Answer:
[tex]\text{Arc Length}=\dfrac{3}{2}=1.5[/tex]
Step-by-step explanation:
The arc length L of a curve y = f(x) over the interval [a, b] is given by the formula:
[tex]\displaystyle L=\int^b_a \sqrt{1+\left(\dfrac{\text{d}y}{\text{d}x}\right)^2}\; \text{d}x[/tex]
where a is the lower limit and b is the upper limit.
Given function:
[tex]y=\dfrac{1}{2}(e^x + e^{-x})[/tex]
To find the length of the curve represented by the given function over the interval [-ln(2), ln(2)], begin by differentiating y with respect to x:
[tex]\dfrac{dy}{dx}=\dfrac{1}{2}(e^x- e^{-x})[/tex]
Now, substitute dy/dx into the arc length formula and integrate over the interval [-ln(2), ln(2)]:
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \sqrt{1+\left(\dfrac{1}{2}(e^x - e^{-x})\right)^2}\; dx[/tex]
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \sqrt{1+\dfrac{1}{4}\left(e^x - e^{-x}\right)^2}\; dx[/tex]
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \sqrt{1+\dfrac{1}{4}(e^{2x}-2+e^{-2x})}\; dx[/tex]
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \sqrt{\dfrac{4}{4}+\dfrac{e^{2x}}{4}-\dfrac{2}{4}+\dfrac{e^{-2x}}{4}}\; dx[/tex]
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \sqrt{\dfrac{e^{2x}+2+e^{-2x}}{4}}\; dx[/tex]
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \sqrt{\dfrac{\left(e^x+e^{-x}\right)^2}{4}}\; dx[/tex]
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \dfrac{e^x+e^{-x}}{2}\; dx[/tex]
[tex]\displaystyle L=\int^{\ln2}_{-\ln2} \dfrac{e^x}{2}+\dfrac{e^{-x}}{2}\; dx[/tex]
[tex]L=\left[\dfrac{e^x}{2}-\dfrac{e^{-x}}{2}\right]^{\ln2}_{-\ln2}[/tex]
[tex]L=\left(\dfrac{e^{\ln2}}{2}-\dfrac{e^{-\ln2}}{2}\right)-\left(\dfrac{e^{-\ln2}}{2}-\dfrac{e^{\ln 2}}{2}\right)[/tex]
[tex]L=\left(\dfrac{2}{2}-\dfrac{\frac12}{2}\right)-\left(\dfrac{\frac12}{2}-\dfrac{2}{2}\right)[/tex]
[tex]L=\left(1-\dfrac{1}{4}\right)-\left(\dfrac{1}{4}-1\right)[/tex]
[tex]L=\dfrac{3}{4}+\dfrac{3}{4}[/tex]
[tex]L=\dfrac{3}{2}=1.5[/tex]
Therefore, the length of the curve of the given function on the interval [-ln(2), ln(2)] is:
[tex]\Large\boxed{\boxed{\dfrac{3}{2}=1.5}}[/tex]