Question

Figure1

Figure 2

(a) Use the rectangles in each graph to approximate the area of the region bounded by y = sin(x), y = 0, x = 0, and x = π. (Round youranswer to three decimal places.)
figure (1)
figure (2)

(b) Describe how you could continue this process to obtain a more accurate approximation of the area.

Continually decrease the height of all rectangles.

Continually increase the height of all rectangles.

Continually increase the number of rectangles.

Continually decrease the number of rectangles.

Answer 1

(a) For Figure 1,

This is a right hand approximation using 6 rectangles of equal width. By taking the function's range [0,π] and dividing by 6 rectangles, wegetπ/6 as our width. As you can see in the figure, the right hand side of the rectangles lie on x = π/6, π/6 2π/6, 3π/6, 4π/6, 5π/6, and 6π/6(the last rectangle has no height). Finding the area of these rectangles gives us the approximation. The first rectangle has a width ofπ/6 andheight of sin(π/6)= .5. Multiplying the width and height of this first rectangle gives us an area ofπ/8. Doing the same thing for the other fiverectangles:

π/6*sin(π/6)=π/8=.393
π/6*sin(2π/6)= .453
π/6*sin(3π/6)= .524
π/6*sin(4π/6)= .453
π/6*sin(5π/6)= .262
π/6*sin(6π/6)=.0

adding these up gives... Area = 2.085

For Figure 2, we do the same thing. There are 4 rectangles (only three visible because the fourth's height equals zero), each with width ofπ/4.

π/4*sin(π/4)= .555
π/4*sin(2π/4)= .785
π/4*sin(3π/4)= .555
π/4*sin(4π/4)= 0

adding these up gives... Area = 1.895

The actual area (using integration) is equal to 2.

(b) By adding more rectangles, the heights become more appropriate. When an infinite amount of rectangles are used, their width goes to zero, and thearea approximation becomes exact.