Riemann sum to approximate exact area

The goal of a Riemann Sum is to approximate the exact area between the curve and the axis by dividing the total area into $n$ vertical rectangles.

As seen in the diagram above, the area between the curve $y = x^2$ and the $x$-axis from $x = 0$ to $x = 1$ can be approximated as above.

\begin{align*} y & = x^2 \\ \\ \text{Height of 1st rectangle} & = (0.25)^2 = 0.0625 \\ \text{Height of 2nd rectangle} & = (0.5)^2 = 0.25 \\ \text{Height of 3rd rectangle} & = (0.75)^2 = 0.5625 \\ \\ \text{Area of three rectangles} & = (0.25 \times 0.0625) + (0.25 \times 0.25) + (0.25 \times 0.5625) \\ & = 0.21875 \text{ units}^2 \\ \\ \text{Actual area} & = \int_0^1 x^2 \phantom{.} dx \\ & = \left[{x^3 \over 3}\right]_0^1 \\ & = {1^3 \over 3} - {0^3 \over 3} \\ & = {1 \over 3} \text{ units}^2 \end{align*}

We can improve the accuracy of the estimate by reducing the width of each rectangles (i.e. dividing into more rectangles. Visually, the top of each rectangle 'hugs' the curve more closely.

\begin{align*} \text{Area of five rectangles} & = \left[{1 \over 6} \times \left(1 \over 6\right)^2\right] + \left[{1 \over 6} \times \left(2 \over 6\right)^2\right] + \left[{1 \over 6} \times \left(3 \over 6\right)^2\right] + \left[{1 \over 6} \times \left(4 \over 6\right)^2\right] + \left[{1 \over 6} \times \left(5 \over 6\right)^2\right] \\ & = 0.2546 \text{ units}^2 \phantom{00000} [\text{Closer to } {1 \over 3} \text{ units}^2 ] \end{align*}

When the number of rectangles $n \rightarrow \infty$, the sum of the areas of the rectangles becomes exactly equal to the area under the curve.

Under-approximation vs. Over-approximation

As seen in the diagram above, the sum of area of rectangles under-approximates the area between the curve $y = x^2$ and the $x$-axis from $x = 0$ to $x = 1$.

\begin{align*} \text{From previous sections, area of five rectangles} & = 0.2546 \text{ units}^2 \end{align*}

When drawn in the way shown above, the sum of area of rectangles over-approximates the area between the curve $y = x^2$ and the $x$-axis from $x = 0$ to $x = 1$.

\begin{align*} \text{Area of six rectangles} & = \left[ {1 \over 6} \times \left( {1 \over 6} \right) \right]^2 + \left[ {1 \over 6} \times \left( {2 \over 6} \right) \right]^2 + ... + \left[ {1 \over 6} \times \left( {6 \over 6} \right) \right]^2 \\ & = 0.4213 \text{ units}^2 \phantom{000000} [\text{More than } {1 \over 3} \text{ units}^2] \end{align*}

For under-approximation or over-approximation, sum of the areas of the rectangles becomes exactly equal to the area under the curve when $n \rightarrow \infty$.

Question

1. The graph of $y = 2^x$ for $0 \le x \le 1$, is shown in the diagram below. Rectangles, each of width ${1 \over n}$, where $n$ is an integer, are drawn under the curve.

(a) Show that the total area of all $n$ rectangles, $A$ is given by

$$ { 2^{1 \over n} \over n(2^{1 \over n} - 1) } $$

Solutions

(b) Explain why $ { 2^{1 \over n} \over n(2^{1 \over n} - 1) } > \int_0^1 2^x \phantom{.} dx $ for $ n \in \mathbb{Z}^+ $.

Solutions

(c) Find the value of $A$ when the number of rectangles $n$ tends to infinity.

Answer: $ 1.44 \text{ units}^2 $

Solutions