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MATH1131 Chap.11 Riemann Integrals and Integration Techniques

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Chapter 11 of 12 · MATH1131

Riemann Integrals and Integration Techniques

The Riemann integral defines area through rectangle sums. A partition of [a, b] gives upper and lower sums, and a function is integrable when these can be made arbitrarily close; the common limit is the definite integral, interpreted as signed area. Every continuous function on a closed interval is integrable, and the integral is linear, additive over adjoining intervals and monotone.

The fundamental theorems connect integration with differentiation: the first says that the integral from a to x of a continuous function is differentiable with derivative equal to the function, and the second evaluates a definite integral as the change in any antiderivative.

Indefinite integrals are found with substitution, which reverses the chain rule and should change the limits along with the variable, and integration by parts, which reverses the product rule. Improper integrals over infinite intervals, or with integrands that blow up, are defined as limits and may converge or diverge; the p-test and the comparison tests decide many cases.

Functions defined by integrals, including those with variable limits, are differentiated using the first fundamental theorem with the chain rule.

In this chapter

What this chapter covers

  • 01

    Partitions, upper sums and lower sums

  • 02

    The Riemann integral as signed area

  • 03

    Properties of the integral

  • 04

    The first fundamental theorem of calculus

  • 05

    The second fundamental theorem of calculus

  • 06

    Integration by substitution

  • 07

    Integration by parts

  • 08

    Improper integrals and comparison tests

Worked example · free

An improper integral by limits and parts

Q [5 marks]. Determine whether the integral of x e^{-x} from 0 to infinity converges, and find its value if it does. The marks shown are a study allocation for practice, not an official University marking scheme.
Two similar curves, opposite verdicts.Two similar curves, opposite verdicts. The area under 1/x² from 1 to R is 1 − 1/R, which tends to 1. The area under 1/x grows without bound, so that integral diverges.xy1481y = 1/x: area divergesy = 1/x²: shaded area = 1
Two similar curves, opposite verdicts. The area under 1/x² from 1 to R is 1 − 1/R, which tends to 1. The area under 1/x grows without bound, so that integral diverges.
  • 1Define the integral as the limit as R tends to infinity of the integral from 0 to R.
  • 1Integrate by parts with u = x and dv = e^{-x} dx: the antiderivative is -x e^{-x} - e^{-x}.
  • 1Evaluate from 0 to R: (-R e^{-R} - e^{-R}) - (0 - 1) = 1 - R e^{-R} - e^{-R}.
  • 1As R tends to infinity, e^{-R} tends to 0 and R e^{-R} tends to 0 because the exponential dominates any power.
  • 1The limit is 1, so the integral converges with value 1.
The integral converges and equals 1.
Sia tip — Write the limit in R explicitly before integrating; an improper integral evaluated directly at infinity loses the convergence argument.
Glossary

Key terms

Riemann sum
A sum of function values times subinterval widths that approximates a definite integral.
Upper sum
A Riemann sum built from the largest function value on each subinterval of a partition.
Signed area
The integral's measure of area, counting regions below the axis as negative.
Integration by parts
The rule that reverses the product rule, trading the integral of u dv for uv minus the integral of v du.
Improper integral
An integral over an unbounded interval or of an unbounded integrand, defined as a limit of proper integrals.
Comparison test
A method that decides convergence of a non-negative improper integral by bounding it with a known integral.
Fundamental theorem of calculus
The pair of results linking differentiation and integration, giving evaluation by antiderivatives.
FAQ

Riemann Integrals and Integration Techniques FAQ

Why is the integral of sin x from 0 to 2 pi equal to zero?

The integral measures signed area. The region above the axis on the first half exactly cancels the region below the axis on the second half, even though the total geometric area is four.

How do I choose u in integration by parts?

Choose u to be the factor that becomes simpler when differentiated, such as a polynomial, and dv a factor you can integrate easily, such as an exponential or trigonometric function.

Do I need to substitute back after a definite substitution?

No, provided you changed the limits to the new variable when you substituted. Evaluate the new integral between the new limits and the answer is complete.

How do I differentiate an integral whose upper limit is a function of x?

Apply the first fundamental theorem with the chain rule: the derivative is the integrand evaluated at the upper limit multiplied by the derivative of that limit.

When does the integral of 1 over x to the p from 1 to infinity converge?

Exactly when p is greater than 1; for p equal to 1 or less it diverges. This benchmark is the usual comparison function for deciding whether other improper integrals converge.

What is the difference between an upper sum and a lower sum?

An upper sum uses the largest value of the function on each subinterval and a lower sum the smallest. Every lower sum is at most every upper sum, and the integral lies between them for integrable functions.

Study strategy

Exam move

Compute at least one Riemann sum limit fully by hand using the formulas for the sums of k and k squared, since that question type recurs. Then build fluency in substitution and parts on mixed lists where you must choose the method yourself. Use the Table of Integrals supplied in assessments to check standard forms quickly.

For improper integrals, always write the limit, and for comparison arguments state the inequality and its direction explicitly before concluding. After each antiderivative, differentiate it to confirm you recover the integrand; this ten-second check catches most sign and constant errors in parts and substitution.

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