9.4 Integrals

Definition of integrals

Integrals solve a fundamental problem: how do you find the total accumulated quantity when something is continuously changing? Think of calculating the area under a curve, the volume of an oddly shaped solid, or the total distance traveled when speed keeps varying. Integration gives you the tools to handle all of these.

This section covers Riemann sums, definite and indefinite integrals, the Fundamental Theorem of Calculus, a range of integration techniques, and applications from area calculations to integral transforms.

Riemann sums

Riemann sums are the conceptual foundation of integration. The idea is straightforward: approximate the area under a curve by chopping it into thin rectangles, then add up their areas.

• Divide the interval $[a, b]$ into $n$ subintervals of width $\Delta x = \frac{b - a}{n}$
• For each subinterval, build a rectangle whose height is the function value at some chosen point in that subinterval
• Sum all the rectangle areas: $S_n = \sum_{i=1}^{n} f(x_i^*) \Delta x$

The "chosen point" determines which type of Riemann sum you're using:

• Left Riemann sum: use the left endpoint of each subinterval
• Right Riemann sum: use the right endpoint
• Midpoint Riemann sum: use the midpoint (often the most accurate of the three for the same $n$)

As $n \to \infty$, the rectangles get thinner and the approximation converges to the exact area. That limit is precisely the definite integral.

Definite vs indefinite integrals

These two types of integrals look similar on paper but answer different questions.

Definite integrals calculate a specific numerical value representing the net area between a curve and the x-axis over an interval:

$\int_a^b f(x)\,dx$

Here $a$ and $b$ are the lower and upper bounds. The result is a number, not a function. Note that area below the x-axis counts as negative.

Indefinite integrals find the general antiderivative of a function:

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

The result is a family of functions differing by the constant $C$. Why the constant? Because many different functions share the same derivative (for example, $x^2 + 5$ and $x^2 - 3$ both have derivative $2x$). The $+ C$ accounts for all of them.

Fundamental Theorem of Calculus

This theorem is the bridge between differentiation and integration. It has two parts, and both are essential.

Part 1 (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 what makes definite integrals practical. Instead of computing a limit of Riemann sums, you just find an antiderivative and plug in the bounds.

Part 2 (Derivative of an Integral Function): If you define $G(x) = \int_a^x f(t)\,dt$, then $G'(x) = f(x)$. In other words, differentiation undoes integration.

Together, these two parts say that differentiation and integration are inverse operations. That single insight powers most of the computational techniques in this section.

Integration techniques

Not every integral yields to a direct antiderivative lookup. The techniques below give you systematic strategies for handling more complex integrands.

Integration by substitution

Substitution reverses the chain rule. When you spot a composite function and its inner derivative lurking in the integrand, substitution simplifies the whole expression.

Steps:

1. Identify an inner function $u = g(x)$ whose derivative $g'(x)$ appears (or nearly appears) in the integrand
2. Compute $du = g'(x)\,dx$
3. Rewrite the entire integral in terms of $u$ and $du$
4. Integrate with respect to $u$
5. Substitute back to express the result in terms of $x$

For definite integrals, you can either convert the bounds to $u$-values or substitute back and then evaluate.

Example: To evaluate $\int 2x \cos(x^2)\,dx$, let $u = x^2$, so $du = 2x\,dx$. The integral becomes $\int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C$.

Integration by parts

Integration by parts reverses the product rule. Use it when the integrand is a product of two functions and substitution doesn't help.

Formula:

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

Steps:

1. Choose $u$ and $dv$ from the integrand. A common guideline is LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential): pick $u$ from whichever category comes first in that list.
2. Compute $du$ by differentiating $u$, and find $v$ by integrating $dv$
3. Apply the formula
4. If the resulting integral is still complex, apply integration by parts again

Example: For $\int x\sin(x)\,dx$, let $u = x$ and $dv = \sin(x)\,dx$. Then $du = dx$ and $v = -\cos(x)$. Applying the formula gives $-x\cos(x) + \int \cos(x)\,dx = -x\cos(x) + \sin(x) + C$.

Partial fractions

Partial fraction decomposition breaks a rational function into simpler pieces that are each easy to integrate.

Steps:

1. Make sure the degree of the numerator is less than the degree of the denominator (if not, do polynomial long division first)

2. Factor the denominator completely

3. Write the decomposition based on the factor types:

   • Distinct linear factors $(x - a)$: use $\frac{A}{x - a}$
   • Repeated linear factors $(x - a)^n$: use $\frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + \cdots$
   • Irreducible quadratic factors $(ax^2 + bx + c)$: use $\frac{Ax + B}{ax^2 + bx + c}$

4. Solve for the unknown coefficients by equating numerators or plugging in strategic values of $x$

5. Integrate each simpler fraction separately

Trigonometric integrals

These integrals involve powers and products of trig functions. The strategy depends on the specific form.

• For $\int \sin^m(x)\cos^n(x)\,dx$: if one exponent is odd, peel off one factor and use a Pythagorean identity to convert the rest. If both are even, use half-angle identities like $\sin^2(x) = \frac{1 - \cos(2x)}{2}$.
• For integrals involving $\tan(x)$ and $\sec(x)$, similar strategies apply using the identity $\sec^2(x) = 1 + \tan^2(x)$.
• The Weierstrass substitution $u = \tan(\frac{x}{2})$ can handle rational expressions in $\sin(x)$ and $\cos(x)$, though it's often a last resort because the algebra gets heavy.

The key with trig integrals is pattern recognition: identify which identity or substitution converts the integrand into something you already know how to handle.

Applications of integrals

Area between curves

To find the area enclosed between two curves $f(x)$ and $g(x)$:

1. Find the intersection points by solving $f(x) = g(x)$
2. Determine which function is on top over the interval
3. Integrate the difference: $A = \int_a^b [f(x) - g(x)]\,dx$, where $f(x) \geq g(x)$ on $[a, b]$

If the curves cross within the interval, split the integral at each crossing point and take absolute values so you don't accidentally subtract area.

Volume of solids

Integration can compute volumes of solids of revolution (shapes formed by rotating a region around an axis). Three common methods:

• Disk method: When the solid has no hole, slice it into thin disks perpendicular to the axis of rotation. $V = \pi \int_a^b [f(x)]^2\,dx$
• Washer method: When there's a gap between the curve and the axis (creating a hollow center), subtract the inner radius from the outer: $V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right)dx$
• Shell method: Slice the solid into thin cylindrical shells parallel to the axis of rotation. $V = 2\pi \int_a^b x \cdot f(x)\,dx$. This is often easier when rotating around a vertical axis but integrating with respect to $x$.

Choosing the right method depends on the axis of rotation and which variable is more convenient to integrate over.

Arc length

The length of a curve $y = f(x)$ from $x = a$ to $x = b$ is:

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

This formula comes from the Pythagorean theorem applied to infinitesimally small segments of the curve. For parametric curves $x = x(t), y = y(t)$, the formula becomes:

$L = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

These integrals are often difficult to evaluate analytically, which is one reason numerical methods matter.

Surface area

When you rotate a curve $y = f(x)$ around the x-axis, the surface area of the resulting solid is:

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

The $2\pi f(x)$ factor represents the circumference of each circular "ring" traced out by the rotation, and the square root factor accounts for the slant of the surface. Be careful about which axis you're rotating around, as the formula changes accordingly.

Improper integrals

Improper integrals extend integration to situations where the interval is infinite or the integrand blows up. They're defined through limits, which means the integral might or might not produce a finite answer.

Infinite limits

When an integration bound is infinite, replace it with a variable and take a limit:

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

If this limit exists and is finite, the integral converges. Otherwise, it diverges.

For doubly infinite integrals, split at any convenient point: $\int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^\infty f(x)\,dx$. Both pieces must converge independently.

Classic example: $\int_1^\infty \frac{1}{x^p}\,dx$ converges when $p > 1$ and diverges when $p \leq 1$. This is the p-test for integrals.

Discontinuous integrands

If $f(x)$ has a discontinuity at some point $c$ in $[a, b]$, split the integral there and use limits:

$\int_a^b f(x)\,dx = \lim_{\epsilon \to 0^+} \int_a^{c - \epsilon} f(x)\,dx + \lim_{\epsilon \to 0^+} \int_{c + \epsilon}^b f(x)\,dx$

Both limits must exist for the integral to converge. A common mistake is integrating straight through a discontinuity without noticing it, which can give a wrong answer.

Convergence vs divergence

Several tests help determine whether an improper integral converges:

• Direct comparison test: If $0 \leq f(x) \leq g(x)$ and $\int g(x)\,dx$ converges, then $\int f(x)\,dx$ converges too. If $\int f(x)\,dx$ diverges, so does $\int g(x)\,dx$.
• Limit comparison test: If $\lim_{x \to \infty} \frac{f(x)}{g(x)} = L$ where $0 < L < \infty$, then both integrals converge or both diverge.
• p-test: $\int_1^\infty \frac{1}{x^p}\,dx$ converges for $p > 1$, diverges for $p \leq 1$.

Absolute convergence (convergence of $\int |f(x)|\,dx$) implies convergence of $\int f(x)\,dx$, but the reverse isn't always true.

Numerical integration

When you can't find a closed-form antiderivative, numerical methods let you approximate the integral to any desired accuracy.

Trapezoidal rule

Instead of rectangles (Riemann sums), use trapezoids to approximate the area under the curve. This captures the slope of the function between sample points.

With $n$ subintervals of width $h = \frac{b - a}{n}$:

$\int_a^b f(x)\,dx \approx \frac{h}{2}\left[f(a) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(b)\right]$

The first and last function values get weight 1; all interior points get weight 2. The error is proportional to $h^2$, meaning doubling the number of subintervals cuts the error roughly by a factor of 4.

Simpson's rule

Simpson's rule fits parabolas through consecutive groups of three points, giving a much better approximation than straight-line segments. It requires an even number of subintervals.

$\int_a^b f(x)\,dx \approx \frac{h}{3}\left[f(a) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(b)\right]$

Notice the alternating pattern of coefficients: 1, 4, 2, 4, 2, ..., 4, 1. The error is proportional to $h^4$, so Simpson's rule converges much faster than the trapezoidal rule for smooth functions. In fact, Simpson's rule gives exact results for any polynomial of degree 3 or less.

Error estimation

Knowing how accurate your approximation is matters as much as the approximation itself.

• Trapezoidal rule error is bounded by $\frac{(b-a)^3}{12n^2} \max|f''(x)|$
• Simpson's rule error is bounded by $\frac{(b-a)^5}{180n^4} \max|f^{(4)}(x)|$

These bounds tell you how many subintervals you need to guarantee a desired accuracy. Adaptive quadrature methods take this further by automatically using more subintervals where the function changes rapidly and fewer where it's smooth. Richardson extrapolation combines results from different step sizes to squeeze out extra accuracy without additional function evaluations.

Multivariable integrals

Multivariable integrals extend everything above to functions of two or more variables. Instead of finding area under a curve, you're now finding volume under a surface, or mass of a 3D region, and so on.

Double integrals

A double integral $\iint_D f(x, y)\,dA$ integrates a function of two variables over a region $D$ in the xy-plane.

You evaluate it as an iterated integral, integrating one variable at a time:

$\iint_D f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$

The order of integration ($dy\,dx$ vs $dx\,dy$) can dramatically affect difficulty. Always sketch the region $D$ to determine the best order and correct limits.

For regions with circular symmetry, converting to polar coordinates ($x = r\cos\theta$, $y = r\sin\theta$, $dA = r\,dr\,d\theta$) often simplifies things considerably.

Triple integrals

Triple integrals $\iiint_E f(x,y,z)\,dV$ work the same way but with three nested integrals. They compute quantities like volume, mass, and moments of inertia for 3D regions.

Coordinate system choice is critical:

• Cartesian $(x, y, z)$: best for rectangular regions
• Cylindrical $(r, \theta, z)$: best for regions with circular cross-sections; $dV = r\,dr\,d\theta\,dz$
• Spherical $(\rho, \theta, \phi)$: best for regions with spherical symmetry; $dV = \rho^2 \sin\phi\,d\rho\,d\phi\,d\theta$

Setting up the limits of integration correctly is usually the hardest part. Draw the region and project it onto coordinate planes to figure out the bounds.

Change of variables

When a region or integrand has a natural symmetry that doesn't align with Cartesian coordinates, a change of variables can transform a difficult integral into a straightforward one.

The key ingredient is the Jacobian determinant, which accounts for how the transformation stretches or compresses area (or volume). For a transformation $(x, y) \to (u, v)$:

$dA = \left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv$

The polar coordinate substitution is the most common example: the Jacobian $r$ in $dA = r\,dr\,d\theta$ comes from this calculation. Cylindrical and spherical coordinate Jacobians work the same way.

Line and surface integrals

These integrals generalize integration to curves and surfaces in space, connecting calculus with vector field analysis.

Vector fields

A vector field assigns a vector to every point in a region of space. In 3D:

$\mathbf{F}(x, y, z) = P(x,y,z)\,\mathbf{i} + Q(x,y,z)\,\mathbf{j} + R(x,y,z)\,\mathbf{k}$

Physical examples include gravitational fields, electric fields, and fluid velocity fields. Two important properties of vector fields are:

• Divergence $\nabla \cdot \mathbf{F}$: measures how much the field "spreads out" from a point (a scalar quantity)
• Curl $\nabla \times \mathbf{F}$: measures the rotational tendency of the field around a point (a vector quantity)

A field is conservative if it's the gradient of some scalar potential function, which happens exactly when its curl is zero (in a simply connected domain).

Green's theorem

Green's theorem connects a line integral around a closed curve in the plane to a double integral over the enclosed region:

$\oint_C (P\,dx + Q\,dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

The curve $C$ must be simple (no self-intersections), closed, and traversed counterclockwise. This theorem is powerful because it lets you convert a potentially difficult line integral into a double integral, or vice versa, depending on which is easier to compute.

Stokes' theorem

Stokes' theorem generalizes Green's theorem to surfaces in 3D:

$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}$

It says the surface integral of the curl of a vector field equals the line integral of the field around the boundary of that surface. The surface must be orientable (it has a well-defined "inside" and "outside"), and the boundary curve's orientation must be consistent with the surface orientation via the right-hand rule.

Green's theorem is actually a special case of Stokes' theorem where the surface is flat and lies in the xy-plane.

Integral transforms

Integral transforms re-express functions in a different "domain" where certain operations become simpler. They're especially useful for solving differential equations.

Laplace transform

The Laplace transform converts a function of time $f(t)$ into a function of complex frequency $s$:

$\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st}f(t)\,dt$

Its main use: it turns differential equations into algebraic equations. You transform the equation, solve the (easier) algebraic problem, then apply the inverse Laplace transform to get back to the time domain. This is particularly effective for linear systems with initial conditions, since initial values get built right into the transformed equation.

Fourier transform

The Fourier transform decomposes a function into its frequency components:

$\mathcal{F}\{f(t)\} = F(\omega) = \int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt$

Where the Laplace transform works with one-sided ($t \geq 0$) signals and complex frequency, the Fourier transform handles signals defined over all of $(-\infty, \infty)$ and decomposes them into real frequencies. It's foundational in signal processing, image analysis, and quantum mechanics. The computational versions (DFT and FFT) make it practical for digital applications.

Applications in differential equations

Both transforms simplify differential equations by converting them to a domain where calculus operations become algebra:

• Laplace transforms are ideal for initial value problems (ODEs). Derivatives become multiplication by $s$, so a second-order ODE becomes a quadratic equation in $s$.
• Fourier transforms excel at boundary value problems and PDEs, especially those on infinite domains (like the heat equation on an infinite rod).
• The convolution theorem states that convolution in the time domain corresponds to multiplication in the transform domain, which simplifies integral equations significantly.
• Green's functions and transfer functions use these transforms to characterize how systems respond to inputs.