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Evaluate and transform a definite integral

Start a set of 8
Step 26 of 29

Apply the integral property before calculating

\[f(u)=4u+4\]
\[\int_{0}^{k}\left(f(u)\right)\,du=\frac{45}{2},\qquad 0\le k\le6\]
(a)3 marks

Choose the primitive with constant term zero

Enter every coefficient in descending powers, including zero for an absent term

\[F(u)=A_2u^{2}+A_1u+A_0\]
(b)1 mark

Select the equation for k when F is a primitive of f

(c)1 mark

Find the unique value of k in the stated domain

Year 12, VCE Units 3 & 4, Chapter 8, 4 difficulty levels, No calculatorBuild a worksheet

Definite integral evaluation and properties

Use a primitive and ordered limits; preserve signs when transforming an integral

\[\int_a^b f(x)\,dx=F(b)-F(a)\]
  1. Reverse or repeat limits Swapping the limits negates the value. Equal limits give zero
  2. Split an interval Add the signed contributions over adjacent intervals
  3. Use linearity Apply each constant to its own integral and add the results
  4. Rename consistently A dummy-variable rename changes neither the integrand rule nor the limits
  5. Check an unknown endpoint Solve the primitive-difference equation and retain only roots in the stated domain. Use monotonicity when claiming uniqueness

Worked example

Evaluate the displayed integral, then reverse its limits

\[\int_1^4(2x+3)\,dx\]
  1. \(\displaystyle F(x)=x^2+3x\)Reverse the power rule term by term
  2. \(\displaystyle F(4)=28,\quad F(1)=4\)Evaluate the chosen primitive at both ordered endpoints
  3. \(\displaystyle \int_1^4(2x+3)\,dx=28-4=24\)Upper value minus lower value; a constant would cancel
  4. \(\displaystyle \int_4^1(2x+3)\,dx=-24\)Reversing the limits changes the sign

Common mistakes

Subtracting in the wrong order
✕\(\displaystyle \int_a^b f(x)\,dx=F(a)-F(b)\)✓\(\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)\)
Use the upper-endpoint primitive value first
Keeping an arbitrary constant
✕\(\displaystyle \int_1^4(2x+3)\,dx=24+C\)✓\(\displaystyle \int_1^4(2x+3)\,dx=24\)
The same C occurs at both endpoints and cancels
Ignoring a negative scale
✕\(\displaystyle 2(24)-3(-5)=33\)✓\(\displaystyle 2(24)-3(-5)=63\)
Subtracting a negative contribution adds its magnitude

Difficulty levels

introForm a primitive and evaluate with nonnegative integer endpoints, or recognise zero width
standardReuse supplied-primitive evaluation with signed/rational endpoints and practise dummy-variable renaming
challengeApply reversal, splitting and linearity, or solve a unique endpoint problem
examSelect a correct property or substitution line, then give the exact value or endpoint

Questions about this topic

Does a correct choice prove full independent integration skill?+

No. Choices show recognition and coefficient blanks show local values. Dependent steps do not establish independent written-method mastery

Is this part of the required Units 1–2 course?+

This placement is Methods Units 3–4. The optional Units 1–2 definite-integration extension is not activated by this packet

Are transcendental and rational-power integrands added here?+

No. This generator uses polynomials. Existing transcendental integration resources retain their own questions and exact-response contracts

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 8: Integration

In this chapter

Approximate area by an endpoint or trapezium ruleEvaluate and transform a definite integralElementary primitives and exact definite integralsFind complete families of polynomial anti-derivativesThe area function and the fundamental theorem of calculusSigned integrals, geometric area and motionFind the area between two curvesAverage value of a functionTotal change and kinematics from a rateDetermine a polynomial from its derivative and a boundary conditionRecover motion from acceleration