GCSE Maths: Areas under graphs (Higher)
Part of the GCSE Maths functions and advanced graphs chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Areas under graphs
In Chapter 13 you used distance–time graphs, where the gradient is the speed. Now look at a speed–time graph. If a car travels at a steady 20 m/s for 10 seconds, it goes 20 × 10 = 200 m. On the graph, 20 is the height and 10 is the width of a rectangle, so the area of the rectangle is the distance travelled. This works for every speed–time graph:
area under a speed–time graph = distance travelled.
In general, the area under any graph is (vertical quantity) × (horizontal quantity): the units multiply. m/s × s = m; litres/min × min = litres; kW × hours = kWh; pounds per hour × hours = pounds.
Straight-line graphs. When the graph is made of straight lines, split the area into rectangles, triangles and trapezia, and add up the areas. For a trapezium, area = ½ × (sum of the parallel sides) × width, where the parallel sides are the two heights.
Curved graphs. When the graph is a curve we cannot find the area exactly with these shapes, but we can estimate it. Cut the area into thin vertical strips of equal width. Treat each strip as a trapezium (the top is the straight line joining two points on the curve) and add the areas. With more strips the estimate gets better. You can also use rectangles, using the height at the left (or right) edge of each strip.
Over or under? If the curve bends upwards (convex), the straight tops of the trapezia lie above the curve, so the estimate is too big (an over-estimate). If the curve bends downwards, the trapezia lie below it and the estimate is too small. For rectangles on an increasing graph, left-hand heights give an under-estimate and right-hand heights give an over-estimate; on a decreasing graph it is the other way round.
Key ideas
- Area under a speed–time graph = distance. In general area = (vertical value) × (horizontal value) and the units multiply.
- Split straight-line graphs into rectangles, triangles and trapezia. Trapezium area = ½ × (sum of parallel sides) × width.
- For a curve, use equal-width strips and trapezia (or rectangles) to estimate the area. More strips give a better estimate.
- Average speed = total distance ÷ total time (area ÷ time).
- A convex (upward-bending) curve: trapezium estimate is too big. A concave curve: too small.
Method. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 10 s): triangle: ½ × 10 × 20 = 100
Part 2 (from 10 to 40 s): rectangle: 20 × 30 = 600
Part 3 (from 40 to 50 s): triangle: ½ × 10 × 20 = 100 - Add the areas. 100 + 600 + 100 = 800
Check. Total time = 50 s, so the average speed is 800 ÷ 50 = 16 m/s. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).
Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.
- Split the area into strips of equal width. The region from time = 0 to 6 is 6 wide. With 3 strips, each strip has width 6 ÷ 3 = 2.
- Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 2 4 6 speed 2 4 10 20 - Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.Strip 1: ½ × (2 + 4) × 2 = 6
Strip 2: ½ × (4 + 10) × 2 = 14
Strip 3: ½ × (10 + 20) × 2 = 30 - Add the strip areas to get the estimate. 6 + 14 + 30 = 50
- Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.(The exact area is 48.00 m, so the estimate is bigger.)
Check. Use the trapezium rule formula: ½ × 2 × [2 + 20 + 2 × (4 + 10)] = 50.00. ✓
Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.
Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.
- Split into strips of equal width. Width of each strip = (6 − 0) ÷ 3 = 2.
- Use the height at the left-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its left edge.
time 0 2 4 6 speed 2 4 10 20 - Area of each rectangle = height × width. Strip 1: 2 × 2 = 4
Strip 2: 4 × 2 = 8
Strip 3: 10 × 2 = 20 - Add. 4 + 8 + 20 = 32
- Say whether it is an over- or under-estimate. The graph is increasing. Using the left-hand heights means each rectangle is below the curve for most of the strip. So the estimate is an under-estimate.estimate 32.00 m < true area 48.00 m
Check. The more strips we use, the closer the estimate gets to the true value (48.00). ✓
Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.
Method. The area under a flow rate–time graph is the volume of water that flows in. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and flow rate on the vertical axis, area = flow rate × time = volume of water that flows in. So the area under the graph is the volume of water that flows in.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 4 minutes): triangle: ½ × 4 × 12 = 24
Part 2 (from 4 to 10 minutes): rectangle: 12 × 6 = 72
Part 3 (from 10 to 14 minutes): triangle: ½ × 4 × 12 = 24 - Add the areas. 24 + 72 + 24 = 120
Check. Total time = 14 minutes, so the average flow rate is 120 ÷ 14 = 607 litres/min. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Method. Write the area under the graph in terms of V (a rectangle and two triangles) and set it equal to 450.
- Sketch the graph. The shape is a trapezium: it rises to height V over 10 s, is flat for 30 s and falls over 20 s.
- Area of the first triangle. ½ × base × height.½ × 10 × V = 5V
- Area of the rectangle. Base 30, height V.30 × V = 30V
- Area of the second triangle. Base 20, height V.½ × 20 × V = 10V
- Add them and set equal to the total distance. The area is the distance.5V + 30V + 10V = 45V = 450
- Solve. Divide both sides by 45.V = 450 ÷ 45 = 10
Check. Total area = ½ × (60 + 30) × 10 = 450 using the trapezium formula with parallel sides 60 and 30. ✓
Watch out. Keep V as a letter until the end. Do not guess a number first.
Watch out
The area under a distance–time graph does not mean anything useful (m × s). Only use the area under a speed–time graph (or other rate graph). Also, when estimating with trapezia, remember the ½ and use the two heights as the parallel sides.Exercise 25I Areas under graphs
Calculator allowedMethod. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 8 s): triangle: ½ × 8 × 16 = 64
Part 2 (from 8 to 20 s): rectangle: 16 × 12 = 192 - Add the areas. 64 + 192 = 256
Check. Total time = 20 s, so the average speed is 256 ÷ 20 = 12.8 m/s. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Method. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 6 s): triangle: ½ × 6 × 12 = 36
Part 2 (from 6 to 14 s): rectangle: 12 × 8 = 96
Part 3 (from 14 to 20 s): triangle: ½ × 6 × 12 = 36 - Add the areas. 36 + 96 + 36 = 168
Check. Total time = 20 s, so the average speed is 168 ÷ 20 = 8.4 m/s. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Method. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 10 s): trapezium: ½ × (5 + 25) × 10 = 150
Part 2 (from 10 to 30 s): rectangle: 25 × 20 = 500
Part 3 (from 30 to 40 s): triangle: ½ × 10 × 25 = 125 - Add the areas. 150 + 500 + 125 = 775
Check. Total time = 40 s, so the average speed is 775 ÷ 40 = 19.375 m/s. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Method. Average speed = total distance ÷ total time. The total distance is the area under the speed–time graph.
- Find the total distance: the area under the graph. Split it into rectangles, triangles and trapezia and add their areas.Part 1: triangle: ½ × 8 × 16 = 64
Part 2: rectangle: 16 × 12 = 192
total = 256 m - Find the total time. Read the last and first values on the time axis.20 − 0 = 20 s
- Divide. Average speed = total distance ÷ total time.256 ÷ 20 = 12.8 m/s
Check. Check that 12.8 lies between the lowest and the highest speed on the graph. ✓
Watch out. Average speed is not the average of the highest and lowest speeds unless the graph is a single straight line.
Method. Average speed = total distance ÷ total time. The total distance is the area under the speed–time graph.
- Find the total distance: the area under the graph. Split it into rectangles, triangles and trapezia and add their areas.Part 1: triangle: ½ × 6 × 12 = 36
Part 2: rectangle: 12 × 8 = 96
Part 3: triangle: ½ × 6 × 12 = 36
total = 168 m - Find the total time. Read the last and first values on the time axis.20 − 0 = 20 s
- Divide. Average speed = total distance ÷ total time.168 ÷ 20 = 8.4 m/s
Check. Check that 8.4 lies between the lowest and the highest speed on the graph. ✓
Watch out. Average speed is not the average of the highest and lowest speeds unless the graph is a single straight line.
Method. Average speed = total distance ÷ total time. The total distance is the area under the speed–time graph.
- Find the total distance: the area under the graph. Split it into rectangles, triangles and trapezia and add their areas.Part 1: trapezium: ½ × (5 + 25) × 10 = 150
Part 2: rectangle: 25 × 20 = 500
Part 3: triangle: ½ × 10 × 25 = 125
total = 775 m - Find the total time. Read the last and first values on the time axis.40 − 0 = 40 s
- Divide. Average speed = total distance ÷ total time.775 ÷ 40 = 19.375 m/s
Check. Check that 19.375 lies between the lowest and the highest speed on the graph. ✓
Watch out. Average speed is not the average of the highest and lowest speeds unless the graph is a single straight line.
Method. Acceleration is the gradient of a speed–time graph: change in speed ÷ time.
- Read the change in speed. The speed goes from 0 to 24 m/s.24 − 0 = 24 m/s
- Divide by the time taken. 8 seconds.24 ÷ 8 = 3 m/s²
Check. After 8 s at 3 m/s² the speed is 3 × 8 = 24 m/s. ✓
Watch out. Acceleration is the gradient, not the area.
Method. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 8 s): triangle: ½ × 8 × 24 = 96
Part 2 (from 8 to 28 s): rectangle: 24 × 20 = 480
Part 3 (from 28 to 34 s): triangle: ½ × 6 × 24 = 72 - Add the areas. 96 + 480 + 72 = 648
Check. Total time = 34 s, so the average speed is 648 ÷ 34 = 32417 m/s. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Method. Average speed = total distance ÷ total time. The total distance is the area under the speed–time graph.
- Find the total distance: the area under the graph. Split it into rectangles, triangles and trapezia and add their areas.Part 1: triangle: ½ × 8 × 24 = 96
Part 2: rectangle: 24 × 20 = 480
Part 3: triangle: ½ × 6 × 24 = 72
total = 648 m - Find the total time. Read the last and first values on the time axis.34 − 0 = 34 s
- Divide. Average speed = total distance ÷ total time.648 ÷ 34 = 32417 m/s
Check. Check that 32417 lies between the lowest and the highest speed on the graph. ✓
Watch out. Average speed is not the average of the highest and lowest speeds unless the graph is a single straight line.
Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).
Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.
- Split the area into strips of equal width. The region from time = 0 to 6 is 6 wide. With 3 strips, each strip has width 6 ÷ 3 = 2.
- Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 2 4 6 speed 2 4 10 20 - Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.Strip 1: ½ × (2 + 4) × 2 = 6
Strip 2: ½ × (4 + 10) × 2 = 14
Strip 3: ½ × (10 + 20) × 2 = 30 - Add the strip areas to get the estimate. 6 + 14 + 30 = 50
- Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.(The exact area is 48.00 m, so the estimate is bigger.)
Check. Use the trapezium rule formula: ½ × 2 × [2 + 20 + 2 × (4 + 10)] = 50.00. ✓
Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.
Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).
Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.
- Split the area into strips of equal width. The region from time = 0 to 4 is 4 wide. With 4 strips, each strip has width 4 ÷ 4 = 1.
- Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 1 2 3 4 speed 1 2 5 10 17 - Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.Strip 1: ½ × (1 + 2) × 1 = 1.5
Strip 2: ½ × (2 + 5) × 1 = 3.5
Strip 3: ½ × (5 + 10) × 1 = 7.5
Strip 4: ½ × (10 + 17) × 1 = 13.5 - Add the strip areas to get the estimate. 1.5 + 3.5 + 7.5 + 13.5 = 26
- Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.(The exact area is 25.33 m, so the estimate is bigger.)
Check. Use the trapezium rule formula: ½ × 1 × [1 + 17 + 2 × (2 + 5 + 10)] = 26.00. ✓
Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.
Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).
Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.
- Split the area into strips of equal width. The region from time = 0 to 4 is 4 wide. With 4 strips, each strip has width 4 ÷ 4 = 1.
- Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 1 2 3 4 speed 16 15 12 7 0 - Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.Strip 1: ½ × (16 + 15) × 1 = 15.5
Strip 2: ½ × (15 + 12) × 1 = 13.5
Strip 3: ½ × (12 + 7) × 1 = 9.5
Strip 4: ½ × (7 + 0) × 1 = 3.5 - Add the strip areas to get the estimate. 15.5 + 13.5 + 9.5 + 3.5 = 42
- Comment on the accuracy. The curve bends downwards (concave), so each trapezium top lies below the curve. So the estimate is an under-estimate. More strips would give a better estimate.(The exact area is 42.67 m, so the estimate is smaller.)
Check. Use the trapezium rule formula: ½ × 1 × [16 + 0 + 2 × (15 + 12 + 7)] = 42.00. ✓
Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.
Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.
- Split into strips of equal width. Width of each strip = (6 − 0) ÷ 3 = 2.
- Use the height at the left-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its left edge.
time 0 2 4 6 speed 2 4 10 20 - Area of each rectangle = height × width. Strip 1: 2 × 2 = 4
Strip 2: 4 × 2 = 8
Strip 3: 10 × 2 = 20 - Add. 4 + 8 + 20 = 32
- Say whether it is an over- or under-estimate. The graph is increasing. Using the left-hand heights means each rectangle is below the curve for most of the strip. So the estimate is an under-estimate.estimate 32.00 m < true area 48.00 m
Check. The more strips we use, the closer the estimate gets to the true value (48.00). ✓
Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.
Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.
- Split into strips of equal width. Width of each strip = (6 − 0) ÷ 3 = 2.
- Use the height at the right-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its right edge.
time 0 2 4 6 speed 2 4 10 20 - Area of each rectangle = height × width. Strip 1: 4 × 2 = 8
Strip 2: 10 × 2 = 20
Strip 3: 20 × 2 = 40 - Add. 8 + 20 + 40 = 68
- Say whether it is an over- or under-estimate. The graph is increasing. Using the right-hand heights means each rectangle is above the curve for most of the strip. So the estimate is an over-estimate.estimate 68.00 m > true area 48.00 m
Check. The more strips we use, the closer the estimate gets to the true value (48.00). ✓
Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.
Method. Split the area into strips, use the height at the left (or right) edge of each as a rectangle, and add the rectangle areas.
- Split into strips of equal width. Width of each strip = (4 − 0) ÷ 4 = 1.
- Use the height at the left-hand edge of each strip. Each strip is treated as a rectangle whose height is the value of the graph at its left edge.
time 0 1 2 3 4 speed 16 15 12 7 0 - Area of each rectangle = height × width. Strip 1: 16 × 1 = 16
Strip 2: 15 × 1 = 15
Strip 3: 12 × 1 = 12
Strip 4: 7 × 1 = 7 - Add. 16 + 15 + 12 + 7 = 50
- Say whether it is an over- or under-estimate. The graph is decreasing. Using the left-hand heights means each rectangle is above the curve for most of the strip. So the estimate is an over-estimate.estimate 50.00 m > true area 42.67 m
Check. The more strips we use, the closer the estimate gets to the true value (42.67). ✓
Watch out. For an increasing graph, left-hand rectangles give an under-estimate and right-hand rectangles give an over-estimate. For a decreasing graph it is the other way round.
Method. The area under a flow rate–time graph is the volume pumped. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and flow rate on the vertical axis, area = flow rate × time = volume pumped. So the area under the graph is the volume pumped.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 5 minutes): triangle: ½ × 5 × 20 = 50
Part 2 (from 5 to 15 minutes): rectangle: 20 × 10 = 200
Part 3 (from 15 to 20 minutes): trapezium: ½ × (20 + 10) × 5 = 75 - Add the areas. 50 + 200 + 75 = 325
Check. Total time = 20 minutes, so the average flow rate is 325 ÷ 20 = 16.25 litres/min. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Method. The area under a speed–time graph is the distance travelled. Split it into rectangles, triangles and trapezia, find each area and add.
- Remember what the area represents. On a graph with time on the horizontal axis and speed on the vertical axis, area = speed × time = distance travelled. So the area under the graph is the distance travelled.
- Split the area under the graph into simple shapes. Draw vertical lines down from each point where the graph changes direction. This makes rectangles, triangles and trapezia whose areas we can find.
- Work out the area of each part. Use rectangle = base × height, triangle = ½ × base × height, trapezium = ½ × (sum of parallel sides) × width.Part 1 (from 0 to 0.5 hours): triangle: ½ × 0.5 × 60 = 15
Part 2 (from 0.5 to 2 hours): rectangle: 60 × 1.5 = 90
Part 3 (from 2 to 2.5 hours): triangle: ½ × 0.5 × 60 = 15 - Add the areas. 15 + 90 + 15 = 120
Check. Total time = 2.5 hours, so the average speed is 120 ÷ 2.5 = 48 km/h. This lies between the lowest and highest values on the graph. ✓
Watch out. A trapezium needs the sum of the two parallel sides (the two heights) halved, then multiplied by the width. Do not forget the half.
Method. The area under a graph is (vertical value) × (horizontal value), so multiply the numbers and the units.
- Write the multiplication. The vertical axis is power in kW and the horizontal axis is time in hours.3 kW × 4 hours
- Multiply the numbers and the units. kW × hours gives kWh (kilowatt-hours of energy used).3 × 4 = 12 kWh (kilowatt-hours of energy used)
Check. Divide back: 12 ÷ 4 = 3 kW. ✓
Watch out. The unit of the area comes from multiplying the two axis units.
Method. The area under a graph is (vertical value) × (horizontal value), so multiply the numbers and the units.
- Write the multiplication. The vertical axis is flow rate in litres per second and the horizontal axis is time in seconds.8 litres/s × 15 s
- Multiply the numbers and the units. litres/s × s gives litres (volume of water).8 × 15 = 120 litres (volume of water)
Check. Divide back: 120 ÷ 15 = 8 litres/s. ✓
Watch out. The unit of the area comes from multiplying the two axis units.
Method. The area under a graph is (vertical value) × (horizontal value), so multiply the numbers and the units.
- Write the multiplication. The vertical axis is pay rate in pounds per hour and the horizontal axis is time in hours.12.5 £/hour × 6 hours
- Multiply the numbers and the units. £/hour × hours gives £ (total pay).12.5 × 6 = £75
Check. Divide back: 75 ÷ 6 = 12.5 £/hour. ✓
Watch out. The unit of the area comes from multiplying the two axis units.
Method. The area under a graph is (vertical value) × (horizontal value), so multiply the numbers and the units.
- Write the multiplication. The vertical axis is speed in m/s and the horizontal axis is time in seconds.15 m/s × 40 s
- Multiply the numbers and the units. m/s × s gives m (distance).15 × 40 = 600 m (distance)
Check. Divide back: 600 ÷ 40 = 15 m/s. ✓
Watch out. The unit of the area comes from multiplying the two axis units.
Key words. An estimate is an approximate answer. The area under a rate graph (like speed) is the total amount (like distance).
Method. Split the area under the curve into strips of equal width. Treat each strip as a trapezium, find its area, and add them. This gives an estimate of the true area.
- Split the area into strips of equal width. The region from time = 0 to 10 is 10 wide. With 5 strips, each strip has width 10 ÷ 5 = 2.
- Read (or calculate) the height of the graph at each strip edge. Each strip is a trapezium with two parallel vertical sides (the heights at its left and right edges).
time 0 2 4 6 8 10 speed 0 1.4 5.6 12.6 22.4 35 - Work out the area of each trapezium. Area of a trapezium = ½ × (sum of the parallel sides) × width.Strip 1: ½ × (0 + 1.4) × 2 = 1.4
Strip 2: ½ × (1.4 + 5.6) × 2 = 7
Strip 3: ½ × (5.6 + 12.6) × 2 = 18.2
Strip 4: ½ × (12.6 + 22.4) × 2 = 35
Strip 5: ½ × (22.4 + 35) × 2 = 57.4 - Add the strip areas to get the estimate. 1.4 + 7 + 18.2 + 35 + 57.4 = 119
- Comment on the accuracy. The curve bends upwards (convex), so each trapezium top lies above the curve. So the estimate is an over-estimate. More strips would give a better estimate.(The exact area is 116.67 m, so the estimate is bigger.)
Check. Use the trapezium rule formula: ½ × 2 × [0 + 35 + 2 × (1.4 + 5.6 + 12.6 + 22.4)] = 119.00. ✓
Watch out. The two end heights are used once; every middle height is used twice (it belongs to two strips). Make sure the strips have equal widths.
Method. Work out trapezium estimates with 1, 2 and 4 strips and compare each with the exact area.
- Say why for a downward-bending curve. Each trapezium top is a straight line joining two points on the curve, and the curve bends upwards above that line (the curve is above its chords). So the trapezia sit below the curve.the estimate is too small
- Work out the estimates. Use width = 4 ÷ number of strips and add the trapezium areas.1 strip: ½ × 4 × (16 + 0) = 32
2 strips: 40
4 strips: 42 - Compare with the exact area. All three are below 42.67, and they get closer as the number of strips grows.32 < 40 < 42 < 42.67
Check. The 4-strip estimate is only 0.67 below the exact area. ✓
Watch out. More strips = better estimate, but a concave curve always gives an under-estimate with the trapezium method.
Method. Multiply the units of the two axes.
- Look at the units of the area under a distance–time graph. Distance is in m and time is in s, so the area is in m × s = metre-seconds.m × s
- Ask whether that is a useful quantity. It is not a distance. A distance is measured in metres only.Ben is wrong.
- Look at a speed–time graph. Speed is in m/s and time in s, so the area is m/s × s = m.area = distance
Check. A car at 10 m/s for 5 s travels 50 m: the area of a 10 by 5 rectangle. ✓
Watch out. For distance–time graphs use the gradient (speed). For speed–time graphs use the area (distance).
Method. The area under the line is a triangle. Write its area in terms of k and set it equal to 54.
- Find the height of the triangle at x = 6. Put x = 6 in y = kx.y = 6k
- Area of the triangle. ½ × base × height.½ × 6 × 6k = 18k
- Set equal to 54. 18k = 54
- Solve. Divide both sides by 18.k = 3
Check. With k = 3 the height at x = 6 is 18, and ½ × 6 × 18 = 54. ✓
Watch out. Draw the shape first, then write its area.