3  Integration and the Fundamental Theorem

3.1 Riemann sums and signed accumulation

Let \(f:[a,b]\to\mathbb R\) be bounded, with \(a<b\). A partition is \(a=x_0<x_1<\cdots<x_n=b\), with widths \(\Delta x_i=x_i-x_{i-1}\) and tags \(\xi_i\in[x_{i-1},x_i]\). Its Riemann sum is \[\sum_{i=1}^n f(\xi_i)\Delta x_i.\] The Riemann integral \(\int_a^b f(x)\,dx=I\) exists if these sums tend to the same \(I\) as the mesh \(\max_i\Delta x_i\to0\), independently of partitions and tags. Positive and negative values contribute with their signs. Geometric area between the curve and the axis is \(\int_a^b|f(x)|\,dx\), when this exists.

Every continuous function on \([a,b]\) is Riemann integrable. A proof uses uniform continuity on the compact interval to make the upper and lower sums arbitrarily close. Boundedness alone does not suffice: the indicator of the rational numbers on \([0,1]\) has upper sums \(1\) and lower sums \(0\) for every partition.

3.1.1 Worked example: evaluating a sum

For \(f(x)=x\) on \([0,1]\), take \(n\) equal subintervals and right endpoints \(i/n\): \[\sum_{i=1}^n\frac in\frac1n =\frac{n(n+1)}{2n^2}\longrightarrow\frac12.\] Continuity guarantees that this limit is the integral, independent of the particular tags. A single sequence of sums would not by itself prove integrability for an arbitrary bounded function.

3.2 Properties and orientation

For integrable \(f,g\) and scalars \(\alpha,\beta\), \[\int_a^b(\alpha f+\beta g)=\alpha\int_a^b f+\beta\int_a^b g.\] Integrals add over adjacent intervals. Define \(\int_a^a f=0\) and \(\int_b^a f=-\int_a^b f\). If \(f\le g\) on \([a,b]\), then \(\int_a^b f\le\int_a^b g\), and \[\left|\int_a^b f\right|\le\int_a^b|f|\le(b-a)\sup_{[a,b]}|f|.\] The order inequalities are stated for \(a\le b\); reversing the limits reverses signs.

3.3 Fundamental Theorem of Calculus

Part I: differentiating accumulation. If \(f\) is continuous on \([a,b]\), define \(F(x)=\int_a^x f(t)\,dt\). Then \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\), with \(F'(x)=f(x)\) there.

Proof sketch. For small nonzero \(h\), \[\frac{F(x+h)-F(x)}h-f(x) =\frac1h\int_x^{x+h}[f(t)-f(x)]\,dt.\] Its absolute value is bounded by the maximum of \(|f(t)-f(x)|\) between \(x\) and \(x+h\), which tends to zero by continuity. The argument covers negative \(h\) using oriented integrals.

Part II: evaluating by antiderivatives. If \(f\) is continuous on \([a,b]\) and \(G\) is continuous on \([a,b]\) and differentiable on \((a,b)\) with \(G'=f\), then \[\int_a^b f(x)\,dx=G(b)-G(a).\] Indeed, Part I constructs \(F\) with \(F'=f\). The Mean Value Theorem shows that \(G-F\) is constant, and continuity extends that identity to the endpoints.

An indefinite integral \(\int f(x)\,dx\) denotes a family of antiderivatives on an interval, written \(G(x)+C\). A definite integral is a number. The constant \(C\) cancels when evaluating a definite integral.

3.3.1 Worked example: signed integral and area

An antiderivative of \(x^2-1\) is \(x^3/3-x\), so \[\int_0^2(x^2-1)\,dx=\left[\frac{x^3}{3}-x\right]_0^2=\frac23.\] The function changes sign at \(1\). Its unsigned area is \[-\int_0^1(x^2-1)\,dx+\int_1^2(x^2-1)\,dx =\frac23+\frac43=2.\]

3.4 Substitution

If \(g\) is continuously differentiable on \([a,b]\) and \(f\) is continuous on an interval containing the range of \(g\), then \[\int_a^b f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)}f(u)\,du.\] Take an antiderivative of \(f\) and apply the chain rule and Part II. For this one-dimensional oriented formula, \(g\) need not be one-to-one.

For example, set \(u=1+x^2\), \(du=2x\,dx\). The endpoints \(x=0,1\) become \(u=1,2\): \[\int_0^1\frac{2x}{1+x^2}\,dx=\int_1^2\frac1u\,du=\log2.\] Changing variables means changing the differential and bounds together.

3.5 Exercises

  1. Evaluate \(\int_0^1(3x^2+2)\,dx\).
  2. Find \(H'(x)\) for \(H(x)=\int_0^{x^2}\cos t\,dt\).
  3. Evaluate \(\int_0^\pi\sin x\,dx\) and \(\int_\pi^0\sin x\,dx\).
  4. Can Part II above be used directly to evaluate \(\int_{-1}^1x^{-2}\,dx\) as \([-1/x]_{-1}^1=-2\)?
  1. \([x^3+2x]_0^1=3\).
  2. \(2x\cos(x^2)\), by Part I and the chain rule.
  3. \(2\) and \(-2\), respectively.
  4. No. The integrand is unbounded and undefined at \(0\), so it is not a proper Riemann integral on this interval. As an improper integral it diverges to \(+\infty\), as each side of the singularity must be considered separately. The antiderivative cannot be evaluated across that singularity using this theorem.

Integration techniques and improper integrals are planned next.