The accumulation function and its uses · Grade 11 · 5 units
In The accumulation function and its uses (Grade 11 · 5 units) you practise 8 skills: Antiderivatives of a root and by the chain rule · The accumulation function from a graph · The properties of the accumulation function · The mean value of a function · Displacement and distance from the velocity · The volume of a solid of revolution · The integral of a moved or stretched function · An amount from its rate. The accumulation function adds up area: its value at a point is the integral of the function from a fixed starting point up to that point. An area above the horizontal axis adds to it, and an area below the axis takes away from it. Its derivative is the function itself, so it increases where the function is positive and has a turning point where the function changes sign.
The integral of a rate is the amount that builds up. The integral of the velocity is the displacement, and the integral of the speed (the size of the velocity) is the distance travelled. The mean value of a function on an interval is its integral divided by the length of the interval. When a region turns about the horizontal axis, the volume of the solid is pi times the integral of the square of the function. A model proof: the volume of a cone. Turn about the horizontal axis the line through the origin, from x equals 0 to x equals the height, so that at the end of the interval the value of the line is the radius. This makes a cone, and the gradient of the line is the radius over the height. The square of the line is the square of the gradient times x squared, and the integral of x squared from 0 to the height is the height cubed over 3. So the volume is pi times the square of the radius over the square of the height, times the height cubed over 3, that is pi times the square of the radius times the height, over 3. A model proof: the volume of a sphere. Turn about the horizontal axis the semicircle with its centre at the origin, from one end of its diameter to the other. This makes a sphere, and the square of the semicircle is the square of the radius minus x squared. Its integral from minus the radius to the radius is twice the radius cubed, minus two thirds of the radius cubed, that is four thirds of the radius cubed. So the volume is four thirds times pi times the radius cubed.
Tip: To find a distance, first find where the velocity changes sign, and split the integral there.
What you learn
Antiderivatives of a root and by the chain rule
For example: Which of these is an antiderivative of f(x) = √x ?
- x√x
- √x ÷ 2
- 2√x
- 2x√x ÷ 3
Answer and solution
The answer: 2x√x ÷ 3
Write 2x√x ÷ 3 as (2 ÷ 3)x^(3/2) . Its derivative is (2 ÷ 3) × (3 ÷ 2)x^(1/2) , that is √x .
The accumulation function from a graph
For example: The picture shows the graph of the function y = f(x) for 0 ≤ x ≤ 10 . The areas of the regions between the graph and the horizontal axis, from left to right, are 11, 8.5, 4 and 6.5.
The function h(x) is the integral of f(t) from t = 0 to t = x . What is h(4) ?Answer and solution
The answer: 11
Up to x = 4 the graph is above the axis, so the value is the area: 11.
The properties of the accumulation function
For example: The function F(x) is the integral of f(t) from t = 1 to t = x . What is F(1) ?
- f(1)
- 1
- Not defined
- 0
Answer and solution
The answer: 0
The value F(1) is the integral from t = 1 to t = 1 : there is no area there, so it is 0.
The mean value of a function
For example: What is the mean value of the function f(x) = 4x − 5 on the interval −3 ≤ x ≤ 0 ?
Answer and solution
The answer: −11
The integral: ∫₋₃⁰ (4x − 5) dx = −33 . The length of the interval is 3, so the mean is −33 ÷ 3 = −11 .
Displacement and distance from the velocity
For example: A body moves along a straight line, and its velocity after t seconds is v(t) = −2t + 10 metres per second. What is the displacement of the body from t = 0 to t = 3 ?
Answer and solution
The answer: 21
The displacement is the integral of the velocity: ∫₀³ (−2t + 10) dt = 21 .
The volume of a solid of revolution
For example: The picture shows the region between the line y = (1 ÷ 6)x , the horizontal axis and the line x = 6 , together with its mirror image in the horizontal axis.
The region turns about the horizontal axis. What is the volume of the solid it makes? Give the answer as a multiple of π .Answer and solution
The answer: 2π
It makes a cone. The square: y² = (1 ÷ 36)x² , so the volume is π ∫₀⁶ (1 ÷ 36)x² dx , that is π × (1 ÷ 36) × 6³ ÷ 3 = 2π .
The integral of a moved or stretched function
For example: The function f is positive on the interval 0 ≤ x ≤ 6 , and the area between its graph and the horizontal axis on that interval is 8. What is the area between the graph of f(x) + 3 and the horizontal axis on the same interval?
Answer and solution
The answer: 26
The graph of f(x) + 3 is the graph of f moved 3 up, so a rectangle of area 3 × 6 = 18 is added. The area: 8 + 18 = 26 .
An amount from its rate
For example: Water flows into a barrel. After t minutes it flows at a rate of 10 − 0.5t litres a minute, until the flow stops. How many litres flow into the barrel in the second minute?
Answer and solution
The answer: 9.25
In the second minute the time runs from t = 1 to t = 2 : ∫₁² (10 − 0.5t) dt = 9.25 .
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).