# How do you find the area bounded by y=(x+1)/x, y=x^2+1 and x=2?

Feb 9, 2017

The area is $\frac{7}{3} - \ln 2 \approx 1.64$ square units.

#### Explanation:

Start by finding the intersection points.

${x}^{2} + 1 = \frac{x + 1}{x}$

${x}^{3} + x = x + 1$

${x}^{3} = 1$

$x = 1$

This means that $y = {1}^{2} + 1 = 2$.

Now complete a rudimentary sketch of the graphs.

So we have bounds of integration of $\left[1 , 2\right]$ (this is why we found the intersection points), and we know the parabola lies above the rational function. The integral to evaluate will therefore be:

${\int}_{1}^{2} {x}^{2} + 1 - \frac{x + 1}{x} \mathrm{dx}$

$= {\int}_{1}^{2} \frac{{x}^{3} + x - x - 1}{x} \mathrm{dx}$

$= {\int}_{1}^{2} \frac{{x}^{3} - 1}{x} \mathrm{dx}$

$= {\int}_{1}^{2} {x}^{2} - \frac{1}{x} \mathrm{dx}$

$= {\left[\frac{1}{3} {x}^{3} - \ln | x |\right]}_{1}^{2}$

Evaluate using ${\int}_{a}^{b} F \left(x\right) = f \left(b\right) - f \left(a\right)$, where function $f \left(x\right)$ is continuous on $\left[a , b\right]$ and $f ' \left(x\right) = F \left(x\right)$.

$= \frac{1}{3} {\left(2\right)}^{3} - \ln | 2 | - \left(\frac{1}{3} {\left(1\right)}^{3} - \ln | 1 |\right)$

$= \frac{8}{3} - \ln 2 - \frac{1}{3} + 0$

$= \frac{7}{3} - \ln 2$

Hopefully this helps!