Area bounded by two curves

Question

What is the area of the region bounded by the curves y=2(x^2)-7 and y=2/x between x=1 and x=3? Please and thank you!

Answer 1

The curves intersect when 2/x = 2x² - 7.  By inspection, this equation has a solution when x = 2.  Between x = 1 and x = 2, the graph of y = 2x² - 7 lies below the graph of y = 2/x, while the graph of y = 2/x is the "lower" curve when x lies between 2 and 3.

 

So, Area = [integral from 1 to 2 of 2/x - (2x² - 7)] +[integral from                         2 to 3 of (2x² - 7) - 2/x]

 

                = [(2ln2 - (2/3)(2)³ + 7(2) - (2ln1 - (2/3)(1)³ + 7(1))] +

                      

                        [(2/3)(3)³ - 7(3) -2ln3 - ((2/3)(2³) - 7(2) -2ln2)]

 

                    ⁼ ln4 - 16/3 + 14 - 0 + 2/3 - 7 + 18 - 21 - ln9 - 16/3 + 14

                       + ln4

 

                = 8 + ln(16/9) ≈ 8.57                              

 

 

 

Answer 2

You need to integrate the difference of the functions defining the curves on the interval [1,3]. The antiderivative for 2 x² - 7 - 2/x is (2/3) x³ -7 x - 2 ln(x). Now evaluate at 3 and 1 and calculate the difference to get

 

[(2/3) 3³ -7 (3) - 2 ln(3)] - [(2/3) 1³ -7 (1) - 2 ln(1)] = (10/3) - 2 ln(3).

 

(Remember that ln(1) = 0!)