GCSE Gradients and areas under curves Practice Questions

Free GCSE Gradients and areas under curves practice questions with full step-by-step worked solutions. Covers chord gradient, average rate of change, tangent gradient, instantaneous rate of change. Practise exam-style problems and check your method.

chord gradientaverage rate of changetangent gradientinstantaneous rate of changerise over runtrapezium rule
GCSE Higher70 questionsStep-by-step solutions
Question 1
2 markseasy
The curve y=x2y = x^2 is drawn for 0≤x≤40 \le x \le 4. Work out the average rate of change of yy between x=1x = 1 and x=3x = 3 by finding the gradient of the chord joining the two points on the curve.
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Worked solution

  1. Find the two points on the curve

    x1=1, y1=1x2=3, y2=9x_1 = 1,\ y_1 = 1 \qquad x_2 = 3,\ y_2 = 9

    Substitute x=1x = 1 and x=3x = 3 into x2x^2 to get the two end points of the chord.

  2. Gradient of the chord = rise ÷ run

    gradient=9−13−1=82=4\text{gradient} = \frac{9 - 1}{3 - 1} = \frac{8}{2} = 4

    The average rate of change over an interval is the gradient of the straight chord joining the two end points.

  3. State the answer

    44

    That is the average rate of change across the interval.

Answer
44
Question 2
2 markseasy
A tangent is drawn to the curve y=6x−x2y = 6x - x^2 at the point where x=2x = 2. The tangent passes through (0, 4)(0,\ 4) and (4, 12)(4,\ 12). Estimate the gradient of the curve at x=2x = 2.
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Worked solution

  1. Use the two points on the tangent

    (0, 4)and(4, 12)(0,\ 4) \quad \text{and} \quad (4,\ 12)

    The tangent is the straight line that just touches the curve at the point. Its gradient is the gradient of the curve there.

  2. Gradient = rise ÷ run

    gradient=12−44−0=84=2\text{gradient} = \frac{12 - 4}{4 - 0} = \frac{8}{4} = 2

    Use the two given points on the straight tangent line, not points on the curve.

  3. State the estimate

    22

    So the gradient of the curve at that point is estimated to be 2. It is an estimate because it depends on how accurately the tangent was drawn.

Answer
22
Question 3
2 marksintermediate
A car's speed, vv m/s, after tt seconds is modelled by v=t2v = t^2 for 0≤t≤40 \le t \le 4. Work out the average acceleration of the car over the first 44 seconds.
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Worked solution

  1. Identify the interval

    t=0tot=4t = 0 \quad \text{to} \quad t = 4

    An average rate of change is measured over an interval, so start by fixing the two ends.

  2. Work out y when t=0t = 0

    y1=0y_1 = 0

    Substitute t=0t = 0 into t2t^2.

  3. Work out y when t=4t = 4

    y2=16y_2 = 16

    Substitute t=4t = 4 into t2t^2.

  4. Find the rise and the run

    rise=16−0=16,run=4−0=4\text{rise} = 16 - 0 = 16, \qquad \text{run} = 4 - 0 = 4

    The rise is the change in the vertical variable; the run is the change in the horizontal one.

  5. Divide rise by run

    gradient=164=4\text{gradient} = \frac{16}{4} = 4

    The gradient of a chord on a speed–time graph is the AVERAGE acceleration over the interval.

  6. State the answer

    4 m/s24\text{ m/s}^2

    The average acceleration over the first 4 seconds is 4 m/s².

Answer
4 m/s24\text{ m/s}^2
Question 4
4 markshard
A car's speed, vv m/s, after tt seconds is modelled by v=10t−t2v = 10t - t^2. The trapezium rule with 44 strips of width 11 is used to estimate the distance travelled in the first 44 seconds. Which statement gives the estimate and correctly explains whether it is an over- or an under-estimate?
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Worked solution

  1. Write down the trapezium rule

    A≈h2[y0+y4+2(y1+⋯+y3)]A \approx \frac{h}{2}\Big[y_0 + y_{4} + 2\big(y_1 + \dots + y_{3}\big)\Big]

    Add the first and last ordinates, add twice every ordinate in between, then multiply by half the strip width.

  2. Find the strip width

    h=4−04=1h = \frac{4 - 0}{4} = 1

    The interval from t=0t = 0 to t=4t = 4 is split into 4 equal strips.

  3. Ordinate y0y_0 at t=0t = 0

    y0=0y_0 = 0

    Substitute t=0t = 0 into 10t−t210t - t^2.

  4. Ordinate y1y_1 at t=1t = 1

    y1=9y_1 = 9

    Substitute t=1t = 1 into 10t−t210t - t^2.

  5. Ordinate y2y_2 at t=2t = 2

    y2=16y_2 = 16

    Substitute t=2t = 2 into 10t−t210t - t^2.

  6. Ordinate y3y_3 at t=3t = 3

    y3=21y_3 = 21

    Substitute t=3t = 3 into 10t−t210t - t^2.

  7. Ordinate y4y_4 at t=4t = 4

    y4=24y_4 = 24

    Substitute t=4t = 4 into 10t−t210t - t^2.

  8. Substitute into the rule

    A≈12[0+24+2(9+16+21)]A \approx \frac{1}{2}\Big[0 + 24 + 2\big(9 + 16 + 21\big)\Big]

    Ends once each, middles twice each.

  9. Work out the estimate

    A≈12×116=58A \approx \frac{1}{2} \times 116 = 58

    The bracket totals 116; multiplying by half the strip width (0.5) gives the estimate.

  10. State the estimate

    A≈58A \approx 58

    The trapezium rule with 4 strips gives approximately 58. It is an under-estimate.

Answer
5858
Question 5
6 markschallenging
A lorry's speed, vv m/s, after tt seconds is modelled by v=40−t2v = 40 - t^2 for 0≤t≤50 \le t \le 5. Use the trapezium rule with 55 strips of width 11 to estimate the distance travelled in the first 55 seconds.
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Worked solution

  1. Write down the trapezium rule

    A≈h2[y0+y5+2(y1+⋯+y4)]A \approx \frac{h}{2}\Big[y_0 + y_{5} + 2\big(y_1 + \dots + y_{4}\big)\Big]

    Add the first and last ordinates, add twice every ordinate in between, then multiply by half the strip width.

  2. Find the strip width

    h=5−05=1h = \frac{5 - 0}{5} = 1

    The interval from t=0t = 0 to t=5t = 5 is split into 5 equal strips.

  3. Ordinate y0y_0 at t=0t = 0

    y0=40y_0 = 40

    Substitute t=0t = 0 into 40−t240 - t^2.

  4. Ordinate y1y_1 at t=1t = 1

    y1=39y_1 = 39

    Substitute t=1t = 1 into 40−t240 - t^2.

  5. Ordinate y2y_2 at t=2t = 2

    y2=36y_2 = 36

    Substitute t=2t = 2 into 40−t240 - t^2.

  6. Ordinate y3y_3 at t=3t = 3

    y3=31y_3 = 31

    Substitute t=3t = 3 into 40−t240 - t^2.

  7. Ordinate y4y_4 at t=4t = 4

    y4=24y_4 = 24

    Substitute t=4t = 4 into 40−t240 - t^2.

  8. Ordinate y5y_5 at t=5t = 5

    y5=15y_5 = 15

    Substitute t=5t = 5 into 40−t240 - t^2.

  9. Add the two end ordinates

    y0+y5=40+15=55y_0 + y_5 = 40 + 15 = 55

    The first and last ordinates are each used once.

  10. Add the middle ordinates

    39+36+31+24=13039 + 36 + 31 + 24 = 130

    Every ordinate strictly between the ends is used twice, so total them first.

  11. Substitute into the rule

    A≈12[40+15+2(39+36+31+24)]A \approx \frac{1}{2}\Big[40 + 15 + 2\big(39 + 36 + 31 + 24\big)\Big]

    Ends once each, middles twice each.

  12. Work out the estimate

    A≈12×315=157.5A \approx \frac{1}{2} \times 315 = 157.5

    The bracket totals 315; multiplying by half the strip width (0.5) gives the estimate.

  13. Is it an over- or under-estimate?

    under-estimate\text{under-estimate}

    The curve is concave here (it bends downwards), so every chord lies BELOW the curve. Each trapezium misses a sliver of area, and the estimate is an UNDER-estimate.

  14. How to improve the estimate

    more strips⇒smaller error\text{more strips} \Rightarrow \text{smaller error}

    Narrower strips hug the curve more closely, so the total sliver of missed or extra area shrinks and the estimate improves. The direction of the bias does not change.

  15. State the estimate

    A≈157.5 mA \approx 157.5\text{ m}

    The estimated distance is 157.5 m. The speed–time curve is concave, so the chords lie below it and the estimate is an under-estimate.

Answer
157.5 m157.5\text{ m}

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