Question

Find the area of the region enclosed by the curves \(y=\frac{x}{x^{2}+1}\) and \(y=m x, \quad 0 < m <1\)

Short answer

The area between the curves \( y=\frac{x}{x^{2}+1}\) and \( y=m x\) is \( \arctan(\frac{1-m}{1+m})-\frac{m*(1-m^{2})}{2*(1+m^{2})}\)

Step-by-step solution

Find Intersection Points

To find the intersection points of \(y=\frac{x}{x^{2}+1}\) and \(y=m x\), set the two equal to each other: \(\frac{x}{x^{2}+1} = mx\) then solve for \(x\). When we solve the equation, \(x=0\) and \(x=\frac{1-m}{1+m}\) are derived.

Set up Integration

We are looking for the area between the two curves from \(x=0\) to \(x=\frac{1-m}{1+m}\). We need to subtract the lower function from the upper one. Checking the two functions at a value between \(0\) and \(\frac{1-m}{1+m}\), say \(x=0.5\), we see that \(\frac{x}{x^{2}+1} > mx\) for \(0 < m <1\). So, we set up our integral as \( \int_{0}^{(1-m)/(1+m)} (\frac{x}{x^{2}+1} - mx) dx\)

Calculate the Integral

Compute the definite integral. The integrals will look like this when broken down: \( \int_{0}^{(1-m)/(1+m)} (\frac{x}{x^{2}+1}) dx - m\int_{0}^{(1-m)/(1+m)} x dx\). The integral \( \int_{0}^{(1-m)/(1+m)} (\frac{x}{x^{2}+1}) dx\) leads to \( [\arctan(x)]_{0}^{(1-m)/(1+m)}\) and the integral \(m\int_{0}^{(1-m)/(1+m)} x dx\) leads to \( [0.5*m*x^{2}]_{0}^{(1-m)/(1+m)}\). Now, we need to evaluate these at the limits of \(0\) and \(\frac{1-m}{1+m}\).

Evaluate at Limits

Evaluating at limits gives us \((\arctan(\frac{1-m}{1+m}))-m*\frac{1}{2}*(\frac{1-m}{1+m})^{2}\) as the area under the curve in the given domain

Simplifying the Expression

We can further simplify the expression to \( \arctan(\frac{1-m}{1+m})- \frac{m*(1-m^{2})}{2*(1+m^{2})}\)