Techniques and Applications of Integration

The Fundamental Theorem of Calculus tells you that a definite integral can be evaluated by finding an antiderivative. The catch is that finding an antiderivative is rarely as simple as reversing a basic differentiation rule. This lesson develops the two most important techniques for actually computing integrals, surveys two more, and then turns to what integrals are for: measuring area, volume, work, and averages ¹.

u-Substitution: Reversing the Chain Rule

The chain rule says \(\frac{d}{dx}F(g(x)) = F'(g(x))\,g'(x)\). Reading this backward gives a method for integrals that contain a function and (a multiple of) its own derivative. Set \(u = g(x)\), so that \(du = g'(x)\,dx\), and the integral collapses into something elementary.

Consider $$\int 2x\sqrt{x^2 + 1}\,dx.$$ Let \(u = x^2 + 1\). Then \(du = 2x\,dx\), which is exactly the \(2x\,dx\) already present. Substituting, $$\int \sqrt{u}\,du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x^2 + 1)^{3/2} + C.$$ You can check the result by differentiating it: the chain rule returns the original integrand. For a definite integral, either convert the limits to \(u\)-values or rewrite the antiderivative in terms of \(x\) before evaluating. The skill is recognizing the inner function whose derivative is also present.

Integration by Parts: Reversing the Product Rule

Where substitution undoes the chain rule, integration by parts undoes the product rule. From \(\frac{d}{dx}(uv) = u\,\frac{dv}{dx} + v\,\frac{du}{dx}\), integrating both sides and rearranging gives $$\int u\,dv = uv - \int v\,du.$$ The art is choosing \(u\) and \(dv\) so that the new integral \(\int v\,du\) is simpler than the original. A useful guideline is to let \(u\) be the factor that gets simpler when differentiated.

Consider $$\int x e^{x}\,dx.$$ Let \(u = x\) and \(dv = e^{x}\,dx\). Then \(du = dx\) and \(v = e^{x}\). Applying the formula, $$\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C = e^{x}(x - 1) + C.$$ Differentiating \(e^{x}(x-1)\) returns \(x e^{x}\), confirming the answer. Parts is the standard tool for integrands that mix a polynomial with an exponential, logarithm, or trigonometric function ¹.

Two Further Techniques

When an integrand is a ratio of polynomials, partial fractions decomposes it into a sum of simpler fractions, each of which integrates to a logarithm or arctangent. For instance, \(\frac{1}{x^2 - 1}\) splits into \(\frac{1/2}{x-1} - \frac{1/2}{x+1}\), and each piece integrates directly. When an integrand contains expressions like \(\sqrt{a^2 - x^2}\) or \(\sqrt{x^2 + a^2}\), trigonometric substitution replaces \(x\) with a sine, tangent, or secant so a Pythagorean identity clears the radical. These extend the same idea: transform the integral into a form you already know how to evaluate ¹.

Area Between Curves

The most direct application of the definite integral is area. If \(f(x) \ge g(x)\) on \([a, b]\), the area of the region between the two curves is the integral of the gap between them: $$A = \int_a^b \big(f(x) - g(x)\big)\,dx.$$ Each thin vertical strip has height \(f(x) - g(x)\) and width \(dx\); the integral sums infinitely many such strips. When the curves cross, split the interval at the intersection points and integrate the positive difference on each piece.

Volumes of Revolution

Rotating a region about an axis sweeps out a solid whose volume an integral can capture. The disk method slices the solid perpendicular to the axis of rotation into thin circular disks. A disk at position \(x\) has radius \(R(x)\) and thickness \(dx\), so its volume is \(\pi R(x)^2\,dx\), and $$V = \int_a^b \pi\,R(x)^2\,dx.$$ The shell method instead slices the region into thin vertical strips parallel to the axis; each strip becomes a cylindrical shell of radius \(r\), height \(h\), and thickness \(dx\), contributing \(2\pi r h\,dx\). Whether disks or shells are easier depends on the geometry, but both rest on the same principle: build the solid from infinitesimal pieces and integrate ¹.

Work and Average Value

Integration also answers physical questions. When a variable force \(F(x)\) acts along a straight path, the work done in moving an object from \(x = a\) to \(x = b\) is $$W = \int_a^b F(x)\,dx.$$ This generalizes the constant-force formula \(W = F\,d\): over each tiny displacement \(dx\), the force is nearly constant, and the integral accumulates the contributions. Compressing a spring with force \(F(x) = kx\), for example, requires work \(\int_0^d kx\,dx = \tfrac{1}{2}kd^2\).

A related quantity is the average value of a function over an interval, $$\bar{f} = \frac{1}{b - a}\int_a^b f(x)\,dx,$$ which divides the accumulated total by the length of the interval — the continuous analog of averaging a list of numbers. Engineers use it for quantities such as the mean value of a fluctuating signal or a varying load ¹.

When Antiderivatives Run Out

These techniques are powerful, but many integrands have no antiderivative expressible in elementary functions — \(e^{-x^2}\) and \(\frac{\sin x}{x}\) are classic examples. In practice, a large share of integrals that arise in engineering are evaluated numerically, by approximating the area with sums of trapezoids or parabolic segments. A later course on computation develops those methods, which let you evaluate any well-behaved integral to whatever precision you need, even when no formula exists.