Class 12 Definite Integrals: Signed Area, Limits and Symmetry Practice
A definite integral can be zero even when a graph encloses visible area. Contributions below the x-axis are negative. Decide whether a question asks for signed accumulation or total geometric area before evaluating anything.
Curriculum checked 2 October 2026 · original practice for selected skills.
PREDICT · CHANGE · EXPLAIN
See how area grows
For y=x from 0 to a, the area is a²/2 square units. Move the endpoint and compare area growth with endpoint growth.
The curve shows this simplified model over the labelled range. Read the exact point below; the sketch is not a measuring instrument.
Area: 2 square units
If a doubles, does the area double or quadruple?
Now close the explanation and try the independent retest. Moving a slider alone does not check your understanding.
Before you start
You will need: Basic antiderivatives, graph signs and interval notation.
Attempt each task before opening a hint. If you need help, reveal the first hint, then the setup, then compare your written steps with the solution. Finish with the retests without referring back.
Three ideas to keep in view
Evaluate an antiderivative at the upper limit and subtract its value at the lower limit. Keep the lower substitution in brackets.
Geometric area is non-negative. If a curve crosses the axis, split the interval and use positive contributions for each region.
Odd functions cancel over symmetric limits. This gives a signed integral of zero, not a claim that there is no geometric area.
Try the worked example first
M12-AREA-W1Signed versus totalFocused practice
Find ∫−22xdx and the total area between y=x and the x-axis on this interval.
Show your working
Need a starting hint?
The two triangular regions lie on different sides of the axis.
Still stuck? Reveal the setup
Each triangle has base 2 and height 2.
Show worked solution+
1
Signed integral
Equal negative and positive contributions cancel.
[x2/2]−22=2−2=0
2
Total area
Add the magnitudes of the two triangular areas.
2+2=4
Answer: Integral 0; total area 4 square units.
Before moving on: can you explain why your method works, as well as give the answer?
Your turn: three different checks
Write a method as well as an answer. The tasks change the reasoning, not just the numbers.
M12-AREA-Q1Evaluate boundsFocused practice
Evaluate ∫132xdx.
Show your working
Need a starting hint?
An antiderivative is x².
Still stuck? Reveal the setup
Subtract the value at 1 from the value at 3.
Show worked solution+
1
Find antiderivative
Differentiate x² to check it.
2
Use bounds
Upper minus lower gives 9−1.
Answer: 8
Before moving on: can you explain why your method works, as well as give the answer?
M12-AREA-Q2Odd symmetryFocused practice
Evaluate ∫−33x3dx without a long calculation.
Show your working
Need a starting hint?
What happens to x³ when x is replaced by −x?
Still stuck? Reveal the setup
The function is odd and the interval is symmetric.
Show worked solution+
1
Identify symmetry
f(−x)=−f(x), so opposite inputs give cancelling contributions.
2
Conclude
The signed integral is zero.
Answer: 0
Before moving on: can you explain why your method works, as well as give the answer?
M12-AREA-Q3Split at the crossingFocused practice
Find the area between y=x−1 and the x-axis from x=0 to x=3.
Show your working
Need a starting hint?
The curve crosses the axis at x=1.
Still stuck? Reveal the setup
Use triangle areas on [0,1] and [1,3].
Show worked solution+
1
Left region
Base 1 and height 1 give area 1/2.
2
Right region
Base 2 and height 2 give area 2. Add positive areas.
Answer: 2.5 square units
Before moving on: can you explain why your method works, as well as give the answer?
Catch a likely mistake
A tempting claim: “Area equals the definite integral in every case.”
A definite integral includes negative contributions below the axis.
Close the examples and try again
These changed questions test whether you can reconstruct the method. If you use a hint, record where you got stuck and retry later. Completing this small set does not establish full chapter mastery.
M12-AREA-R1New boundsIndependent retest
Evaluate ∫243x2dx.
Show your working
Need a starting hint?
Use x³ as the antiderivative.
Still stuck? Reveal the setup
Compute 4³−2³.
Show worked solution+
1
Integrate
The power rule gives x³.
2
Evaluate
64−8=56.
Answer: 56
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
M12-AREA-R2New crossingIndependent retest
Find total area between y=x−2 and the axis on [0,4].
Show your working
Need a starting hint?
Split at x=2.
Still stuck? Reveal the setup
Two triangles have base 2 and height 2.
Show worked solution+
1
Each region
Each area equals 2.
2
Add magnitudes
Their signs differ as integrals, but areas add.
Answer: 4 square units
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Choose the next useful step
If a retest exposed the same error, rewrite the first incorrect step and explain its correction aloud. If both were independent, return to a mixed exercise or a missed paper question. A parent can ask what changed in the method rather than only asking for the answer.
You may use an antiderivative with a constant, but the same constant cancels between the two limits. The final definite integral is a number, not a family with +C.
Can I print the questions and worked solutions separately?+
Yes. Use the two print buttons above. The question sheet leaves working space and hides solutions; the worked-solutions option includes the solution steps. Your browser can save either view as a PDF.
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