Find the area between the graphs of $f(x)=e^{.25x}$ and $g(x)=\frac{-1}{x}$ between $x=1$ and $x=2$

Question

5.5

Can somebody verify this solution for me?

Find the area between the graphs of $f(x)=e^{.25x}$ and $g(x)=\frac{-1}{x}$ between $x=1$ and $x=2$

Since $f(x) > g(x)$ on $(1,2)$, the area between the graphs is:

$\int_1^2 e^{.25x} - (\frac{-1}{x})dx$

$=\int_1^2 e^{.25x} + (\frac{1}{x})dx$

$= \frac{e^{.25x}}{.25} + ln(|x|)|_1^2dx$

$= \frac{e^{.25(2)}}{.25} + ln(2) - \frac{e^{.25(1)}}{.25} - ln(1)$

$= \frac{e^{.25(2)}}{.25} + ln(2) - \frac{e^{.25(1)}}{.25} - ln(1)$

Since $ln(1)=0$, we get:

$= 4e^{.5} + ln(2) - 4e^{.25}$

Answer 1

Well, area is an absolute thing. So we know that the total area $\mathcal{A}$ is given by:

$$\mathcal{A}=\int_1^2\exp\left(\frac{x}{4}\right)\space\text{d}x+\left|\int_1^2-\frac{1}{x}\space\text{d}x\right|=$$ $$\left[4\exp\left(\frac{x}{4}\right)\right]_1^2+\left|\left[-\ln\left|x\right|\right]_1^2\right|=$$ $$4\exp\left(\frac{2}{4}\right)-4\exp\left(\frac{1}{4}\right)+\left|-\ln\left|2\right|-\left(-\ln\left|1\right|\right)\right|=$$ $$4\exp\left(\frac{1}{2}\right)-4\exp\left(\frac{1}{4}\right)+\left|-\ln\left|2\right|-\left(-0\right)\right|=$$ $$4\exp\left(\frac{1}{2}\right)-4\exp\left(\frac{1}{4}\right)+\left|-\ln\left|2\right|\right|=$$ $$4\exp\left(\frac{1}{2}\right)-4\exp\left(\frac{1}{4}\right)+\ln\left(2\right)\approx2.15193\tag1$$

So, yes you're right.