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MATH1061 Chap.3 Integration: ∫, Σ, Δ

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Chapter 3 of 7 · MATH1061

Integration: ∫, Σ, Δ

Integration has two linked meanings: reversing differentiation and accumulating signed contributions across an interval. This chapter begins with antiderivative families and the constant of integration, then develops definite integrals through Riemann sums and the Fundamental Theorem of Calculus. Substitution is framed as a structural reversal of the chain rule.

Area questions keep geometric area separate from signed integral so that regions below the axis are handled deliberately rather than silently discarded. The Riemann-sum picture explains why interval width, sample height and summation belong together. As the partition becomes finer, those rectangle contributions approach a definite integral under suitable continuity conditions.

The Fundamental Theorem then converts the limiting accumulation into endpoint evaluation of an antiderivative. That result makes computation efficient without erasing interpretation. A complete solution still states the interval, preserves or converts bounds consistently under substitution, and checks whether the requested quantity is signed accumulation, net change or total geometric area.

In this chapter

What this chapter covers

  • 01

    Recover antiderivative families

  • 02

    Explain Riemann sums as accumulation

  • 03

    Evaluate definite integrals with endpoints

  • 04

    Apply the Fundamental Theorem of Calculus

  • 05

    Choose substitutions from derivative structure

  • 06

    Separate signed integral from geometric area

Worked example · free

Use substitution on a definite integral

Q [4 marks]. Evaluate the integral from 0 to 1 of 2x times (x squared + 1) cubed with respect to x. This weighting is an AskSia revision device, not a published University mark allocation.
For an increasing curve, left-endpoint rectangles lie below the graph, so their total gives a lower sum that approaches the definite integral as widths shrink.For an increasing curve, left-endpoint rectangles lie below the graph, so their total gives a lower sum that approaches the definite integral as widths shrink. aby=f(x)equal subintervals
For an increasing curve, left-endpoint rectangles lie below the graph, so their total gives a lower sum that approaches the definite integral as widths shrink.
  • 1Choose u = x squared + 1, so du = 2x dx.
  • 1Convert the bounds: x = 0 gives u = 1 and x = 1 gives u = 2.
  • 1Integrate u cubed from 1 to 2 to obtain one quarter u to the fourth at the bounds.
  • 1The value is (16 - 1)/4 = 15/4.
The definite integral equals 15/4.
Sia tip — For a definite integral, either convert the bounds immediately or return to x before applying them; do not mix the two variables.
Glossary

Key terms

Antiderivative
A function whose derivative is the given integrand.
Integration constant
The arbitrary constant representing all antiderivatives of an indefinite integral.
Signed area
Accumulation counted positively above the axis and negatively below it.
Substitution
A change of variable that simplifies an integral by matching derivative structure.
Partition
A division of an interval into smaller subintervals used for finite accumulation.
Sample point
The chosen input whose function value supplies a rectangle height in a Riemann sum.
Net change
The final value minus the initial value obtained by integrating a rate.
Geometric area
The nonnegative size of regions, requiring negative signed contributions to be reversed.
FAQ

Integration: ∫, Σ, Δ FAQ

Why does an indefinite integral need a constant?

Differentiation removes additive constants, so many functions have the same derivative. The constant records that full family of antiderivatives. A definite integral does not retain an arbitrary constant because the endpoint subtraction cancels it. Omitting the constant from an indefinite result loses valid solutions and can break a later initial-value condition.

How is an integral different from geometric area?

A definite integral is signed accumulation. Portions below the horizontal axis contribute negatively, while geometric area is nonnegative. To find total geometric area, locate every sign change, split the interval and reverse the sign of negative contributions or integrate the absolute value carefully. A quick sketch often reveals whether a single integral can represent the requested quantity.

How can I check an integration answer?

Differentiate an indefinite result and confirm that the original integrand returns exactly. For a definite integral, also check sign and scale from the graph or rough rectangle bounds. Under substitution, verify that every occurrence of the old variable, including the differential and bounds, was converted consistently before evaluating.

When should the bounds change under substitution?

If the integral remains in the new variable, convert each original endpoint by the substitution before evaluation. If you instead return the antiderivative to the original variable, keep the original bounds. Both routes are valid, but mixing new-variable expressions with old-variable endpoints is not. Writing the variable beside each bound prevents that inconsistency.

Study strategy

Exam move

Pair every integration technique with its derivative pattern. Practise identifying a likely inner expression and checking whether its derivative appears as a factor before calculating. Build a sorting exercise with power forms, composite forms and definite integrals before attempting full solutions. On substitution problems, write a conversion ledger containing the new variable, differential and transformed endpoints.

For definite integrals, draw a quick axis sketch and predict whether the answer should be positive, negative or near zero. Separate intervals at every axis crossing when total area is requested. After solving, differentiate the antiderivative and compare the numerical size with simple upper and lower rectangle estimates. If the curve is increasing, use left and right sums to bracket the answer and confirm its scale.

Classify mistakes as an antiderivative rule error, a missing constant, inconsistent bounds or a sign interpretation error. Correcting the category, rather than merely recopying the solution, makes the check available on the next mixed problem.

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