Find out the area of the region bounded by y = 2 x² + 10 and y = 4 x + 16 .

Solution

In this case the intersection points (that we'll required eventually) are not going to be simply identified from the graph therefore let's go ahead and get them now.  Note as well that for most of these problems you'll not be capable to accurately recognize the intersection points from the graph and therefore you'll have to be able to find out them by hand.  In this particular case we can get the intersection points by setting the two equations equivalent.

2x² + 10 = 4x + 16

2 x² - 4 x - 6 = 0

2 ( x + 1) ( x - 3) = 0

Therefore it looks like the two curves will intersect at x= -1 and x = 3 .  If we required them we can get the y values corresponding to each by plugging the values back into either of the equations.  We'll leave it to you to check out that the coordinates of the two intersection points on the graph are (-1,12) & (3,28).

Following is a graph of the region.

Along with the graph now we can identify the upper and lower function and therefore we can now find the enclosed area.

                           A = ∫^b_a (Upper function) - (lower function) dx

            = ∫³ _(-1) 4x + 16 -  (2x² +10)dx

= ∫³ _(-1) (-2x² + 4x + 6) dx

           = - (2/3) x³ + 2 x² + 6x )|³_(-1)

            =(64/3)