mathrm{ m}$$

Problem 7 (HL)

Details

Determine whether $\displaystyle\int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx$ converges, and evaluate if it does.

Let $u = \ln x$, $du = \dfrac{1}{x}\,dx$. When $x = 2$, $u = \ln 2$; when $x \to \infty$, $u \to \infty$.

\begin`\{aligned}` \int_2^{\infty} \frac{1}{x(\ln x)^2}\,dx &= \int_{\ln 2}^{\infty} \frac{1}{u^2}\,du = \lim_{t \to \infty} \int_{\ln 2}^t u^{-2}\,du \\[6pt] &= \lim_{t \to \infty} \left[-\frac{1}{u}\right]_{\ln 2}^t = \lim_{t \to \infty} \left(-\frac{1}{t} + \frac{1}{\ln 2}\right) = \frac{1}{\ln 2} \end`\{aligned}`

The integral converges to $\dfrac{1}{\ln 2}$.

Problem 8 (HL)

Details

Evaluate $\int \frac{x^2}{x^2 - 1}\,dx$ using polynomial long division and partial fractions.

Since $\deg(\mathrm{numerator}) = \deg(\mathrm{denominator})$, perform long division first:

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

Partial fractions: $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$

$1 = A(x+1) + B(x-1)$

$x = 1$: $A = \dfrac{1}{2}$, $x = -1$: $B = -\dfrac{1}{2}$

\begin`\{aligned}` \int \frac{x^2}{x^2 - 1}\,dx &= \int \left(1 + \frac{1}{2(x-1)} - \frac{1}{2(x+1)}\right)\,dx \\[6pt] &= x + \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C \end`\{aligned}`

Problem 9

Details

Evaluate $\int_0^1 \frac{e^x}{1 + e^x}\,dx$.

Let $u = 1 + e^x$, $du = e^x\,dx$. When $x = 0$, $u = 2$; when $x = 1$, $u = 1 + e$.

$$\int_0^1 \frac{e^x}{1 + e^x}\,dx = \int_2^{1+e} \frac{1}{u}\,du = \Big[\ln u\Big]_2^{1+e} = \ln(1 + e) - \ln 2$$

Problem 10 (HL)

Details

Find $\int e^x \cos x\,dx$ using integration by parts twice.

Let $I = \displaystyle\int e^x \cos x\,dx$.

First application: $u = e^x$, $dv = \cos x\,dx$, so $du = e^x\,dx$, $v = \sin x$.

$$I = e^x \sin x - \int e^x \sin x\,dx$$

Second application on $\int e^x \sin x\,dx$: $u = e^x$, $dv = \sin x\,dx$, so $du = e^x\,dx$, $v = -\cos x$.

$$\int e^x \sin x\,dx = -e^x \cos x + \int e^x \cos x\,dx = -e^x \cos x + I$$

Substituting back:

$$I = e^x \sin x - (-e^x \cos x + I) = e^x \sin x + e^x \cos x - I$$$$2I = e^x (\sin x + \cos x) \implies I = \frac{e^x (\sin x + \cos x)}{2} + C$$

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Cross-References

• Differentiation -- Integration is the inverse operation of differentiation. See the Number and Algebra notes for function fundamentals including derivative rules that motivate integration techniques.
• Functions -- Domain and range considerations determine when antiderivatives are valid. See Number and Algebra for function fundamentals.
• Complex Numbers -- The exponential form $e^{i\theta} = \cos\theta + i\sin\theta$ provides an elegant derivation of trigonometric integral results. See Complex Numbers.
• Logic -- Proof techniques (direct proof, contradiction) are used to justify properties of integrals. See Logic.