Integrals

Divide the interval $[a,b]$ into $n$ thin strips of width $\Delta x = (b-a)/n$. In each strip, build a rectangle with height $f(x_i)$.

The total area of all rectangles is approximately $\sum_{i=1}^{n} f(x_i)\Delta x$. This is a Riemann sum.

As $n \to \infty$ (strips get infinitely thin), the Riemann sum converges to the definite integral: $\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\Delta x$.

This is why integrals represent accumulation: you are adding up infinitely many infinitesimally small contributions.

FTC Part 1: if $F(x) = \int_a^x f(t)\,dt$, then $F'(x) = f(x)$. Differentiation and integration are inverse operations.

FTC Part 2: $\int_a^b f(x)\,dx = F(b) - F(a)$ where $F$ is any antiderivative of $f$ (meaning $F'=f$).

Part 2 is the computational powerhouse: instead of computing limits of Riemann sums, just find an antiderivative and evaluate at the bounds.

The connection to derivatives: if position is $s(t)$ and velocity is $v(t) = s'(t)$, then total displacement is $\int_a^b v(t)\,dt = s(b) - s(a)$. Integration recovers position from velocity.

The definite integral computes net (signed) area: regions above the $x$-axis contribute positively, regions below contribute negatively.

If you want total area (all positive), integrate the absolute value: $\int_a^b |f(x)|\,dx$.

In practice, find where $f(x) = 0$, split the interval at those zeros, and flip the sign on intervals where $f < 0$.

An indefinite integral $\int f(x)\,dx$ represents the family of all antiderivatives of $f$.

Since the derivative of a constant is $0$, any two antiderivatives differ by a constant. That's why we write $+C$.

Never forget $+C$ on indefinite integrals. It's not optional: it represents an infinite family of solutions.

If the integrand looks like $f(g(x)) \cdot g'(x)$, substitute $u = g(x)$ so $du = g'(x)\,dx$.

The integral becomes $\int f(u)\,du$, which is usually simpler.

Example: $\int 2x \cos(x^2)\,dx$. Let $u = x^2$, $du = 2x\,dx$. Integral becomes $\int \cos u\,du = \sin u + C = \sin(x^2) + C$.

For definite integrals, either convert the bounds to $u$-values, or back-substitute and use the original bounds.

Tip: look for an "inner function" whose derivative (or a constant multiple of it) appears as a factor in the integrand.

Formula: $\int u\,dv = uv - \int v\,du$. This comes from the product rule run backwards.

Choose $u$ and $dv$ using the LIATE heuristic: Logarithms, Inverse trig, Algebraic ($x^n$), Trig, Exponentials. Pick $u$ from higher in this list.

Example: $\int x e^x\,dx$. Let $u = x$ (algebraic), $dv = e^x\,dx$. Then $du = dx$, $v = e^x$.

Result: $x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1) + C$.

Sometimes you need to apply by-parts twice (e.g., $\int x^2 e^x\,dx$) or use the tabular method for repeated applications.

A special case: $\int e^x \sin x\,dx$ requires by-parts twice, then solving for the integral algebraically.

Powers of sin and cos: for $\int \sin^m x \cos^n x\,dx$, use the following strategies:

If one power is odd, peel off one factor, convert the rest using $\sin^2 x + \cos^2 x = 1$, and substitute.

If both powers are even, use the half-angle identities: $\sin^2 x = \frac{1-\cos(2x)}{2}$ and $\cos^2 x = \frac{1+\cos(2x)}{2}$.

Example: $\int \sin^2 x\,dx = \int \frac{1-\cos(2x)}{2}\,dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C$.

Example: $\int \cos^3 x\,dx = \int \cos^2 x \cos x\,dx = \int (1-\sin^2 x)\cos x\,dx$. Let $u = \sin x$: result is $\sin x - \frac{\sin^3 x}{3} + C$.

For integrands involving $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$, use trig substitutions:

$\sqrt{a^2 - x^2}$: let $x = a\sin\theta$, so $\sqrt{a^2-x^2} = a\cos\theta$.

$\sqrt{a^2 + x^2}$: let $x = a\tan\theta$, so $\sqrt{a^2+x^2} = a\sec\theta$.

$\sqrt{x^2 - a^2}$: let $x = a\sec\theta$, so $\sqrt{x^2-a^2} = a\tan\theta$.

After substituting, the square root disappears and you get a trig integral. Evaluate, then convert back to $x$ using a reference triangle.

To integrate a rational function $\frac{P(x)}{Q(x)}$ where $\deg P < \deg Q$, decompose it into simpler fractions.

Factor the denominator, then write: $\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$.

Solve for $A$ and $B$ by clearing denominators and comparing coefficients or plugging in strategic $x$ values.

Repeated factors: $\frac{1}{(x-1)^2}$ requires terms $\frac{A}{x-1} + \frac{B}{(x-1)^2}$.

Irreducible quadratics: for $(x^2+1)$ in the denominator, the numerator is $Ax+B$, not just $A$.

If $\deg P \geq \deg Q$, perform polynomial long division first, then decompose the remainder.

An improper integral has an infinite bound or an integrand with an infinite discontinuity.

Type 1 (infinite bound): $\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b\to\infty} \int_1^b \frac{1}{x^2}\,dx = \lim_{b\to\infty} [-1/x]_1^b = \lim_{b\to\infty}(-1/b+1) = 1$. This converges.

Type 2 (discontinuity): $\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{a\to 0^+} \int_a^1 x^{-1/2}\,dx = \lim_{a\to 0^+} [2\sqrt{x}]_a^1 = 2$. This also converges.

If the limit is finite, the improper integral converges. If the limit is $\pm\infty$ or does not exist, it diverges.

Key result: $\int_1^{\infty} \frac{1}{x^p}\,dx$ converges if $p > 1$ and diverges if $p \leq 1$. This mirrors the p-series test.

1. Simplify first: can algebra, trig identities, or long division reduce the integral?

2. Check the table: is it a standard form you already know?

3. Substitution: is there an inner function whose derivative appears in the integrand?

4. Trig integrals: powers of $\sin$ and $\cos$? Use identities.

5. Trig substitution: see $\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$?

6. Integration by parts: product of two different types of functions?

7. Partial fractions: rational function with a factorable denominator?

Practice is the only way to develop intuition for which technique to try first.

Area between $f(x)$ and $g(x)$ on $[a,b]$: $\int_a^b |f(x) - g(x)|\,dx$. Determine which function is on top on each sub-interval.

Volume by disks/washers: revolve a region around the $x$-axis: $V = \pi\int_a^b [f(x)]^2\,dx$.

Volume by shells: revolve around the $y$-axis: $V = 2\pi\int_a^b x\,f(x)\,dx$.

These are applications of integration, extending the concept of 'summing up infinitely many small pieces' to 2D and 3D geometry.