Definite Integration
Pinning the integral to two numbers — turning the area under a curve into an exact value, the theorem that ties that area to the antiderivative, and the symmetry tricks that collapse a fearsome integral to a single line
- The definite integral as a limit of a sum — the exact area swept out under a curve.
- The Fundamental Theorem of Calculus, which makes the antiderivative compute that area.
- The basic properties — reversal of limits and additivity over adjoining intervals.
- Symmetry for even and odd functions, and the king property of reflection.
- Substitution with changed limits, and the Leibniz rule for variable limits.
- Evaluating infinite series as integrals, and the meaning of an improper integral.
The Definite Integral as a Limit of a Sum
To measure the area under a curve between \(x=a\) and \(x=b\), slice the strip into \(n\) thin rectangles of width \(\Delta x\), add their areas, and let the rectangles grow infinitely thin. That limiting sum is the definite integral — a single number, not a family of functions.
The numbers \(a\) and \(b\) are the limits of integration. Where the indefinite integral returned a function, the definite integral returns a value. The variable of integration is a dummy: \(\int_a^b f(x)\,dx=\int_a^b f(t)\,dt\).
Rectangles below the \(x\)-axis carry a negative height, so the integral counts area there as negative. For \(\int_0^{2\pi}\sin x\,dx=0\) the area above the axis exactly cancels the area below — a true geometric area would need \(\int |f|\) instead.
The Fundamental Theorem of Calculus
Computing that limit by hand for every function would be hopeless. The Fundamental Theorem rescues us: it says the area-so-far function \(A(x)=\int_a^x f(t)\,dt\) has the original \(f\) as its derivative — so accumulating area and taking a slope are inverse acts. Definite integrals therefore collapse to a subtraction of antiderivative values.
The first part says differentiation undoes integration; the second turns any antiderivative \(F\) (with \(F'=f\)) into an evaluation, written \(\big[F(x)\big]_a^b\). Crucially, the constant \(C\) cancels in the subtraction — which is why a definite integral needs no \(+C\).
Basic Properties
A few properties follow straight from the limit-of-a-sum picture and save a great deal of labour. Swapping the limits flips the sign; a zero-width interval contributes nothing; and an integral can be broken at any interior point and reassembled.
The splitting point \(c\) need not lie between \(a\) and \(b\) — the signed-area bookkeeping makes it work anywhere. Additivity is exactly what lets you handle a piecewise integrand, such as \(\int_0^2 |x-1|\,dx\), by breaking the interval at the corner.
Symmetry & the King Property
When the interval is symmetric about the origin, the parity of the function does half your work. An even function repeats its left half on the right, so the area doubles; an odd function mirrors it upside-down, so the two halves cancel. A reflection of the variable — the king property — turns many stubborn integrals into trivial ones.
The second identity is the king property: reflecting \(x\mapsto a+b-x\) leaves the integral unchanged. Adding the integral to its reflected twin often replaces a complicated integrand with the constant \(1\) — the standard trick for ratios like \(\tfrac{\sin x}{\sin x+\cos x}\) on \([0,\tfrac\pi2]\).
Substitution in Definite Integrals
Substitution works just as it did for indefinite integrals, with one welcome simplification: instead of converting the answer back to \(x\), you convert the limits to the new variable and evaluate there directly.
When \(x=a\), the new variable is \(u=g(a)\); when \(x=b\), it is \(u=g(b)\). Change the limits along with the variable and there is no back-substitution at the end — a common slip is to switch the variable but forget to switch the limits.
The Leibniz Rule
Sometimes the limits of integration themselves depend on \(x\), and we want the derivative of the whole integral. The first part of the Fundamental Theorem, combined with the chain rule, gives a clean formula — no need to integrate at all.
Evaluate the integrand at each limit and multiply by that limit's derivative; the lower limit enters with a minus sign. If a limit is constant its derivative is zero and the term drops — recovering \(\tfrac{d}{dx}\int_a^x f=f(x)\) as the special case \(v=x,\ u=a\).
Evaluating Series as Integrals
The limit-of-a-sum definition runs both ways. A sum that no algebraic trick can collapse may, once you spot the pattern \(\tfrac1n\sum f(\tfrac rn)\), be read as a Riemann sum on \([0,1]\) — and an intractable infinite sum becomes a one-line integral.
To use it, pull out a factor of \(\tfrac1n\) and write every remaining \(r\) as the ratio \(\tfrac rn\), which becomes the variable \(x\). The running index \(\tfrac rn\) sweeps from \(0\) to \(1\), fixing the limits. General limits follow from \(\lim\tfrac1n\sum f(a+\tfrac{r}{n}(b-a))=\tfrac{1}{b-a}\int_a^b f\).
Improper Integrals
What if the region stretches to infinity, or the curve shoots up to infinity inside the interval? The integral is then improper, and we define it as the limit of an ordinary integral with a finite cut-off. Sometimes that limit settles on a finite number (the integral converges); sometimes it grows without bound (it diverges).
The benchmark is the p-test: \(\displaystyle\int_1^{\infty}\frac{dx}{x^{p}}\) converges precisely when \(p>1\) and diverges otherwise. So \(\int_1^\infty x^{-2}\,dx=1\) is finite, while the seemingly similar \(\int_1^\infty x^{-1}\,dx\) diverges — a surprisingly delicate boundary.
Putting It to Work
Problem. Evaluate \(\displaystyle\int_0^{\pi/2}\sin x\,dx\).
Solution. An antiderivative is \(-\cos x\); evaluate it at the two limits:
Problem. Evaluate \(\displaystyle\int_0^{1} 2x\,(x^{2}+1)^{3}\,dx\).
Solution. Let \(u=x^{2}+1\), so \(du=2x\,dx\); the limits become \(u:1\to2\):
Problem. Evaluate \(\displaystyle\int_{-1}^{1} x^{3}\,dx\).
Solution. Here \(f(-x)=-f(x)\), so the integrand is odd on a symmetric interval:
Problem. Evaluate \(\displaystyle\int_{-2}^{2} x^{2}\,dx\).
Solution. Since \(x^{2}\) is even, fold the interval and double:
Problem. Evaluate \(\displaystyle I=\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx\).
Solution. Reflect \(x\mapsto\tfrac\pi2-x\), which swaps sine and cosine, then add the two forms:
Problem. Evaluate \(\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{n}{n^{2}+r^{2}}\).
Solution. Factor out \(\tfrac1n\) to expose the Riemann sum with \(x=\tfrac rn\):
Chapter Summary
\(\int_a^b f\,dx=\lim_{n\to\infty}\sum f(x_i)\Delta x\) — a signed area.
\(\int_a^b f=F(b)-F(a)\); the constant cancels, no \(+C\).
Reverse limits flip the sign; split at any point with additivity.
Even doubles, odd vanishes; reflect \(x\mapsto a+b-x\).
Change the limits with the variable; no back-substitution.
Differentiate variable limits; take a limit out to \(\infty\).
Problems
Decide the route first — direct evaluation, symmetry, substitution, the king property, Leibniz, a series in disguise, or an improper limit — then apply the matching tool. Difficulty rises down the list.
- Evaluate \(\displaystyle\int_0^{3}(2x+1)\,dx\).
- Evaluate \(\displaystyle\int_1^{e}\frac{1}{x}\,dx\).
- Evaluate \(\displaystyle\int_0^{\pi}\sin x\,dx\).
- Use symmetry to evaluate \(\displaystyle\int_{-2}^{2} x^{3}\,dx\).
- Use symmetry to evaluate \(\displaystyle\int_{-1}^{1}(x^{2}+1)\,dx\).
- Use substitution to evaluate \(\displaystyle\int_0^{1} x\,e^{x^{2}}\,dx\).
- Evaluate \(\displaystyle\int_0^{\pi/2}\cos^{2}x\,dx\).
- Use the king property to evaluate \(\displaystyle\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}\,dx\).
- Evaluate \(\displaystyle\int_1^{4}\frac{1}{\sqrt{x}}\,dx\).
- Find \(\displaystyle\frac{d}{dx}\int_0^{x^{2}}\sin t\,dt\) using the Leibniz rule.
- Evaluate \(\displaystyle\lim_{n\to\infty}\frac1n\sum_{r=1}^{n}\sqrt{\frac{r}{n}}\).
- Determine whether \(\displaystyle\int_1^{\infty}\frac{dx}{x^{2}}\) converges, and find its value.
- Evaluate \(\displaystyle\int_0^{2}|x-1|\,dx\) by splitting at the corner.