Q. 15 - Calculus

Question

Consider the area between the graph of $f\left(x\right)=4-x^2$ and the x-axis on [0, 2]. 

(a) Use the four rectangles shown on the left to approximate the given area, and then use the eight rectangles shown on the right to obtain another approximation of that area. Be sure to use the fact that the graph shown is that of the function $f\left(x\right)=4-x^2$ in your calculations. 

(b) Describe what would happen if we did similar approximations with more and more rectangles, and make a guess for the resulting limit. 

Step-by-Step Solution

Answer

Part (a). 

Area from 4 rectangles = 6.25

Area from 8 rectangles = 5.8125

Part (b). Area will be more accurate. 

Part (a) Step 1. Given information.

We have been given the area between the graph of $f\left(x\right)=4-x^2$ and the x-axis on [0, 2].

We have to approximate the given area using the four rectangles shown on the left, and then obtain another approximation for the area using the eight rectangles shown on the right. Be sure to use the fact that the graph shown is that of the function $f\left(x\right)=4-x^2$ in the calculations. 

Part (a) Step 2. Area of the rectangle.

The four rectangles in the graph have a width of 1.

Height is given by :

$f\left(0\right),f\left(\frac{1}{2}\right),f\left(1\right),f\left(\frac{3}{2}\right)$

Therefore, Area is :

$\begin{array}{r}A=f\left(0\right)\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+f\left(1\right)\left(\frac{1}{2}\right)+f\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)\\ =(4-0)\left(\frac{1}{2}\right)+\left(4-\frac{1}{4}\right)\left(\frac{1}{2}\right)+(4-1)\frac{1}{2}+\left(4-\frac{9}{4}\right)\left(\frac{1}{2}\right)\\ =2+\frac{15}{8}+\frac{3}{2}+\frac{7}{8}\\ =\frac{16+15+12+7}{8}\\ =\frac{50}{8}\\ =6.25\end{array}$

Part (a) Step 3. The area from 8 rectangles.

The 8 rectangles in the graph have a width of $\frac{1}{4}$

Their height is given by :

$f\left(0\right),f\left(\frac{1}{4}\right),f\left(\frac{1}{2}\right),f\left(\frac{3}{4}\right),f\left(1\right),f\left(\frac{5}{4}\right),f\left(\frac{3}{2}\right),f\left(\frac{7}{4}\right)$

Therefore, the area is :
$\begin{array}{r}A=f\left(0\right)\frac{1}{4}+f\left(\frac{1}{4}\right)\frac{1}{4}+f\left(\frac{1}{2}\right)\frac{1}{4}+f\left(\frac{3}{4}\right)\frac{1}{4}+f\left(1\right)\frac{1}{4}\\ +f\left(\frac{5}{4}\right)\frac{1}{4}+f\left(\frac{3}{2}\right)\frac{1}{4}+f\left(\frac{7}{4}\right)\frac{1}{4}\\ =(4-0)\frac{1}{4}+\left(4-\frac{1}{16}\right)\frac{1}{4}+\left(4-\frac{9}{16}\right)\frac{1}{4}+(4-1)\frac{1}{4}+\left(4-\frac{25}{16}\right)\frac{1}{4}\\ +\left(4-\frac{9}{4}\right)\frac{1}{4}+\left(4-\frac{49}{16}\right)\frac{1}{4}\\ =1+\frac{63}{64}+\frac{55}{64}+\frac{3}{4}+\frac{39}{64}+\frac{7}{16}+\frac{15}{64}\\ =5.8125\end{array}$

Part (b) Step 1. Guess for resulting limit

If we take smaller and smaller rectangles, the area will be more accurate.