current attempt in progress the graph of the function f(x)=(x^2 - 2)/x is shown on the right. compute the…

Explanation:

Step1: Simplify the integrand

First, simplify $\frac{x^{2}-2}{x}$ to $x-\frac{2}{x}$. So the integral becomes $\int_{0.7}^{5}(x - \frac{2}{x})dx$.

Step2: Integrate term - by - term

Using the power rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$) and $\int\frac{1}{x}dx=\ln|x|+C$, we have $\int(x-\frac{2}{x})dx=\frac{x^{2}}{2}-2\ln|x|+C$.

Step3: Apply the fundamental theorem of calculus

$\left[\frac{x^{2}}{2}-2\ln|x|\right]_{0.7}^{5}=\left(\frac{5^{2}}{2}-2\ln(5)\right)-\left(\frac{(0.7)^{2}}{2}-2\ln(0.7)\right)$. $=\frac{25}{2}-2\ln(5)-\frac{0.49}{2}+2\ln(0.7)$. $=\frac{25 - 0.49}{2}-2\ln(5)+2\ln(0.7)$. $=\frac{24.51}{2}-2(\ln(5)-\ln(0.7))$. $ = 12.255-2\ln(\frac{5}{0.7})$. $=12.255 - 2\ln(\frac{50}{7})\approx12.255-2\times1.9606\approx12.255 - 3.9212=8.3338$.

Answer:

(a) $8.3338$ (b) The net signed area between the curve $y = f(x)=\frac{x^{2}-2}{x}$, the $x$ - axis, and the vertical lines $x = 0.7$ and $x = 5$. That is, the area above the $x$ - axis minus the area below the $x$ - axis in the interval $[0.7,5]$.