What is the net area between #f(x)=(x^2-5)/(x^3-5x# in #x in[1,2] # and the x-axis?

Question

What is the net area  between #f(x)=(x^2-5)/(x^3-5x# in #x  in[1,2]    # and the x-axis?

Samantha Adkison · Accepted Answer

Area #=ln 2=0.693147" "#square unit

Given #y=(x^2-5)/(x^3-5x)# from #x=1# to #x=2# and the x-axis #y=(x^2-5)/(x^3-5x)=1/x# Area #int_1^2 (1/x)dx=[ln x]_1^2=ln 2- ln 1=ln 2# Area #=ln 2=0.693147" "#square unit graph{(y-(x^2-5)/(x^3-5x))(y+10000x-10000)(y+10000x-20000)(y-0)=0[-0.1,4,-2,2]} God bless...I hope the explanation is useful

Emily Ammerman · Answer

undefined To find the net area between the curve $f(x) = \frac{x^2 - 5}{x^3 - 5x}$ and the x-axis over the interval $[1, 2]$, you need to integrate the absolute value of the function from 1 to 2. This is because the function may be above or below the x-axis over that interval, resulting in areas both above and below the x-axis.

1. Find the points of intersection between the curve and the x-axis by setting $f(x) = 0$ and solving for $x$.
2. Determine the intervals where the function is above or below the x-axis over the interval $[1, 2]$.
3. Integrate the absolute value of the function over each interval.
4. Add up the individual areas to find the net area.

It's important to note that the integral of the absolute value function may result in multiple integrals to consider, depending on the behavior of the function over the interval.