What is the net area between #f(x) = 3-xsqrt(x^2-1) # and the x-axis over #x in [2, 3 ]#?

Question

Ezra Adams · Accepted Answer

The area #~~2.81#

Here is a graph of the function #f(x) = 3-xsqrt(x^2-1)#:

graph{3-xsqrt(x^2-1) [-1, 5, -7, 3]}

Please observe that the #f(x)# is negative over the region #[2,3]#, therefore, to obtain an positive area, we shall evaluate from 3 to 2.

To integrate #-xsqrt(x^2-1)dx# let #u =x^2-1# and #du = 2xdx#

#intx(x^2-1)^(1/2)dx = int1/2u^(1/2)du = 2/3(1/2)u^(3/2) = 1/3u^(3/2)#:

The integration of the constant term is trivial:

#int_3^2 3-xsqrt(x^2-1)dx = (3x - 1/3(x^2-1)^(3/2)]_3^2=#

#3(2)-1/3(2^2-1)^(3/2)-3(3)+1/3(3^2-1)^(3/2) ~~2.81#

Christopher Agresta · Answer

undefined To find the net area between $ f(x) = 3 - x\sqrt{x^2 - 1} $ and the x-axis over $ x $ in the interval $ [2, 3] $, you need to evaluate the definite integral of $ f(x) $ over that interval. The net area is given by the absolute value of the integral:

$$ \text{Net Area} = \left| \int_{2}^{3} (3 - x\sqrt{x^2 - 1}) \, dx \right| $$

Using calculus techniques, you can integrate the function $ f(x) $ over the interval $ [2, 3] $ to find the net area.