How do you use integrals to find the area bounded by the curve y = (x^2 - 9) and the x-axis for x = 0 to x = 5?

1 Answer
Jul 27, 2015

Answer:

I found: area#=-10/3#

Explanation:

You write it as:
#int_0^5(x^2-9)dx=#
you can break it into two as:
#int_0^5(x^2)dx-int_0^5(9)dx=#
integrating you get:
#x^3/3-9x|_0^5#

now you use you extremes of integration substituting them and subtracting the resulting expressions as:
#(5^3/3-9*5)-(0^3/3-9*0)=#
#=125/3-45=-10/3#

You may wonder about the negative sign but looking at your area:
enter image source here
you can see that quite a big chunk (pink) is below the #x# axis giving a negative contribution (this are is equal to #-18#) while the blue part (positive) gives you a contribution of only #14.6#.