Expert: Steve Holleran - 12/2/2007

Question
hi, i submitted these questions before but there was a typo error for question 2, so here's an updated one. thanks!

1. Find the area under the curve y = 5 x - 1/2 x^2 between x = 2, x = 8 and the x-axis.

2. find the area under the curve y = x^3 �ãx + 2x^2 - 2/ x^2 between x = 1, x = 4 and the x-axis leaving your answer as fraction.


Answer
Hi Lill,

Okay, on #1, if you graph the curve, you'll see the area is above the x-axis, so we can just do a straightforward integration:

  A = INT (x = 2, x =8) [5x - 1/2 * x^2]

    = 5x^2 / 2  - 1/2 * x^3 / 3

evaluating from x = 2 t x = 8, you should get 66

#2:  Again, if you graph the area, it is above the x-axis, so you can just do the integration without splitting it up:

 A = INT (x=1 to x=4) [x^3 + 2x^2 - 2x^-2]

   = [x^4/4 + 2x^3/3 + 2/x](x=1 to x=4) = 104.25

Steve