Integral is square root of 36-x^2 with bounds from -6 to 6 .Find and approximation of the area of the region using 6 subintervals of equal width and the mid

Question

Integral is the square root of 36-x^2 with bounds from -6 to 6 . b) Find and approximation of the area of the region using 6 subintervals of equal width and the midpoints of each interval. You may round to nearest hundredth.

Answer 1

b)

 

The graph is symmetrical about the y-axis.  Since the distance x=-6 and y-axis is the same distance between y-axis and x=6, we take the area under the curve for half of the section.  Then multiply it by 2 to get the full area.

 

We should sketch the graph of this curve to see what it looks like.   It is actually a circle with centerpoint (0, 0) and radius of 6.

 

We will take the area under the curve between x=0 and x=6. Using left-hand corners.  Each rectangle will have a base length of 1.  The height is the f(x) value at a point.

 

1st rectangle = f(0)

2nd rectangle = f(1)

3rd rectangle = f(2)

4th rectangle = f(3)

5th rectangle = f(4)

6th rectangle = f(5)

 

Evaluate each of these f(x) values then add them up.  Finally, multiply the sum by 2 to get the full area.  This area is known as an overestimate because some portions of the rectangles are above the curve.

 

Then compare this area under the curve with the integral that another tutor gave you earlier today.  Do they match?  Do them come close?

 

 

Remember:

Rectangles give estimations of area under the curve.  Integral give exact area under the curve.  And taking 2 the area of this semi-circle will also give the exact area under the curve.