A-Level Numerical integration Practice Questions

Free A-Level Numerical integration practice questions with full step-by-step worked solutions. Covers numerical-integration, trapezium-rule, strip-width, ordinates. Practise exam-style problems and check your method.

numerical-integrationtrapezium-rulestrip-widthordinatesestimateover-under
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The trapezium rule is used to estimate ∫13x dx\int_{1}^{3} \sqrt{x} \, dx with 2 strips. Find the strip width hh.
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Worked solution

  1. Identify the limits and number of strips

    a=1, b=3, n=2a=1,\ b=3,\ n=2

    Read these from the integral.

  2. Apply the strip-width formula

    h=b−an=3−12h=\frac{b-a}{n}=\frac{3-1}{2}

    The interval is split into n equal strips.

  3. State the strip width

    h=22=1h=\frac{2}{2}=1

    This is the width of each strip.

Answer
11
Question 2
2 markseasy
Which of the following correctly states the trapezium rule for nn strips of width hh with ordinates y0, y1, …, yny_0,\,y_1,\,\ldots,\,y_n?
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Worked solution

  1. Recall the trapezium rule

    T≈h2[y0+yn+2(y1+⋯+yn−1)]T\approx\frac{h}{2}\left[y_0+y_n+2\left(y_1+\cdots+y_{n-1}\right)\right]

    The standard statement of the rule.

  2. Identify the structure of the formula

    12×width×(ends+2×interior)\tfrac{1}{2}\times\text{width}\times(\text{ends}+2\times\text{interior})

    Half the width multiplies the ends plus twice the interior ordinates.

  3. Select the correct formula

    h2[y0+yn+2(y1+⋯+yn−1)]\frac{h}{2}\left[y_0+y_n+2\left(y_1+\cdots+y_{n-1}\right)\right]

    This matches the trapezium rule exactly.

Answer
h2[y0+yn+2(y1+⋯+yn−1)]\frac{h}{2}\left[y_0+y_n+2\left(y_1+\cdots+y_{n-1}\right)\right]
Question 3
3 marksintermediate
Which formula gives the trapezium-rule estimate from ordinates y0,…,yny_0,\ldots,y_n and strip width hh?
Show worked solution

Worked solution

  1. Recall the trapezium rule

    T≈h2[y0+yn+2(y1+⋯+yn−1)]T\approx\frac{h}{2}\left[y_0+y_n+2\left(y_1+\cdots+y_{n-1}\right)\right]

    The standard statement of the rule.

  2. Identify the structure of the formula

    12×width×(ends+2×interior)\tfrac{1}{2}\times\text{width}\times(\text{ends}+2\times\text{interior})

    Half the width multiplies the ends plus twice the interior ordinates.

  3. Count the ordinates

    n strips⇒n+1 ordinatesn\text{ strips}\Rightarrow n+1\text{ ordinates}

    There is one more ordinate than the number of strips.

  4. Explain the factor of one half

    Each trapezium area=12(sides)×width.\text{Each trapezium area}=\tfrac{1}{2}(\text{sides})\times\text{width}.

    The rule sums the areas of trapezia.

  5. Explain the factor of two

    Interior ordinates are shared by two strips.\text{Interior ordinates are shared by two strips.}

    So they are counted twice.

  6. Select the correct formula

    h2[y0+yn+2(y1+⋯+yn−1)]\frac{h}{2}\left[y_0+y_n+2\left(y_1+\cdots+y_{n-1}\right)\right]

    This matches the trapezium rule exactly.

Answer
h2[y0+yn+2(y1+⋯+yn−1)]\frac{h}{2}\left[y_0+y_n+2\left(y_1+\cdots+y_{n-1}\right)\right]
Question 4
5 markshard
Use the trapezium rule with 4 strips to estimate ∫131x dx\int_{1}^{3} \frac{1}{x} \, dx to 4 decimal places. Which value is correct?
Show worked solution

Worked solution

  1. Identify the limits and number of strips

    a=1, b=3, n=4a=1,\ b=3,\ n=4

    These fix the interval and how many strips to use.

  2. Find the strip width

    h=b−an=3−14=0.5h=\frac{b-a}{n}=\frac{3-1}{4}=0.5

    This is the width of each strip.

  3. List the x-values

    x=1, 1.5, 2, 2.5, 3x=1,\ 1.5,\ 2,\ 2.5,\ 3

    The ordinates are taken at these x-values.

  4. Calculate the ordinate y_0

    y0=f(1)=1.0000y_{0}=f(1)=1.0000

    Evaluate the function at x=1.

  5. Calculate the ordinate y_1

    y1=f(1.5)=0.6667y_{1}=f(1.5)=0.6667

    Evaluate the function at x=1.5.

  6. Calculate the ordinate y_2

    y2=f(2)=0.5000y_{2}=f(2)=0.5000

    Evaluate the function at x=2.

  7. Calculate the ordinate y_3

    y3=f(2.5)=0.4000y_{3}=f(2.5)=0.4000

    Evaluate the function at x=2.5.

  8. Calculate the ordinate y_4

    y4=f(3)=0.3333y_{4}=f(3)=0.3333

    Evaluate the function at x=3.

  9. Apply the trapezium rule

    ∫131x dx≈0.52[1.0000+0.3333+2(0.6667+0.5000+0.4000)]\int_{1}^{3} \frac{1}{x} \, dx\approx \frac{0.5}{2}\left[1.0000+0.3333+2\left(0.6667+0.5000+0.4000\right)\right]

    Substitute the ordinates into the formula.

  10. Select the correct estimate

    ∫131x dx≈1.1167\int_{1}^{3} \frac{1}{x} \, dx\approx 1.1167

    This is the trapezium estimate to the stated accuracy.

Answer
1.11671.1167
Question 5
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

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