Hard A-Level Numerical integration Questions

Challenging, exam-style A-Level Numerical integration questions with worked solutions. Stretch yourself on the hardest numerical-integration, trapezium-rule, estimate, error problems.

numerical-integrationtrapezium-ruleestimateerrorover-underpercentage-error
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Use the trapezium rule with 5 strips to estimate ∫05log⁡(x+1) dx\int_{0}^{5} \log{\left(x + 1 \right)} \, dx to 4 decimal places. Which value is correct?
Show worked solution

Worked solution

  1. Identify the limits and number of strips

    a=0, b=5, n=5a=0,\ b=5,\ n=5

    These fix the interval and how many strips to use.

  2. Find the strip width

    h=b−an=5−05=1h=\frac{b-a}{n}=\frac{5-0}{5}=1

    This is the width of each strip.

  3. List the x-values

    x=0, 1, 2, 3, 4, 5x=0,\ 1,\ 2,\ 3,\ 4,\ 5

    The ordinates are taken at these x-values.

  4. Calculate the ordinate y_0

    y0=f(0)=0.0000y_{0}=f(0)=0.0000

    Evaluate the function at x=0.

  5. Calculate the ordinate y_1

    y1=f(1)=0.6931y_{1}=f(1)=0.6931

    Evaluate the function at x=1.

  6. Calculate the ordinate y_2

    y2=f(2)=1.0986y_{2}=f(2)=1.0986

    Evaluate the function at x=2.

  7. Calculate the ordinate y_3

    y3=f(3)=1.3863y_{3}=f(3)=1.3863

    Evaluate the function at x=3.

  8. Calculate the ordinate y_4

    y4=f(4)=1.6094y_{4}=f(4)=1.6094

    Evaluate the function at x=4.

  9. Calculate the ordinate y_5

    y5=f(5)=1.7918y_{5}=f(5)=1.7918

    Evaluate the function at x=5.

  10. Add the first and last ordinates

    y0+y5=0.0000+1.7918=1.7918y_0+y_{5}=0.0000+1.7918=1.7918

    These end ordinates carry weight one.

  11. Add twice the interior ordinates

    2(0.6931+1.0986+1.3863+1.6094)=9.57502\left(0.6931+1.0986+1.3863+1.6094\right)=9.5750

    The interior ordinates carry weight two.

  12. Substitute into the trapezium rule

    ∫05log⁡(x+1) dx≈12[0.0000+1.7918+2(0.6931+1.0986+1.3863+1.6094)]\int_{0}^{5} \log{\left(x + 1 \right)} \, dx\approx \frac{1}{2}\left[0.0000+1.7918+2\left(0.6931+1.0986+1.3863+1.6094\right)\right]

    Insert the weighted ordinates into the formula.

  13. Evaluate the bracket

    [1.7918+9.5750]=11.3667\left[1.7918+9.5750\right]=11.3667

    Combine the weighted ordinates.

  14. Multiply by half the strip width

    12×11.3667=5.6834\frac{1}{2}\times 11.3667=5.6834

    This gives the trapezium estimate.

  15. Select the correct estimate

    ∫05log⁡(x+1) dx≈5.6834\int_{0}^{5} \log{\left(x + 1 \right)} \, dx\approx 5.6834

    This is the trapezium estimate to the stated accuracy.

Answer
5.68345.6834
Question 2
8 markschallenging
Use the trapezium rule with 5 strips to estimate ∫05x dx\int_{0}^{5} \sqrt{x} \, dx to 4 decimal places. Which value is correct?
Show worked solution

Worked solution

  1. Identify the limits and number of strips

    a=0, b=5, n=5a=0,\ b=5,\ n=5

    These fix the interval and how many strips to use.

  2. Find the strip width

    h=b−an=5−05=1h=\frac{b-a}{n}=\frac{5-0}{5}=1

    This is the width of each strip.

  3. List the x-values

    x=0, 1, 2, 3, 4, 5x=0,\ 1,\ 2,\ 3,\ 4,\ 5

    The ordinates are taken at these x-values.

  4. Calculate the ordinate y_0

    y0=f(0)=0.0000y_{0}=f(0)=0.0000

    Evaluate the function at x=0.

  5. Calculate the ordinate y_1

    y1=f(1)=1.0000y_{1}=f(1)=1.0000

    Evaluate the function at x=1.

  6. Calculate the ordinate y_2

    y2=f(2)=1.4142y_{2}=f(2)=1.4142

    Evaluate the function at x=2.

  7. Calculate the ordinate y_3

    y3=f(3)=1.7321y_{3}=f(3)=1.7321

    Evaluate the function at x=3.

  8. Calculate the ordinate y_4

    y4=f(4)=2.0000y_{4}=f(4)=2.0000

    Evaluate the function at x=4.

  9. Calculate the ordinate y_5

    y5=f(5)=2.2361y_{5}=f(5)=2.2361

    Evaluate the function at x=5.

  10. Add the first and last ordinates

    y0+y5=0.0000+2.2361=2.2361y_0+y_{5}=0.0000+2.2361=2.2361

    These end ordinates carry weight one.

  11. Add twice the interior ordinates

    2(1.0000+1.4142+1.7321+2.0000)=12.29252\left(1.0000+1.4142+1.7321+2.0000\right)=12.2925

    The interior ordinates carry weight two.

  12. Substitute into the trapezium rule

    ∫05x dx≈12[0.0000+2.2361+2(1.0000+1.4142+1.7321+2.0000)]\int_{0}^{5} \sqrt{x} \, dx\approx \frac{1}{2}\left[0.0000+2.2361+2\left(1.0000+1.4142+1.7321+2.0000\right)\right]

    Insert the weighted ordinates into the formula.

  13. Evaluate the bracket

    [2.2361+12.2925]=14.5286\left[2.2361+12.2925\right]=14.5286

    Combine the weighted ordinates.

  14. Multiply by half the strip width

    12×14.5286=7.2643\frac{1}{2}\times 14.5286=7.2643

    This gives the trapezium estimate.

  15. Select the correct estimate

    ∫05x dx≈7.2643\int_{0}^{5} \sqrt{x} \, dx\approx 7.2643

    This is the trapezium estimate to the stated accuracy.

Answer
7.26437.2643
Question 3
8 markschallenging
The trapezium rule is used to estimate ∫04x+1 dx\int_{0}^{4} \sqrt{x + 1} \, dx. Does it give an over-estimate or an under-estimate, and why?
Show worked solution

Worked solution

  1. Recall the concavity test for the trapezium rule

    f′′>0⇒over-estimate; f′′<0⇒under-estimatef''>0\Rightarrow\text{over-estimate};\ f''<0\Rightarrow\text{under-estimate}

    The sign of the second derivative decides the direction of the error.

  2. Differentiate the function twice

    f′′(x)=−14(x+1)32f''(x)=- \frac{1}{4 \left(x + 1\right)^{\frac{3}{2}}}

    We examine the concavity of the curve.

  3. Determine the sign of f''(x) on the interval

    f′′(2)=−0.0481 < 0f''(2)=-0.0481\ <\ 0

    One sign throughout means the concavity does not change.

  4. Write the first derivative

    f′(x)=12x+1f'(x)=\frac{1}{2 \sqrt{x + 1}}

    A stepping stone to the second derivative.

  5. Describe the shape of the curve

    The curve is concave here.\text{The curve is concave here.}

    Concavity has a clear geometric meaning.

  6. Evaluate f'' at the left endpoint

    f′′(0)=−0.2500f''(0)=-0.2500

    Checking an endpoint confirms the sign.

  7. Evaluate f'' at the right endpoint

    f′′(4)=−0.0224f''(4)=-0.0224

    Checking the other endpoint confirms the sign.

  8. Locate each chord relative to the curve

    Each chord lies below the curve.\text{Each chord lies below the curve.}

    The trapezia are built from these chords.

  9. Compare trapezium area with the true area

    Trapezium area is less than the true area.\text{Trapezium area is less than the true area.}

    This is the origin of the error.

  10. Consider one representative strip

    On each strip the arc is replaced by a chord.\text{On each strip the arc is replaced by a chord.}

    The rule is linear on every strip.

  11. Confirm the concavity does not change

    f′′(x)=−14(x+1)32 (one sign)f''(x)=- \frac{1}{4 \left(x + 1\right)^{\frac{3}{2}}}\ (\text{one sign})

    No inflection point lies inside the interval.

  12. Recall the meaning of convex

    Convex means f′′>0.\text{Convex means }f''>0.

    A convex curve bends upwards.

  13. Recall the meaning of concave

    Concave means f′′<0.\text{Concave means }f''<0.

    A concave curve bends downwards.

  14. Relate the concavity to the estimate

    concave⇒under-estimate\text{concave}\Rightarrow\text{under-estimate}

    The shape determines the direction of the error.

  15. State whether it is an over- or under-estimate

    under-estimate\text{under-estimate}

    This follows directly from the concavity of the curve.

Answer
Under-estimate; the curve is concave on the interval
Question 4
8 markschallenging
The trapezium rule is used to estimate ∫141x dx\int_{1}^{4} \frac{1}{x} \, dx. Does it give an over-estimate or an under-estimate, and why?
Show worked solution

Worked solution

  1. Recall the concavity test for the trapezium rule

    f′′>0⇒over-estimate; f′′<0⇒under-estimatef''>0\Rightarrow\text{over-estimate};\ f''<0\Rightarrow\text{under-estimate}

    The sign of the second derivative decides the direction of the error.

  2. Differentiate the function twice

    f′′(x)=2x3f''(x)=\frac{2}{x^{3}}

    We examine the concavity of the curve.

  3. Determine the sign of f''(x) on the interval

    f′′(2.5)=0.1280 > 0f''(2.5)=0.1280\ >\ 0

    One sign throughout means the concavity does not change.

  4. Write the first derivative

    f′(x)=−1x2f'(x)=- \frac{1}{x^{2}}

    A stepping stone to the second derivative.

  5. Describe the shape of the curve

    The curve is convex here.\text{The curve is convex here.}

    Concavity has a clear geometric meaning.

  6. Evaluate f'' at the left endpoint

    f′′(1)=2.0000f''(1)=2.0000

    Checking an endpoint confirms the sign.

  7. Evaluate f'' at the right endpoint

    f′′(4)=0.0312f''(4)=0.0312

    Checking the other endpoint confirms the sign.

  8. Locate each chord relative to the curve

    Each chord lies above the curve.\text{Each chord lies above the curve.}

    The trapezia are built from these chords.

  9. Compare trapezium area with the true area

    Trapezium area is greater than the true area.\text{Trapezium area is greater than the true area.}

    This is the origin of the error.

  10. Consider one representative strip

    On each strip the arc is replaced by a chord.\text{On each strip the arc is replaced by a chord.}

    The rule is linear on every strip.

  11. Confirm the concavity does not change

    f′′(x)=2x3 (one sign)f''(x)=\frac{2}{x^{3}}\ (\text{one sign})

    No inflection point lies inside the interval.

  12. Recall the meaning of convex

    Convex means f′′>0.\text{Convex means }f''>0.

    A convex curve bends upwards.

  13. Recall the meaning of concave

    Concave means f′′<0.\text{Concave means }f''<0.

    A concave curve bends downwards.

  14. Relate the concavity to the estimate

    convex⇒over-estimate\text{convex}\Rightarrow\text{over-estimate}

    The shape determines the direction of the error.

  15. State whether it is an over- or under-estimate

    over-estimate\text{over-estimate}

    This follows directly from the concavity of the curve.

Answer
Over-estimate; the curve is convex on the interval
Question 5
8 markschallenging
The trapezium rule is used to estimate ∫14log⁡(x) dx\int_{1}^{4} \log{\left(x \right)} \, dx. Does it give an over-estimate or an under-estimate, and why?
Show worked solution

Worked solution

  1. Recall the concavity test for the trapezium rule

    f′′>0⇒over-estimate; f′′<0⇒under-estimatef''>0\Rightarrow\text{over-estimate};\ f''<0\Rightarrow\text{under-estimate}

    The sign of the second derivative decides the direction of the error.

  2. Differentiate the function twice

    f′′(x)=−1x2f''(x)=- \frac{1}{x^{2}}

    We examine the concavity of the curve.

  3. Determine the sign of f''(x) on the interval

    f′′(2.5)=−0.1600 < 0f''(2.5)=-0.1600\ <\ 0

    One sign throughout means the concavity does not change.

  4. Write the first derivative

    f′(x)=1xf'(x)=\frac{1}{x}

    A stepping stone to the second derivative.

  5. Describe the shape of the curve

    The curve is concave here.\text{The curve is concave here.}

    Concavity has a clear geometric meaning.

  6. Evaluate f'' at the left endpoint

    f′′(1)=−1.0000f''(1)=-1.0000

    Checking an endpoint confirms the sign.

  7. Evaluate f'' at the right endpoint

    f′′(4)=−0.0625f''(4)=-0.0625

    Checking the other endpoint confirms the sign.

  8. Locate each chord relative to the curve

    Each chord lies below the curve.\text{Each chord lies below the curve.}

    The trapezia are built from these chords.

  9. Compare trapezium area with the true area

    Trapezium area is less than the true area.\text{Trapezium area is less than the true area.}

    This is the origin of the error.

  10. Consider one representative strip

    On each strip the arc is replaced by a chord.\text{On each strip the arc is replaced by a chord.}

    The rule is linear on every strip.

  11. Confirm the concavity does not change

    f′′(x)=−1x2 (one sign)f''(x)=- \frac{1}{x^{2}}\ (\text{one sign})

    No inflection point lies inside the interval.

  12. Recall the meaning of convex

    Convex means f′′>0.\text{Convex means }f''>0.

    A convex curve bends upwards.

  13. Recall the meaning of concave

    Concave means f′′<0.\text{Concave means }f''<0.

    A concave curve bends downwards.

  14. Relate the concavity to the estimate

    concave⇒under-estimate\text{concave}\Rightarrow\text{under-estimate}

    The shape determines the direction of the error.

  15. State whether it is an over- or under-estimate

    under-estimate\text{under-estimate}

    This follows directly from the concavity of the curve.

Answer
Under-estimate; the curve is concave on the interval

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