Integrals

Antiderivatives and Indefinite Integrals (CED Unit 6)

An antiderivative of $f$ is a function $F$ such that $F' = f$ . The indefinite integral is:

{/prettier-ignore/}

Adjust the parameters in the graph above to explore the relationships between variables.

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

Where $C$ is the constant of integration.

Basic Integration Rules

Function $f(x)$	Antiderivative $F(x)$
$x^n$ ( $n \ne -1$ )	$\frac{x^{n+1}}{n+1} + C$
$\frac{1}{x}$	$\ln \\| x \\| + C$
$e^x$	$e^x + C$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$
$\csc^2 x$	$-\cot x + C$
$\sec x \tan x$	$\sec x + C$
$\csc x \cot x$	$-\csc x + C$
$\frac{1}{1+x^2}$	$\arctan x + C$
$\frac{1}{\sqrt{1-x^2}}$	$\arcsin x + C$

The Power Rule for Integration

\int x^n\, dx = \frac\{x^\{n+1\}\}\{n+1\} + C, \quad n \ne -1

Why the power rule excludes $n = -1$ . Substituting $n = -1$ gives $\frac{x^0}{0}$ Which is Undefined. The antiderivative of $\frac{1}{x}$ is $\ln|x|$ A fundamental result with deep Connections to the natural logarithm.

Why the absolute value in $\ln|x|$ . The derivative of $\ln x$ is $\frac{1}{x}$ for $x \gt 0$ . For $x \lt 0$ The derivative of $\ln(-x)$ is $\frac{1}{-x} \cdot (-1) = \frac{1}{x}$ . So the Antiderivative of $\frac{1}{x}$ on any interval not containing $0$ is $\ln|x| + C$ .

Evaluate $\displaystyle\int \frac{2x^3 - x + 4}{\sqrt{x}}\, dx$ .

Rewrite: $\displaystyle\int (2x^{5/2} - x^{1/2} + 4x^{-1/2})\, dx = \frac{4x^{7/2}}{7} - \frac{2x^{3/2}}{3} + 8x^{1/2} + C$ .

Riemann Sums and the Definite Integral (CED Unit 6)

Riemann Sums

A Riemann sum approximates the area under a curve by dividing the region into rectangles:

\sum_\{i=1\}^\{n\} f(x_i^*) \Delta x

Where $\Delta x = \frac{b - a}{n}$ and $x_i^*$ is a sample point in the $i$ Th subinterval.

Type	Sample Point $x_i^*$
Left Riemann sum	Left endpoint
Right Riemann sum	Right endpoint
Midpoint sum	Midpoint of subinterval
Trapezoidal sum	Average of endpoints (trapezoids, not rectangles)

Why Riemann sums matter. They are the foundation of the definite integral. As $n \to \infty$ The approximation becomes exact (for continuous functions).

Theorem. If $f$ is continuous on $[a, b]$ Then the Riemann sum converges to the same value Regardless of the choice of sample points $x_i^*$ .

The Definite Integral

The definite integral of $f$ from $a$ to $b$ is:

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

Provided this limit exists. When it does, $f$ is said to be integrable on $[a, b]$ .

Properties of Definite Integrals

$\displaystyle\int_a^a f(x)\, dx = 0$
$\displaystyle\int_a^b f(x)\, dx = -\int_b^a f(x)\, dx$
$\displaystyle\int_a^b [f(x) \pm g(x)]\, dx = \int_a^b f(x)\, dx \pm \int_a^b g(x)\, dx$
$\displaystyle\int_a^b c \cdot f(x)\, dx = c \int_a^b f(x)\, dx$
$\displaystyle\int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx$ (Additivity)

Property 5 (Additivity) is powerful. It allows splitting integrals at discontinuities. For Example, if $f$ has a jump at $c \in (a, b)$ You can split:

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

Comparison properties. If $f(x) \ge g(x)$ on $[a, b]$ Then $\displaystyle\int_a^b f(x)\, dx \ge \int_a^b g(x)\, dx$ .

In particular, if $m \le f(x) \le M$ on $[a, b]$ Then $m(b-a) \le \displaystyle\int_a^b f(x)\, dx \le M(b-a)$ .

Integral as Net Area

If $f(x) \ge 0$ on $[a, b]$ Then $\displaystyle\int_a^b f(x)\, dx$ equals the area under the curve. If $f$ changes sign, the integral gives the net (signed) area.

The total area between $f$ and the $x$ -axis on $[a, b]$ is:

\mathrm\{Total Area = \int_a^b |f(x)|\, dx

Find the total area between $f(x) = x^2 - 4$ and the $x$ -axis on $[-3, 3]$ .

Find the zeros: $x^2 - 4 = 0 \implies x = \pm 2$ .

\mathrm\{Total Area = \int_\{-3\}^\{-2\} (x^2 - 4)\, dx + \int_\{-2\}^\{2\} (4 - x^2)\, dx + \int_\{2\}^\{3\} (x^2 - 4)\, dx

= \left[\frac\{x^3\}\{3\} - 4x\right]_\{-3\}^\{-2\} + \left[4x - \frac\{x^3\}\{3\}\right]_\{-2\}^\{2\} + \left[\frac\{x^3\}\{3\} - 4x\right]_\{2\}^\{3\}

= \frac\{-8 - (-27)\}\{3\} - \frac\{-8\}\{3\} + \frac\{8 - (-8)\}\{3\} + \frac\{27 - 8\}\{3\} = \frac\{19\}\{3\} + \frac\{16\}\{3\} + \frac\{19\}\{3\} = \frac\{54\}\{3\} = 18

The Fundamental Theorem of Calculus (CED Unit 6)

FTC Part 1

If $f$ is continuous on $[a, b]$ Then the function $g$ defined by:

G(x) = \int_a^x f(t)\, dt

Is differentiable on $(a, b)$ And:

G'(x) = f(x)

More generally, by the chain rule:

\frac\{d\}\{dx\}\!\left[\int_a^\{u(x)\} f(t)\, dt\right] = f(u(x)) \cdot u'(x)

Intuition. FTC Part 1 says: the rate at which the accumulated area changes is just the height of The curve at that point. This connects the two halves of calculus: the derivative and the integral are inverse operations.

Find $\displaystyle\frac{d}{dx}\!\left[\int_x^{x^2} \frac{1}{1+t^2}\, dt\right]$ .

Split the integral at a constant (say $0$ ):

\int_x^\{x^2\} \frac\{1\}\{1+t^2\}\, dt = \int_0^\{x^2\} \frac\{1\}\{1+t^2\}\, dt - \int_0^x \frac\{1\}\{1+t^2\}\, dt

Differentiating:

\frac\{1\}\{1+x^4\} \cdot 2x - \frac\{1\}\{1+x^2\} = \frac\{2x\}\{1+x^4\} - \frac\{1\}\{1+x^2\}

FTC 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)

This is the evaluation theorem. We write $F(x)\Big|_a^b$ to denote $F(b) - F(a)$ .

Evaluate $\displaystyle\int_1^3 \left(x^2 + \frac{1}{x}\right) dx$ .

\int_1^3 \left(x^2 + \frac\{1\}\{x\}\right) dx = \left[\frac\{x^3\}\{3\} + \ln x\right]_1^3 = \left(\frac\{27\}\{3\} + \ln 3\right) - \left(\frac\{1\}\{3\} + 0\right) = \frac\{26\}\{3\} + \ln 3

Proof of FTC Part 2 using the MVT

Let $F$ be an antiderivative of $f$ . Define $g(x) = \int_a^x f(t)\, dt$ .

By FTC Part 1, $g'(x) = f(x) = F'(x)$ So $g(x) = F(x) + C$ for some constant $C$ .

At $x = a$ : $g(a) = 0 = F(a) + C \implies C = -F(a)$ .

Therefore, $g(x) = F(x) - F(a)$ And:

\int_a^b f(t)\, dt = g(b) = F(b) - F(a)

$\blacksquare$

Integration Techniques

$u$ -Substitution

If $\displaystyle\int f(g(x)) \cdot g'(x)\, dx$ Let $u = g(x)$ , $du = g'(x)\, dx$ :

\int f(u)\, du = F(u) + C = F(g(x)) + C

Strategy for choosing $u$ . Look for a function and its derivative in the integrand. If you can Spot $f(g(x))$ and $g'(x)$ Set $u = g(x)$ .

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

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

\int_0^1 \frac\{2x\}\{\sqrt\{1 + x^2\}\}\, dx = \int_1^2 u^\{-1/2\}\, du = \left[2\sqrt\{u\}\right]_1^2 = 2\sqrt\{2\} - 2

Integration by Parts (CED BC Unit 6.11)

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

Choose $u$ using LIATE priority: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.

Why LIATE works. The antiderivative of $u$ should be simpler than $u$ itself. Logarithmic and Inverse trig functions simplify upon differentiation. Algebraic functions require integration by Parts to reduce their degree.

Tabular integration (DI method). For integrals of the form $\displaystyle\int f(x)g(x)\, dx$ Where $f(x)$ is a polynomial and $g(x)$ has an repeatable derivative pattern, use a table. Label columns D (derivatives of $f$ ) and I (integrals of $g$ ), alternating signs $+$$-$$+$$-$ . The result is the sum of diagonal products.

Evaluate $\displaystyle\int x^2 e^x\, dx$ .

Let $u = x^2$$dv = e^x\, dx$ . Then $du = 2x\, dx$$v = e^x$ .

= x^2 e^x - \int 2x e^x\, dx

Apply integration by parts again for $\int 2x e^x\, dx$ . Let $u = 2x$$dv = e^x\, dx$ $du = 2\, dx$$v = e^x$ :

\int 2x e^x\, dx = 2xe^x - 2e^x + C

Therefore:

\int x^2 e^x\, dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C

Tabular method check:

Sign	D (derivatives)	I (integrals)
$+$	$x^2$	$e^x$
$-$	$2x$	$e^x$
$+$	$2$	$e^x$

Result: $x^2 e^x - 2xe^x + 2e^x = e^x(x^2 - 2x + 2)$ . Confirmed.

Evaluate $\displaystyle\int \frac{1}{x^2 - 1}\, dx$ .

Factor: $x^2 - 1 = (x-1)(x+1)$ .

\frac\{1\}\{(x-1)(x+1)\} = \frac\{A\}\{x-1\} + \frac\{B\}\{x+1\}

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

At $x = 1$ : $1 = 2A \implies A = \frac{1}{2}$ . At $x = -1$ : $1 = -2B \implies B = -\frac{1}{2}$ .

\int \frac\{1\}\{x^2 - 1\}\, dx = \frac\{1\}\{2\}\int \frac\{1\}\{x-1\}\, dx - \frac\{1\}\{2\}\int \frac\{1\}\{x+1\}\, dx = \frac\{1\}\{2\}\ln|x-1| - \frac\{1\}\{2\}\ln|x+1| + C = \frac\{1\}\{2\}\ln\!\left|\frac\{x-1\}\{x+1\}\right| + C

Improper Integrals

An improper integral involves either an infinite limit of integration or an infinite discontinuity in the interval.

Type 1: Infinite interval:

\int_a^\{\infty\} f(x)\, dx = \lim_\{b \to \infty\} \int_a^b f(x)\, dx

Type 2: Infinite discontinuity at $a$ :

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

Evaluate $\displaystyle\int_0^{\infty} xe^{-x^2}\, dx$ .

Let $u = x^2$$du = 2x\, dx$ :

\int_0^\{\infty\} xe^\{-x^2\}\, dx = \frac\{1\}\{2\}\int_0^\{\infty\} e^\{-u\}\, du = \frac\{1\}\{2\}\left[-e^\{-u\}\right]_0^\{\infty\} = \frac\{1\}\{2\}(0 + 1) = \frac\{1\}\{2\}

The Gaussian integral. The integral $\displaystyle\int_0^{\infty} e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2}$ is a celebrated result that cannot be evaluated by elementary methods. The standard technique uses a double integral in polar Coordinates. The full Gaussian integral from $-\infty$ to $\infty$ equals $\sqrt{\pi}$ .

Note that $\displaystyle\int_0^{\infty} xe^{-x^2}\, dx = \frac{1}{2}$ (computed above via $u$ -substitution) is a different integral from $\displaystyle\int_0^{\infty} e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2}$ .

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

Find intersections: $x^2 = 2x$ So $x^2 - 2x = 0$ Giving $x = 0$ and $x = 2$ .

Between $x = 0$ and $x = 2$$2x \ge 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\}

Volumes of Solids of Revolution

Disk method (rotating about the $x$ -axis):

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

Washer method (rotating region between $f(x)$ and $g(x)$ about the $x$ -axis):

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

Shell method (rotating about the $y$ -axis):

V = 2\pi \int_a^b x \cdot f(x)\, dx

When to use which method. Use the disk/washer method when integrating perpendicular to the axis Of rotation. Use the shell method when integrating parallel to the axis of rotation.

Find the volume of the solid obtained by rotating the region bounded by $y = e^{-x}$$y = 0$ $x = 0$$x = 1$ about the $x$ -axis.

Using the disk method:

V = \pi \int_0^1 (e^\{-x\})^2\, dx = \pi \int_0^1 e^\{-2x\}\, dx = \pi\left[-\frac\{e^\{-2x\}\}\{2\}\right]_0^1 = \pi\left(-\frac\{e^\{-2\}\}\{2\} + \frac\{1\}\{2\}\right) = \frac\{\pi\}\{2\}\!\left(1 - \frac\{1\}\{e^2\}\right)

Average Value of a Function

The average value of $f$ on $[a, b]$ is:

F_\{\mathrm\{avg\} = \frac\{1\}\{b - a\}\int_a^b f(x)\, dx

By the MVT for integrals, there exists $c \in [a, b]$ such that $f(c) = f_{\mathrm{avg}}$ .

Common Pitfalls

Forgetting the constant of integration in indefinite integrals. Without it, the answer is incomplete.
Incorrect $u$ -substitution limits. When using $u$ -substitution for definite integrals, always change the limits or substitute back to the original variable.
Forgetting to change $dx$ to $du$ when performing $u$ -substitution.
Applying FTC Part 2 to discontinuous functions. $f$ must be continuous on $[a, b]$ .
Confusing net area with total area. Use absolute values for total area.
Incorrectly choosing $u$ in integration by parts. Use LIATE: prioritize Logarithmic over Inverse trig over Algebraic over Trig over Exponential.
Sign errors with FTC Part 1 chain rule. The derivative of $\int_a^{g(x)} f(t)\, dt$ is $f(g(x)) \cdot g'(x)$ Not $f(g(x))$ .
Forgetting to split improper integrals at the discontinuity when a singularity is in the interval.
Choosing the disk vs. Washer vs. Shell method incorrectly. Disk: rotate around $x$ -axis using radius. Shell: rotate around $y$ -axis using height as the integrand. Washer: region between two curves rotated about an axis.
Dropping the absolute value in $\ln|x|$ . The antiderivative of $\frac{1}{x}$ is $\ln|x| + C$ Not $\ln x + C$ . On intervals where $x \lt 0$ The integral is well-defined and equals $\ln(-x) + C$ .
Confusing the Gaussian integrals. $\displaystyle\int_0^{\infty} xe^{-x^2}\, dx = \frac{1}{2}$ (evaluated via $u$ -substitution), but $\displaystyle\int_0^{\infty} e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2}$ (requires advanced techniques).

Practice Questions

Evaluate $\displaystyle\int \frac{x}{x^2 + 1}\, dx$ .
Evaluate $\displaystyle\int_0^{\pi/2} \sin^2 x\, dx$ . (Hint: use the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ .)
Use integration by parts to evaluate $\displaystyle\int x^2 e^x\, dx$ .
Find the area between $y = x^2$ and $y = 2x$ .
Find the volume of the solid obtained by rotating the region bounded by $y = x^2$$y = 0$ $x = 1$ about the $y$ -axis. (Use the shell method.)
Determine whether $\displaystyle\int_0^{\infty} e^{-x}\, dx$ converges, and find its value if it does.
Find the average value of $f(x) = \sin x$ on $[0, \pi]$ .
Given $\displaystyle g(x) = \int_0^{x^3} \cos(t^2)\, dt$ Find $g'(x)$ .
Evaluate $\displaystyle\int_0^1 \frac{2x}{\sqrt{1 + x^2}}\, dx$ .
Find the arc length of $y = \frac{x^2}{8} + 2$ from $x = 0$ to $x = 4$ .
Evaluate $\displaystyle\int_1^e \frac{\ln x}{x}\, dx$ .
Find the volume when the region bounded by $y = e^{-x}$$y = 0$$x = 0$$x = 1$ is rotated about the $x$ -axis.
Evaluate $\displaystyle\int_0^1 \frac{1}{1 + x^2}\, dx$ and explain the result geometrically.
Use the substitution $x = \tan\theta$ to evaluate $\displaystyle\int \frac{1}{1 + x^2}\, dx$ .
Find the area between the curves $y = \sin x$ and $y = \cos x$ for $0 \le x \le \pi/2$ .
Evaluate $\displaystyle\int_0^1 x^2 e^{-x}\, dx$ using the tabular method.

Practice Problems

Question 1: Integration by parts (tabular method)

Evaluate $\displaystyle\int x^3 e^x \, dx$ using the tabular method.

Answer

Differentiate $x^3$ (column 1) and integrate $e^x$ (column 2) alternately with signs $+,-,+,-$ :

Signs	$x^3$	$e^x$
+	$x^3$	$e^x$
-	$3x^2$	$e^x$
+	$6x$	$e^x$
-	$6$	$e^x$
+	$0$	$e^x$

Result: $x^3 e^x - 3x^2 e^x + 6xe^x - 6e^x + C = e^x(x^3 - 3x^2 + 6x - 6) + C$ .

Question 2: Trigonometric substitution

Evaluate $\displaystyle\int \frac{x^2}{\sqrt{9 - x^2}} \, dx$ .

Answer

Let $x = 3\sin\theta$$dx = 3\cos\theta \, d\theta$$\sqrt{9 - x^2} = 3\cos\theta$ .

$= \int \frac{9\sin^2\theta}{3\cos\theta} \cdot 3\cos\theta \, d\theta = 9\int \sin^2\theta \, d\theta = 9\int \frac{1 - \cos 2\theta}{2} \, d\theta$

$= \frac{9}{2}\left(\theta - \frac{\sin 2\theta}{2}\right) + C = \frac{9}{2}\theta - \frac{9}{4}\sin 2\theta + C$ .

Since $\theta = \arcsin(x/3)$ and $\sin 2\theta = 2\sin\theta\cos\theta = \frac{2x\sqrt{9-x^2}}{9}$ :

$= \frac{9}{2}\arcsin\!\left(\frac{x}{3}\right) - \frac{x\sqrt{9-x^2}}{2} + C$ .

Question 3: Area between curves

Find the area enclosed by the curves $y = x^2$ and $y = 2x + 3$ .

Answer

Intersection: $x^2 = 2x + 3$$x^2 - 2x - 3 = 0$$(x-3)(x+1) = 0$ . $x = -1$ and $x = 3$ .

On $[-1, 3]$$2x + 3 \ge x^2$ .

Area $= \int_{-1}^{3} [(2x + 3) - x^2] \, dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3}$

$= (9 + 9 - 9) - (1 - 3 + 1/3) = 9 - (-5/3) = 9 + 5/3 = 32/3$ square units.

Question 4: Improper integral

Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x^p} \, dx$ converges or diverges, and find its value when it converges.

Answer

$\displaystyle\int_1^{\infty} \frac{1}{x^p} \, dx = \lim_{b \to \infty} \int_1^b x^{-p} \, dx$ .

For $p \ne 1$ : $= \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p}$ .

If $p \gt 1$ : $1 - p \lt 0$ So $b^{1-p} \to 0$ . Integral converges to $\frac{1}{p - 1}$ .
If $p \lt 1$ : $1 - p \gt 0$ So $b^{1-p} \to \infty$ . Integral diverges.

For $p = 1$ : $\int_1^b \frac{1}{x} \, dx = \ln b \to \infty$ . Diverges.

The integral converges if and only if $p \gt 1$ With value $\frac{1}{p-1}$ .

Question 5: Volume of revolution

Find the volume obtained by rotating the region bounded by $y = \sqrt{x}$$y = 0$ And $x = 4$ about the x-axis.

Answer

Using the disk method:

$V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi(8) = 8\pi$ cubic units.

Summary

This topic covers the mathematical techniques and concepts related to integrals, including key theorems, methods, and problem-solving approaches.

Key concepts include:

differentiation from first principles
product, quotient, and chain rules
integration techniques (by parts, substitution)
differential equations
applications to kinematics

Worked Examples

Intuition

The water filling analogy: Integration is like measuring how much water fills a container with an irregular shape. You add up infinitely many tiny slices to find the total volume.

Why it matters: Integrals allow us to calculate areas, volumes, work, and other quantities that accumulate continuously. They’re essential in physics, engineering, economics, and medicine.

The key insight: Integration sums up infinitely many infinitesimal contributions to find a total quantity.

Cross-References

Derivatives — Integration is the reverse of differentiation, and the fundamental theorem of calculus connects the two.
Differential Equations — Solving separable differential equations requires integrating both sides of the equation.
Sequences and Series — Improper integrals and convergence of series share the same mathematical framework.
AP Physics, Work, Energy, and Power: Work is defined as the integral of force over displacement, making integration essential for energy analysis.
AP Physics, Circuits: Charge flow in circuits is the integral of current over time, connecting integral calculus to electrical analysis.

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