A-Level Definite integration and areas Practice Questions

Free A-Level Definite integration and areas practice questions with full step-by-step worked solutions. Covers definite-integral, power-rule, linear, constant. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Evaluate ∫12(3x2) dx\displaystyle \int_{1}^{2} \left( 3 x^{2} \right) \, dx.
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Worked solution

  1. Write down the definite integral

    ∫12(3x2) dx\int_{1}^{2} \left( 3 x^{2} \right) \, dx

    Identify the function to integrate and note the lower limit (bottom number) and upper limit (top number).

  2. Integrate using the reverse power rule

    =[x3]12= \left[ x^{3} \right]_{1}^{2}

    Add one to each power and divide by the new power, then keep the limits attached to the square bracket.

  3. Substitute the limits (top minus bottom)

    =(8)−(1)= \left(8\right) - \left(1\right)

    Put in the upper limit first, then subtract the value you get from the lower limit.

  4. Work out the value

    =7= 7

    Finish the arithmetic to get the single number.

Answer
77
Question 2
2 markseasy
Why is the constant of integration +c+c not needed when evaluating a definite integral?
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Worked solution

  1. Write the integral with the constant

    [F(x)+c]ab\left[ F(x) + c \right]_a^b

    Include the constant to see what happens.

  2. Substitute both limits

    (F(b)+c)−(F(a)+c)(F(b)+c) - (F(a)+c)

    Put in the top and bottom limits.

  3. Simplify

    F(b)−F(a)F(b) - F(a)

    The two +c+c terms cancel, so the constant never affects a definite integral.

Answer
It cancels when we subtract the value at the lower limit from the value at the upper limit.
Question 3
3 marksintermediate
Which diagram correctly shades the region whose area is ∫13(x2+1) dx\displaystyle\int_{1}^{3}(x^2+1)\,dx?
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Worked solution

  1. Identify the boundaries of the region

    x=1 to x=3x = 1 \text{ to } x = 3

    The shaded region must start and end at the correct xx-values given by the limits.

  2. Identify the correct function

    y=x2+1y = x^{2} + 1

    The region is measured against this curve, so the picture must show this shape.

  3. Describe the correct shaded region

    ∫13(x2+1)dx\int_{1}^{3} \left(x^{2} + 1\right) dx

    The right diagram shades the area between the curve and the xx-axis (or between the two curves) over exactly this interval.

  4. Reject the picture with the wrong limits

    wrong interval shaded\text{wrong interval shaded}

    Any diagram shading a different interval cannot represent this integral.

  5. Reject the reflected or shifted curve

    wrong curve\text{wrong curve}

    A diagram with the curve flipped in the xx-axis or shifted sideways shows a different function.

  6. Select the matching diagram

    correct region between the correct limits\text{correct region between the correct limits}

    The remaining diagram is the one that matches both the correct curve and the correct limits.

Answer
Correct shaded region\text{Correct shaded region}
Question 4
5 markshard
The curve has equation y=x3+1y = x^{3} + 1. Find the exact area of the region bounded by the curve, the xx-axis, and the lines x=0x = 0 and x=2x = 2.
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Worked solution

  1. Understand the area we need

    Area=∫02(x3+1)dx\text{Area} = \int_{0}^{2} \left(x^{3} + 1\right) dx

    The area between a curve and the xx-axis is found by integrating the curve between the two xx-values. Here the curve is above the axis, so the integral gives the area directly.

  2. Recall the reverse power rule

    ∫xn dx=xn+1n+1  (n≠−1)\int x^{n}\,dx = \frac{x^{n+1}}{n+1}\;(n\neq -1)

    Add one to the power and divide by the new power; this is the reverse of differentiating.

  3. Integrate the term 11

    ∫1 dx=x\int 1\,dx = x

    Apply the power rule to this single term on its own.

  4. Integrate the term x3x^{3}

    ∫x3 dx=x44\int x^{3}\,dx = \frac{x^{4}}{4}

    Apply the power rule to this single term on its own.

  5. Collect the integrated terms into one antiderivative

    x44+x\frac{x^{4}}{4} + x

    Adding the integrated terms together gives the full antiderivative.

  6. Write in square-bracket form with the limits

    =[x44+x]02= \left[ \frac{x^{4}}{4} + x \right]_{0}^{2}

    Keeping the limits attached stops us forgetting to substitute both of them.

  7. Substitute the upper limit x=2x=2

    upper=6\text{upper} = 6

    Put the top limit into the antiderivative and simplify.

  8. Substitute the lower limit x=0x=0

    lower=0\text{lower} = 0

    Put the bottom limit into the antiderivative and simplify.

  9. Subtract lower from upper

    =(6)−(0)= \left(6\right) - \left(0\right)

    The definite integral is the top value minus the bottom value.

  10. Simplify to the final value

    =6= 6

    Finish the arithmetic to reach the answer.

  11. State the area

    Area=6 square units\text{Area} = 6 \text{ square units}

    Because the region lies above the xx-axis the integral is positive and equals the area. Areas are written in square units.

Answer
Area=6\text{Area} = 6
Question 5
7 markschallenging
The curve y=x3−9xy = x^{3} - 9 x crosses the xx-axis between x=−4x=-4 and x=4x=4. Find the total area enclosed between the curve and the xx-axis over this interval.
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Worked solution

  1. Sketch and understand the region

    y=x3−9xy = x^{3} - 9 x

    The curve dips above and below the xx-axis between the limits. Any part below the axis gives a negative integral, so we must handle the pieces separately to get a true area.

  2. Find where the curve meets the xx-axis

    x3−9x=0  ⇒  x=−3,  0,  3x^{3} - 9 x = 0 \;\Rightarrow\; x = -3,\; 0,\; 3

    These crossing points split the interval into pieces that are each entirely above or entirely below the axis.

  3. Integrate the curve once

    ∫x3−9x dx=x44−9x22\int x^{3} - 9 x\,dx = \frac{x^{4}}{4} - \frac{9 x^{2}}{2}

    We find the antiderivative a single time and reuse it for every piece. Add one to each power and divide by the new power.

  4. Integrate from x=−4x=-4 to x=−3x=-3

    ∫−4−3x3−9x dx=−494\int_{-4}^{-3} x^{3} - 9 x\,dx = - \frac{49}{4}

    Substitute the two ends of this piece into the antiderivative and subtract.

  5. Interpret the sign of this piece

    −494  ⇒  region below the axis- \frac{49}{4} \;\Rightarrow\; \text{region below the axis}

    A negative integral means this piece lies below the xx-axis, so for area we will use its size (modulus).

  6. Integrate from x=−3x=-3 to x=0x=0

    ∫−30x3−9x dx=814\int_{-3}^{0} x^{3} - 9 x\,dx = \frac{81}{4}

    Substitute the two ends of this piece into the antiderivative and subtract.

  7. Interpret the sign of this piece

    814  ⇒  region above the axis\frac{81}{4} \;\Rightarrow\; \text{region above the axis}

    A positive integral means this piece lies above the xx-axis, so for area we will use its size (modulus).

  8. Integrate from x=0x=0 to x=3x=3

    ∫03x3−9x dx=−814\int_{0}^{3} x^{3} - 9 x\,dx = - \frac{81}{4}

    Substitute the two ends of this piece into the antiderivative and subtract.

  9. Interpret the sign of this piece

    −814  ⇒  region below the axis- \frac{81}{4} \;\Rightarrow\; \text{region below the axis}

    A negative integral means this piece lies below the xx-axis, so for area we will use its size (modulus).

  10. Integrate from x=3x=3 to x=4x=4

    ∫34x3−9x dx=494\int_{3}^{4} x^{3} - 9 x\,dx = \frac{49}{4}

    Substitute the two ends of this piece into the antiderivative and subtract.

  11. Interpret the sign of this piece

    494  ⇒  region above the axis\frac{49}{4} \;\Rightarrow\; \text{region above the axis}

    A positive integral means this piece lies above the xx-axis, so for area we will use its size (modulus).

  12. Explain why we cannot just add the signed integrals

    ∑∫≠Area if signs differ\sum \int \neq \text{Area if signs differ}

    If we simply added the signed integrals, the negative pieces would cancel part of the positive pieces and we would understate the true area.

  13. Rule out the other options

    eliminate each distractor\text{eliminate each distractor}

    Check the remaining choices against your working: each wrong option matches a common slip (a sign error, wrong limits, or forgetting to subtract the lower limit), so they can be discarded.

  14. Check the answer is reasonable

    compare with a quick estimate\text{compare with a quick estimate}

    Estimate the size of the region or value roughly and compare it with your exact answer. If they are wildly different, go back and look for an arithmetic slip.

  15. Add the sizes of the pieces

    Area=∣−494∣+∣814∣+∣−814∣+∣494∣=65 square units\text{Area} = \left|- \frac{49}{4}\right| + \left|\frac{81}{4}\right| + \left|- \frac{81}{4}\right| + \left|\frac{49}{4}\right| = 65 \text{ square units}

    Taking the modulus of each piece before adding guarantees a positive total, which is the real area enclosed.

Answer
Area=65\text{Area} = 65

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