Approximating the accumulation with rectangles

When the rate is not constant, no single rectangle has the accumulation as its area. So split the interval into equal pieces, and in each piece choose a constant value close to the function’s values there. The function is replaced by a step function, and the sum of the rectangles is an approximation of the accumulation.

The example below approximates the area under the graph of $\displaystyle {f(x)}={x^2}$ for $\displaystyle 1\le x\le2$, with one rectangle:

• The height of the rectangle – the value at the left end: $\displaystyle {f(1)}={1}$. The area: $\displaystyle {1\cdot1}={1}$ – too small.
• The value at the right end: $\displaystyle {f(2)}={4}$. The area: $\displaystyle {4\cdot1}={4}$ – too big.
• The mean of the two values: $\displaystyle {\frac{1+4}{2}}={2.5}$. The area: $\displaystyle 2.5$.
• The exact area, $\displaystyle \frac{7}{3}\approx2.33$, lies between 1 and 4. (How to calculate it – in the next lessons.)

The example below splits the same interval, $\displaystyle 1\le x\le2$, into four pieces, each $\displaystyle 0.25$ wide, and the height of each rectangle is the value at the left end of its piece:

• The heights: $\displaystyle {f(1)}={1}$, $\displaystyle {f(1.25)}={1.5625}$, $\displaystyle {f(1.5)}={2.25}$, $\displaystyle {f(1.75)}={3.0625}$.
• The sum of the rectangles: $\displaystyle {0.25\cdot(1+1.5625+2.25+3.0625)}={1.96875}$.

• With the right end: $\displaystyle {0.25\cdot(1.5625+2.25+3.0625+4)}={2.71875}$.
• These approximations are closer to the exact area, $\displaystyle 2.33$, than those of the single rectangle.
• The shorter the pieces, the smaller the gap between the two approximations, and both approach the exact area. With 100 pieces: $\displaystyle 2.31835$ and $\displaystyle 2.34835$.

When the function is linear, the mean of the values at the two ends gives the exact area: the area under the line is a trapezium, and the area of a trapezium is the mean of its parallel sides times its height. In the example below $\displaystyle {f(x)}={x+1}$ for $\displaystyle 0\le x\le2$:

• The values at the ends: $\displaystyle {f(0)}={1}$ and $\displaystyle {f(2)}={3}$.
• The area: $\displaystyle {\frac{1+3}{2}\cdot2}={4}$ – exactly.
• The triangle the rectangle adds above the line equals the triangle it leaves out below it.

The function $\displaystyle {f(x)}={x^2}$ is given for $\displaystyle 0\le x\le2$.
The interval is split into four pieces, each $\displaystyle 0.5$ wide.

Calculate the sum of the rectangles when the height of each is the value at the left end of its piece.

Calculate the sum of the rectangles when the height of each is the value at the right end of its piece.

The exact area is $\displaystyle \frac{8}{3}$. Which of the two approximations is closer to it?

A runner starts running from rest.
Her speed was measured every 10 seconds:

Suppose that in each 10 seconds she ran at the speed at their start. How many metres did she run in the 40 seconds?

Suppose that in each 10 seconds she ran at the speed at their end. How many metres did she run?

Her speed rose all the time. What can you say about the real distance?

The function $\displaystyle {f(x)}={2x+1}$ is given.
Calculate the exact area under its graph for $\displaystyle 1\le x\le3$, using the mean of the values at the ends.