Calculus II: Integration

Two questions, one answer#

Integration answers two apparently unrelated questions. Question one: what is the area under a curve? Chop it into rectangles, add them up, refine forever:

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

Question two: which function has derivative f? Such an F (with F' = f) is an antiderivative, written \int f(x)\,dx = F(x) + C — the constant C because derivatives kill constants.

That these two questions have the same answer is the miracle at the center of calculus:

The Fundamental Theorem of Calculus. If F' = f, then \int_a^b f(x)\,dx = F(b) - F(a). Area — a global, accumulated quantity — is computed by evaluating an antiderivative at just the two endpoints. Accumulation and rate-of-change are inverse operations: the odometer is the integral of the speedometer.

The basic antiderivatives are the derivative table read in reverse:

\begin{aligned}\int x^n dx = \frac{x^{n+1}}{n+1} + C \;(n \ne -1), \\ \int \frac{dx}{x} = \ln|x| + C, \\ \int e^x dx = e^x + C,\end{aligned} \begin{aligned}\int \cos x\, dx = \sin x + C, \\ \int \sin x\, dx = -\cos x + C.\end{aligned}

The two great techniques#

Substitution is the chain rule run backwards. The chain rule says \frac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x — so if we meet \int 2x\cos(x^2)\,dx, the answer must be \sin(x^2) + C. Substitution is the bookkeeping that makes this recognition systematic. The five-step ritual, fully narrated:

Evaluate \int 2x\cos(x^2)\,dx.

1. Choose u: pick the inner function. u = x^2

\begin{aligned}&\text{2. Compute } du: \text{ differentiate } u. && \frac{du}{dx} = 2x \ \Rightarrow\ du = 2x\,dx \\ &\text{3. Rewrite the integral entirely in } u: && \int \cos(\underbrace{x^2}_{u})\,\underbrace{2x\,dx}_{du} = \int \cos u\,du \\ &\phantom{3.}\ \text{— the } 2x\,dx \text{ we "happened" to have is } du.\end{aligned}

(that coincidence is the whole method)

\begin{aligned}&\text{4. Integrate in } u: && \int \cos u\,du = \sin u + C \\ &\text{5. Substitute back } u = x^2: && \boxed{\sin(x^2) + C}\end{aligned}

Verify by differentiating — the habit that makes integration self-correcting: \frac{d}{dx}\sin(x^2) = \cos(x^2)\cdot 2x ✓, exactly the integrand.

What if the constant doesn’t match perfectly? Then balance it by hand. For \int x\,e^{x^2}dx: choose u = x^2, so du = 2x\,dx — but the integrand has only x\,dx, which is \frac{1}{2}du. So \begin{gathered}\int x e^{x^2} dx \\ = \int e^u \cdot \tfrac{1}{2}\,du \\ = \tfrac{1}{2}e^u + C \\ = \tfrac{1}{2}e^{x^2} + C,\end{gathered} and differentiating the answer gives \frac{1}{2}e^{x^2}\!\cdot 2x = xe^{x^2} ✓. Substitution succeeds precisely when the integrand contains a function and (up to a constant) its derivative; training the eye to spot that pair is the skill.

Integration by parts is the product rule run backwards. Start from the product rule and integrate both sides: \begin{gathered}(uv)' \\ = u'v + uv' \\ \Rightarrow uv \\ = \int u'v\,dx + \int uv'\,dx \\ \Rightarrow \int u\,dv \\ = uv - \int v\,du.\end{gathered} It is a trade: swap the integral you have for a (hopefully easier) one. The craft is choosing which factor to call u — pick the factor that gets simpler when differentiated (a useful priority order: Logs, Inverse trig, Algebraic powers, Trig, Exponentials — differentiate the earlier, integrate the later). Fully worked:

Evaluate \int x e^x\,dx.

\begin{aligned}&\text{1. Choose: } u = x \text{ (a power — differentiating kills it)}, && dv = e^x dx \\ &\text{2. Then: } du = dx, && v = \int e^x dx = e^x \\ &\text{3. Apply the formula: } \int u\,dv = uv - \int v\,du && \int xe^x dx = xe^x - \int e^x\,dx\end{aligned}

4. The new integral is elementary: = xe^x - e^x + C = (x-1)e^x + C.

Verify: \frac{d}{dx}\big[(x-1)e^x\big] = 1\cdot e^x + (x-1)e^x = xe^x ✓. (Try the other choice, u = e^x, dv = x\,dx: you get \int \frac{x^2}{2}e^x dx — harder than what you started with. Parts punishes bad casting immediately, which at least makes the mistake obvious.)

What integrals compute#

Everything that accumulates. The area between curves f (above) and g (below) is \int_a^b (f - g)\,dx. A solid of revolution about the x-axis has volume \pi\int_a^b f(x)^2\,dx (a stack of discs). Distance is the integral of speed; total rainfall the integral of rainfall rate; probability the integral of a density — the normal curve of Chapter 9 encloses total area exactly 1.

Worked example, start to finish. Find the area enclosed between y = x^2 and y = x + 2.

1. Where do they cross? Set them equal: x^2 = x + 2

\begin{aligned}&\phantom{1.}\ \text{Rearrange and factor:} && (x-2)(x+1) = 0 \;\Rightarrow\; x = -1,\ 2 \\ &\text{2. Which is on top? Test } x = 0: && x + 2 = 2 > 0 = x^2 \text{ — the line} \\ &\text{3. Area} = \int_{-1}^{2}\big[\text{top} - \text{bottom}\big]\,dx && = \int_{-1}^{2}\big[(x+2) - x^2\big]dx\end{aligned}

4. Antidifferentiate term by term: \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}

\begin{aligned}&\text{5. Evaluate at } x = 2: && 2 + 4 - \frac{8}{3} = \frac{10}{3} \\ &\text{6. Evaluate at } x = -1: && \frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}\end{aligned}

7. Subtract the two: \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \boxed{\frac{9}{2}}.

A positive answer is a sanity check in itself — an enclosed area must be positive; getting a negative number almost always means top and bottom were swapped in step 3.

If you keep one thing from this chapter: Integration accumulates, differentiation reads rates, and the Fundamental Theorem says they are inverse operations: area from just two endpoint values of an antiderivative.