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?

Question

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?

Emilia Aguilar · Accepted Answer

I found: area#=-10/3#

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:

Ezekiel Alexander · Answer

undefined To find the area bounded by the curve $ y = x^2 - 9 $ and the x-axis for $ x = 0 $ to $ x = 5 $, you integrate the absolute value of the function $ y = x^2 - 9 $ with respect to $ x $ from $ x = 0 $ to $ x = 5 $. The integral is:

$$ \int_{0}^{5} |x^2 - 9| \, dx $$

Split the integral at the points where $ x^2 - 9 = 0 $, which are $ x = -3 $ and $ x = 3 $. Then integrate each portion separately and take the absolute value since the function changes sign between $ x = -3 $ and $ x = 3 $. Calculate the integral for each portion and sum them to find the total area.