What is the area of the region in the first quadrant that is enclosed by the graphs of #y=x^3+8# and #y=x+8#?

Question

13038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-04T16:34:44

The area is #1/4#.

Th two graphs intersect where
#x^3+8 = x+8# which is at #x=0# and #x=1#.

If we test at #x=1/2# we see that #x+8 > x^3+8# in the interval #[0,1]#

So,

The area is

#int_0^1 ((x+8)-(x^3+8)) dx = int_0^1 (x-x^3) dx = 1/4#

Although I didn't do it, you can draw the graph if you like.

#y=x+8# is a line with slope #1#, through #(0,8)#

#y = x^3+8# is the graph of the cubic translated up #8#.