Hard A-Level Definite integration and areas Questions

Challenging, exam-style A-Level Definite integration and areas questions with worked solutions. Stretch yourself on the hardest area-between-curves, signed-area, cubic, expand-then-integrate problems.

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A-Level34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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
Question 2
7 markschallenging
The curve y=x3−5x2+6xy = x^{3} - 5 x^{2} + 6 x crosses the xx-axis between x=−1x=-1 and x=4x=4. Find the total area enclosed between the curve and the xx-axis over this interval.
Show worked solution

Worked solution

  1. Sketch and understand the region

    y=x3−5x2+6xy = x^{3} - 5 x^{2} + 6 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

    x3−5x2+6x=0  ⇒  x=0,  2,  3x^{3} - 5 x^{2} + 6 x = 0 \;\Rightarrow\; x = 0,\; 2,\; 3

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

  3. Integrate the curve once

    ∫x3−5x2+6x dx=x44−5x33+3x2\int x^{3} - 5 x^{2} + 6 x\,dx = \frac{x^{4}}{4} - \frac{5 x^{3}}{3} + 3 x^{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=−1x=-1 to x=0x=0

    ∫−10x3−5x2+6x dx=−5912\int_{-1}^{0} x^{3} - 5 x^{2} + 6 x\,dx = - \frac{59}{12}

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

  5. Interpret the sign of this piece

    −5912  ⇒  region below the axis- \frac{59}{12} \;\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=0x=0 to x=2x=2

    ∫02x3−5x2+6x dx=83\int_{0}^{2} x^{3} - 5 x^{2} + 6 x\,dx = \frac{8}{3}

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

  7. Interpret the sign of this piece

    83  ⇒  region above the axis\frac{8}{3} \;\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=2x=2 to x=3x=3

    ∫23x3−5x2+6x dx=−512\int_{2}^{3} x^{3} - 5 x^{2} + 6 x\,dx = - \frac{5}{12}

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

  9. Interpret the sign of this piece

    −512  ⇒  region below the axis- \frac{5}{12} \;\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

    ∫34x3−5x2+6x dx=3712\int_{3}^{4} x^{3} - 5 x^{2} + 6 x\,dx = \frac{37}{12}

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

  11. Interpret the sign of this piece

    3712  ⇒  region above the axis\frac{37}{12} \;\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=∣−5912∣+∣83∣+∣−512∣+∣3712∣=13312 square units\text{Area} = \left|- \frac{59}{12}\right| + \left|\frac{8}{3}\right| + \left|- \frac{5}{12}\right| + \left|\frac{37}{12}\right| = \frac{133}{12} \text{ square units}

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

Answer
Area=13312\text{Area} = \frac{133}{12}
Question 3
6 markschallenging
Find the exact area of the region enclosed between the curves y=10−x2y = 10 - x^{2} and y=1y = 1.
Show worked solution

Worked solution

  1. Sketch the region and find where the curves meet

    10−x2=110 - x^{2} = 1

    The curves cross where their yy-values are equal. Those crossing points are the left and right edges of the region we want.

  2. Move everything to one side

    9−x2=09 - x^{2} = 0

    Bringing all terms to one side turns the problem into a single equation we can solve, just like eliminating in simultaneous equations.

  3. Factorise the equation

    −(x−3)(x+3)=0- \left(x - 3\right) \left(x + 3\right) = 0

    Factorising exposes the roots directly. (If it did not factorise nicely we would use the quadratic formula instead.)

  4. State the limits of integration

    x=−3andx=3x = -3 \quad\text{and}\quad x = 3

    The two solutions are the xx-coordinates where the curves meet, so they become the limits of the integral.

  5. Check a limit by substitution

    x=−3:  10−x2=1x=-3:\; 10 - x^{2} = 1

    Substituting the limit back into both curves gives the same yy-value, confirming the crossing point is correct.

  6. Decide which curve is on top

    at x=0:  10>1\text{at } x=0:\; 10 > 1

    Testing an xx-value between the limits shows which curve has the larger yy-value there; that is the upper curve.

  7. Set up the area as an integral of (upper −- lower)

    Area=∫−33[(10−x2)−(1)]dx\text{Area} = \int_{-3}^{3} \left[\left(10 - x^{2}\right)-\left(1\right)\right] dx

    Integrating the gap between the curves gives the enclosed area and automatically copes with any part below the axis.

  8. Simplify the integrand

    =∫−33(9−x2)dx= \int_{-3}^{3} \left(9 - x^{2}\right) dx

    Subtracting the lower curve from the upper curve and tidying up leaves a single expression to integrate.

  9. 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.

  10. Integrate the term 99

    ∫9 dx=9x\int 9\,dx = 9 x

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

  11. Integrate the term −x2- x^{2}

    ∫−x2 dx=−x33\int - x^{2}\,dx = - \frac{x^{3}}{3}

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

  12. Collect the integrated terms into one antiderivative

    −x33+9x- \frac{x^{3}}{3} + 9 x

    Adding the integrated terms together gives the full antiderivative.

  13. Write in square-bracket form with the limits

    =[−x33+9x]−33= \left[ - \frac{x^{3}}{3} + 9 x \right]_{-3}^{3}

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

  14. Substitute the upper limit x=3x=3

    upper=18\text{upper} = 18

    Put the top limit into the antiderivative and simplify.

  15. Substitute the lower limit x=−3x=-3

    lower=−18\text{lower} = -18

    Put the bottom limit into the antiderivative and simplify.

  16. Subtract lower from upper

    =(18)−(−18)= \left(18\right) - \left(-18\right)

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

  17. Simplify to the final value

    =36= 36

    Finish the arithmetic to reach the answer.

  18. State the area

    Area=36 square units\text{Area} = 36 \text{ square units}

    This positive number is the exact area trapped between the two curves.

Answer
Area=36\text{Area} = 36
Question 4
7 markschallenging
Find the exact area of the region enclosed between the curves y=−x2+6xy = - x^{2} + 6 x and y=x2−2xy = x^{2} - 2 x.
Show worked solution

Worked solution

  1. Sketch the region and find where the curves meet

    −x2+6x=x2−2x- x^{2} + 6 x = x^{2} - 2 x

    The curves cross where their yy-values are equal. Those crossing points are the left and right edges of the region we want.

  2. Move everything to one side

    −2x2+8x=0- 2 x^{2} + 8 x = 0

    Bringing all terms to one side turns the problem into a single equation we can solve, just like eliminating in simultaneous equations.

  3. Factorise the equation

    −2x(x−4)=0- 2 x \left(x - 4\right) = 0

    Factorising exposes the roots directly. (If it did not factorise nicely we would use the quadratic formula instead.)

  4. State the limits of integration

    x=0andx=4x = 0 \quad\text{and}\quad x = 4

    The two solutions are the xx-coordinates where the curves meet, so they become the limits of the integral.

  5. Check a limit by substitution

    x=0:  −x2+6x=0x=0:\; - x^{2} + 6 x = 0

    Substituting the limit back into both curves gives the same yy-value, confirming the crossing point is correct.

  6. Decide which curve is on top

    at x=2:  8>0\text{at } x=2:\; 8 > 0

    Testing an xx-value between the limits shows which curve has the larger yy-value there; that is the upper curve.

  7. Set up the area as an integral of (upper −- lower)

    Area=∫04[(−x2+6x)−(x2−2x)]dx\text{Area} = \int_{0}^{4} \left[\left(- x^{2} + 6 x\right)-\left(x^{2} - 2 x\right)\right] dx

    Integrating the gap between the curves gives the enclosed area and automatically copes with any part below the axis.

  8. Simplify the integrand

    =∫04(−2x2+8x)dx= \int_{0}^{4} \left(- 2 x^{2} + 8 x\right) dx

    Subtracting the lower curve from the upper curve and tidying up leaves a single expression to integrate.

  9. 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.

  10. Integrate the term −2x2- 2 x^{2}

    ∫−2x2 dx=−2x33\int - 2 x^{2}\,dx = - \frac{2 x^{3}}{3}

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

  11. Integrate the term 8x8 x

    ∫8x dx=4x2\int 8 x\,dx = 4 x^{2}

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

  12. Collect the integrated terms into one antiderivative

    −2x33+4x2- \frac{2 x^{3}}{3} + 4 x^{2}

    Adding the integrated terms together gives the full antiderivative.

  13. Write in square-bracket form with the limits

    =[−2x33+4x2]04= \left[ - \frac{2 x^{3}}{3} + 4 x^{2} \right]_{0}^{4}

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

  14. Substitute the upper limit x=4x=4

    upper=643\text{upper} = \frac{64}{3}

    Put the top limit into the antiderivative and simplify.

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

    lower=0\text{lower} = 0

    Put the bottom limit into the antiderivative and simplify.

  16. Subtract lower from upper

    =(643)−(0)= \left(\frac{64}{3}\right) - \left(0\right)

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

  17. Simplify to the final value

    =643= \frac{64}{3}

    Finish the arithmetic to reach the answer.

  18. State the area

    Area=643 square units\text{Area} = \frac{64}{3} \text{ square units}

    This positive number is the exact area trapped between the two curves.

Answer
Area=643\text{Area} = \frac{64}{3}
Question 5
7 markschallenging
Find the exact area of the region enclosed between the curves y=10−x2y = 10 - x^{2} and y=x2−8y = x^{2} - 8.
Show worked solution

Worked solution

  1. Sketch the region and find where the curves meet

    10−x2=x2−810 - x^{2} = x^{2} - 8

    The curves cross where their yy-values are equal. Those crossing points are the left and right edges of the region we want.

  2. Move everything to one side

    18−2x2=018 - 2 x^{2} = 0

    Bringing all terms to one side turns the problem into a single equation we can solve, just like eliminating in simultaneous equations.

  3. Factorise the equation

    −2(x−3)(x+3)=0- 2 \left(x - 3\right) \left(x + 3\right) = 0

    Factorising exposes the roots directly. (If it did not factorise nicely we would use the quadratic formula instead.)

  4. State the limits of integration

    x=−3andx=3x = -3 \quad\text{and}\quad x = 3

    The two solutions are the xx-coordinates where the curves meet, so they become the limits of the integral.

  5. Check a limit by substitution

    x=−3:  10−x2=1x=-3:\; 10 - x^{2} = 1

    Substituting the limit back into both curves gives the same yy-value, confirming the crossing point is correct.

  6. Decide which curve is on top

    at x=0:  10>−8\text{at } x=0:\; 10 > -8

    Testing an xx-value between the limits shows which curve has the larger yy-value there; that is the upper curve.

  7. Set up the area as an integral of (upper −- lower)

    Area=∫−33[(10−x2)−(x2−8)]dx\text{Area} = \int_{-3}^{3} \left[\left(10 - x^{2}\right)-\left(x^{2} - 8\right)\right] dx

    Integrating the gap between the curves gives the enclosed area and automatically copes with any part below the axis.

  8. Simplify the integrand

    =∫−33(18−2x2)dx= \int_{-3}^{3} \left(18 - 2 x^{2}\right) dx

    Subtracting the lower curve from the upper curve and tidying up leaves a single expression to integrate.

  9. 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.

  10. Integrate the term 1818

    ∫18 dx=18x\int 18\,dx = 18 x

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

  11. Integrate the term −2x2- 2 x^{2}

    ∫−2x2 dx=−2x33\int - 2 x^{2}\,dx = - \frac{2 x^{3}}{3}

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

  12. Collect the integrated terms into one antiderivative

    −2x33+18x- \frac{2 x^{3}}{3} + 18 x

    Adding the integrated terms together gives the full antiderivative.

  13. Write in square-bracket form with the limits

    =[−2x33+18x]−33= \left[ - \frac{2 x^{3}}{3} + 18 x \right]_{-3}^{3}

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

  14. Substitute the upper limit x=3x=3

    upper=36\text{upper} = 36

    Put the top limit into the antiderivative and simplify.

  15. Substitute the lower limit x=−3x=-3

    lower=−36\text{lower} = -36

    Put the bottom limit into the antiderivative and simplify.

  16. Subtract lower from upper

    =(36)−(−36)= \left(36\right) - \left(-36\right)

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

  17. Simplify to the final value

    =72= 72

    Finish the arithmetic to reach the answer.

  18. State the area

    Area=72 square units\text{Area} = 72 \text{ square units}

    This positive number is the exact area trapped between the two curves.

Answer
Area=72\text{Area} = 72

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