Is the actual area of the shaded region more or less than the value given by computer program? Explain

The shaded region is bounded by the graph of$y=x^2$, by the x-axis , and by the vertical line through the points$(1,0)$and$(1,1)$.

The program for calculating the area of shaded region using 10 rectangles :$\begin{array}{l}10\text{LET}X=0.1\\ 20\text{FOR}N=1\text{TO10}\\ \text{30LET}Y=X\uparrow 2\\ 40\text{LET}A=A+Y*0.1\\ 50\text{LET}X=X+0.1\\ 60\text{NEXT}N\\ 70\text{PRINT"AREAISAPPROXIMATELY";}A\\ 80\text{END}\end{array}$

We can approximate the area of the shaded region by drawing 1000 rectangles having base vertices a$x=0,0.001,0.002,0.003,\dots ,1.000,$tand computing the sum of the areas of the 1000 rectangles. The base of each rectangle is$0.001$, and the height of each rectangle is given by$y=x^2$

The following comuter program will comute and add the areas of the 1000 rectangles.In line$30,Y$is the height of each rectangle.In line40 , A gives the current total of all the areas.

$\begin{array}{l}10\text{LET}X=0.001\\ 20\text{FOR}N=1\text{TO1000}\\ \text{30LET}Y=X\uparrow 2\\ 40\text{LET}A=A+Y*0.001\\ 50\text{LET}X=X+0.001\\ 60\text{NEXT}N\\ 70\text{PRINT"AREAISAPPROXIMATELY";}A\\ 80\text{END}\end{array}$

If the program is run, the computer will print AREA IS APPROXIAMTELY $0.333833$.

We find the area of the shaded region by integration:

$\begin{array}{l}A=\underset{0}{\overset{1}{\int }}{x}^{2}dx\\ \Rightarrow A={\left[\frac{x^3}{3}\right]}_0^1\\ \Rightarrow A=\frac{1^3}{3}-\frac{0^3}{3}\\ \Rightarrow A=\frac{1}{3}=0.\overline{3}<0.333833\end{array}$

So, the actual area of the shaded region is less than the value computed by the computer program.