This problem makes sense if the domain of the two functions is something like [0.3]. Then the area enclosed by the two curves is a small region specified by the two values of x for which the functions have the same value. These two values are x =1 and
x =2.
On the interval (1,2) the cubic function is greater than the linear one. Therefore
Area = integral from 1 to 2 of x^{3} -8 x^{2} +18 x -5 - (x +5) =
integral from 1 to 2 of x^{3} -8 x^{2} + 17 x -10
The antiderivative can be found term by term
[ (1/4) x^{4} - (8/3) x^{3} + (17/2) x^{2} - 10x ]
area = (this expression evaluated at x = 2) - (this expression evaluated at x =1 )
area = 7/12
Comments