Question

Find the absolute value of the resulting error if the value of the integral from 0 to 3 of x cubed, dx is estimated with 3 circumscribed rectangles of equal width.

Answer 1

For a better understanding of the solution provided here, please check the attached file which contains the required diagram pertaining to the question.

The three circumscribed rectangles are shown in the diagram in light green color.

The width of each rectangle is 1 unit.

The height of each rectangle is determined by the equation of the curve under consideration, which is: $y=x^3$. Thus the height of each rectangle is the value of the y-coordinate of the rectangle which, as can be clearly seen from the diagram, is 1 for the first rectangle, 8 for the second and 27 for the third rectangle. These values are derived by plugging in x=1,2,3 in the equation $y=x^3$ and finding out the corresponding value of y.

As can be clearly seen from the diagram, the area estimated by the circumscribed rectangles is an overestimation and thus will present an inherent error. Let us now proceed to calculate that error in absolute value as required by the question.

We know that the total area of the rectangles is the sum of the areas of each circumscribed rectangle. Thus:

Area of Rectangles=Area of 1st rectangle+Area of 2nd rectangle+Area of 3rd rectangle=$1\times 1+8\times 1+27\times 1=1+8+27=36$

From integration we know that the actual area will be:

Actual Area =$\int\limits^3_ 0{x^3} \,dx=[\frac{x^4}{4}]_{0}^{3}=\frac{1}{4}(3^4-0^4)=\frac{81}{4}=20.25$

Thus, the absolute value of the resulting error is:

Absolute Error=|Area of Rectangles - Actual Area|=$\left | 36-20.25 \right |=\left | 15.75 \right |=15.75$

Thus, the Absolute Error is 15.75