The Timpson Lyceum · Calculus 1 (2024)

Integration by Substitution, and Area Between Curves - Lecture 16 (Calculus 1)

Overview

This lecture introduces the first and most important method of integration: integration by substitution (u-substitution). While derivatives are mechanical to compute, antiderivatives are genuinely hard — most algebraic functions have no antiderivative expressible in elementary terms — so we need special techniques for the special forms we can handle. U-substitution is essentially the chain rule run in reverse: it lets us recognize an integrand as $f(g(x))\,g'(x)$ and collapse it back to a simple integral in a new variable $u = g(x)$ . The lecture applies the method to indefinite integrals, proves why it works, uses the power-reduction identities to integrate $\sin^2$ and $\cos^2$ , then extends substitution to definite integrals (changing the limits) and finishes with even/odd symmetry shortcuts and the setup for finding the area between two curves.

Key Concepts

Why integration is hard

Differentiation is algorithmic; integration is not. Many simple-looking functions (e.g. $\sqrt{x^3+1}$ ) have no elementary antiderivative.
We can only integrate functions that match one of a limited set of recognizable forms. U-substitution is the tool for spotting the "chain-rule" form.

The substitution rule (reversing the chain rule)

The chain rule gives $\dfrac{d}{dx}F(g(x)) = F'(g(x))\,g'(x) = f(g(x))\,g'(x)$ , where $F' = f$ .
Reading this backward yields the substitution theorem: if $u = g(x)$ , then $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du.$
How to apply it: Let $u$ be the "inside" function $g(x)$ . Compute $du = g'(x)\,dx$ . Rewrite the entire integrand — including $dx$ — in terms of $u$ and $du$ , integrate, then substitute $x$ back.

The key trick — manufacturing $du$

The whole challenge is rewriting the integrand as $f(u)\,du$ . If the integrand is off by a constant factor, fix it by multiplying and dividing by that constant (you may pull a constant outside the integral, e.g. $\int g(7\theta+3)\,d\theta = \tfrac17\int g(7\theta+3)\cdot 7\,d\theta$ ).
If a leftover $x$ remains after substituting, solve $u = g(x)$ for $x$ and substitute that in too (e.g. $\int x\sqrt{2x+1}\,dx$ with $x = \tfrac{u-1}{2}$ ).

Power-reduction for $\sin^2$ and $\cos^2$

These come from the double-angle identity $\cos 2\theta = \cos^2\theta - \sin^2\theta$ combined with $\sin^2\theta + \cos^2\theta = 1$ : $\sin^2\theta = \frac{1-\cos 2\theta}{2}, \qquad \cos^2\theta = \frac{1+\cos 2\theta}{2}.$
Each then integrates term-by-term, with $\int\cos 2\theta\,d\theta = \tfrac12\sin 2\theta$ (a quick mental substitution $u = 2\theta$ ).

Substitution in definite integrals

Two equivalent options. Either (a) change the limits: when $u = g(x)$ , replace $x = a, b$ with $u = g(a), g(b)$ and never return to $x$ ; or (b) keep the $x$ -limits, back-substitute $u \to g(x)$ , and evaluate at the original $x$ -values. Both give the same answer. $\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.$

Even/odd symmetry shortcuts

Even ( $f(-x)=f(x)$ , symmetric about the $y$ -axis): $\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ .
Odd ( $f(-x)=-f(x)$ , symmetric about the origin): $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$ .

Area between two curves

To find the area enclosed by two curves, first find their intersection points (set the expressions equal and solve) to get the limits of integration, then integrate the (top curve − bottom curve) across that interval. A sketch is almost always needed to decide which curve is on top.

Worked Example

Indefinite u-substitution. To evaluate $\displaystyle\int (x^3+x)^5\,(3x^2+1)\,dx$ , let $u = x^3+x$ so $du = (3x^2+1)\,dx$ :

\begin{aligned} \int (x^3+x)^5\,(3x^2+1)\,dx &= \int u^5\,du \\ &= \frac{u^6}{6} + C \\ &= \frac{(x^3+x)^6}{6} + C. \end{aligned}

Definite u-substitution (changing the limits). For $\displaystyle\int_{-1}^{1} 3x^2\sqrt{x^3+1}\,dx$ , let $u = x^3+1$ , $du = 3x^2\,dx$ . When $x=-1,\ u=0$ ; when $x=1,\ u=2$ :

\begin{aligned} \int_{-1}^{1} 3x^2\sqrt{x^3+1}\,dx &= \int_{0}^{2} u^{1/2}\,du \\ &= \left[\frac{2}{3}u^{3/2}\right]_{0}^{2} \\ &= \frac{2}{3}\,2^{3/2} - 0 \;=\; \frac{4\sqrt{2}}{3}. \end{aligned}

Summary

U-substitution undoes the chain rule: spot an integrand of the form $f(g(x))\,g'(x)$ , set $u=g(x)$ , and reduce it to $\int f(u)\,du$ .
Always rewrite everything (including $dx \to du$ ) in terms of $u$ ; fix a missing constant by multiplying and dividing, and solve for $x$ if a stray $x$ remains.
Integrate $\sin^2\theta$ and $\cos^2\theta$ with the power-reduction identities $\frac{1\mp\cos 2\theta}{2}$ .
For definite integrals you may either change the limits to $u$ -values (then never return to $x$ ) or back-substitute and use the original $x$ -limits — both yield the same result.
Use symmetry: an even function over $[-a,a]$ integrates to twice its $[0,a]$ integral; an odd function over $[-a,a]$ integrates to $0$ .
To find the area between curves, locate the intersection points for the limits, then integrate top minus bottom — sketch the region first.