Integrals

Integration reverses differentiation: $F' = f \Rightarrow \int f\,dx = F + C$. The definite integral $\int_a^b f\,dx$ is the signed area under $f$ — and by the fundamental theorem, $F(b) - F(a)$.

Basic rules

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \;\;(n \neq -1) \qquad \int \frac{dx}{x} = \ln|x| + C$

$\int e^x\,dx = e^x + C \qquad \int \sin x\,dx = -\cos x + C \qquad \int \cos x\,dx = \sin x + C$

Linearity as with derivatives. Always check by differentiating your result — it's free.

Worked example.

$\int \left(3x^2 - \frac{4}{x} + 5e^x\right) dx = x^3 - 4\ln|x| + 5e^x + C$

Exercise 2.1

a) $\displaystyle\int \left(6x^2 - 2x + 7\right) dx$

b) $\displaystyle\int \frac{3}{\sqrt{x}}\,dx$

c) $\displaystyle\int \left(2\cos x - \frac{1}{2}e^x + \frac{5}{x}\right) dx$

Substitution

Reverse chain rule: set $u = g(x)$, then $du = g'(x)\,dx$ and

$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$

Pick $u$ so that its derivative is sitting in the integrand (up to a constant).

Worked example. $\displaystyle\int x\,e^{x^2}\,dx$

$u = x^2$, $du = 2x\,dx \Rightarrow x\,dx = \tfrac12 du$:

$\int e^u \cdot \tfrac12\,du = \tfrac12 e^u + C = \tfrac12 e^{x^2} + C$

Exercise 2.2

Solve by substitution; state your $u$ explicitly:

a) $\displaystyle\int (2x+1)^5\,dx$

b) $\displaystyle\int \frac{x}{x^2 + 3}\,dx$

c) $\displaystyle\int \cos^3 x \,\sin x\,dx$

d) $\displaystyle\int \frac{\ln x}{x}\,dx$

Integration by parts

Reverse product rule:

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

Choose $u$ as the factor that gets simpler when differentiated (rough priority: $\ln$, polynomial, exponential/trig).

Worked example. $\displaystyle\int x\,e^x\,dx$

$u = x$, $dv = e^x dx \Rightarrow du = dx$, $v = e^x$:

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

Exercise 2.3

a) $\displaystyle\int x \cos x\,dx$

b) $\displaystyle\int \ln x\,dx$ (hint: $u = \ln x$, $dv = dx$)

c) $\displaystyle\int x^2 e^x\,dx$ (parts twice)

Definite integrals

$\int_a^b f(x)\,dx = F(b) - F(a)$

No $+C$. Area below the axis counts negative. With substitution, transform the bounds too — then you never substitute back.

Worked example. $\displaystyle\int_0^2 (3x^2 - 2x)\,dx = \left[x^3 - x^2\right]_0^2 = 8 - 4 = 4$

Worked example (substitution with bounds). $\displaystyle\int_0^{\sqrt{\pi}} x \sin(x^2)\,dx$ with $u = x^2$: bounds $0 \to 0$, $\sqrt{\pi} \to \pi$:

$\tfrac12 \int_0^{\pi} \sin u\,du = \tfrac12\left[-\cos u\right]_0^\pi = \tfrac12(1 + 1) = 1$

See it

Riemann sums converging to $\int_0^2 x^2\,dx = \tfrac83$. Crank N up.

import numpy as np
import matplotlib.pyplot as plt

f = lambda x: x**2
a, b, N = 0.0, 2.0, 8      # try N = 8, 20, 100

xs = np.linspace(a, b, 400)
edges = np.linspace(a, b, N + 1)
mids = (edges[:-1] + edges[1:]) / 2
w = (b - a) / N

plt.bar(mids, f(mids), width=w, alpha=0.4, edgecolor='k')
plt.plot(xs, f(xs), 'r')
plt.title(f'midpoint sum = {np.sum(f(mids)*w):.5f}   exact = {8/3:.5f}')
plt.grid(True)
plt.show()

Exercise 2.4

a) $\displaystyle\int_1^3 \left(2x - \frac{1}{x^2}\right) dx$

b) $\displaystyle\int_0^{\pi/2} \sin(2x)\,dx$

c) $\displaystyle\int_1^e \frac{(\ln x)^2}{x}\,dx$