Solution for Explore! on Page 431

The numerical integration feature can be found using option 7 of the CALC menu (2nd TRACE). You must specify the lower and upper limits of the area desired. Visualizing a triangle drawn from (2,0) to (2,2) to (4,0), containing area 2 square units, the area under the curve can be roughly estimated to be more than 3 sq units. The single triangle is not a good method for estimation. More vertical estimate lines are needed to form rectangles and triangles of known areas whose areas can be summed up. Using widths of one unit to form a rectangle and a triangle yields an better estimate of slightly more than 3 sq units. By the numerical integration feature, the area can be seen to be 3 1/3 exactly.

The function f(x) = 8x(x² + 1)³ has values 0 and 64 at x = 0 and 1, respectively. A viewing window [-.2, 1.2].2 by [-10, 70]10 shows a concave downward graph, whose area is sizably less than the triangle formed by the points (0,0), (1,64) and (0,64). A table starting at x = 0 with 0.2 increments shows f(x) values every .2 widths. The sum of these f(x) values from .2 to 1.0 add up to 122.10. Rectangles of heights f(x) with widths 0.2 total in area to 122.10(.2) = 22.22. Try using widths 0.1. The actual area is 15 sq. units.

Intersections can be found at x = -4, 0 and 1. Treating y = 4x as a horizontal baseline, the area can be viewed as the difference curve Y2 – Y1 using the prescribed window. The numerical integrator applied to this curve between x = -4 and 1 yields the desired areas between two curves. Note that there is positive area between x = -4 and 0 and negative area from x = 0 to 1.

Graphs of the two cost rate functions and the revenue rate function on the same screen show that the new cost rate function allows for slightly more years of machine profitability, specifically 10.741723 years. The new net earnings can be calculated to be $21,483.45, a bit more than the original $20,000.

Visually D₁ (q) is less than D(q) for all q values, supporting the conjecture that the area under the curve of D₁ (q) will be less than that of D(q). This area is calculated to be $240 versus the value of $264 for D(q).

The area under f(x) from x = 1 to 4 is 9.75 sq units, which equals the area of the rectangular portion under y = 9.75/3 = 3.25 of length 3. It is as though the area under f(x) over [1,4] turned to water and became a level surface of height 3.25, the average f(x) value. This value is attained at x = 1.874 and x = 3.473 for the function f(x).

The numerical integration feature of the graphing calculator is exemplified below. The table displays the area under f(x) from x = 1 to the designated X value. It appears that the integral values converge slowly to to the value 1.

For (a) see above. One way for (b) is to calculate 1 - P(x < 2) , where P(x< 2) = .6321, as above. Another way is to compute P(x ³ 2) via the improper integral. A tabular display of the numerical integrator (fnInt(Y1, x , 2, x)) starting at x = 10 in increments of 10 show a quick convergence to .36788.