Find the arc length of the curve y = 1/2 e^2x between

Find the arc length of the curve $y = \dfrac{1}{2}e^{2x}$ between the points with abscissas $x = 0$ and $x = 1$.

We reduce the integral to a standard form via substitution.

Step 1: Differentiate

\[y'=e^{2x} \implies 1+(y')^2 = 1+e^{4x}\]

Step 2: Set up the integral

\[L=\int_0^1\sqrt{1+e^{4x}}\,dx\]

Step 3: Substitute $u=e^{2x}$

Then $du=2e^{2x}dx$, i.e. $dx=\dfrac{du}{2u}$, and $e^{4x}=u^2$. The limits $x=0,1$ become $u=1,\,e^2$:

\[L=\frac12\int_1^{e^2}\frac{\sqrt{1+u^2}}{u}\,du\]

Step 4: Standard antiderivative

\[\int\frac{\sqrt{1+u^2}}{u}\,du = \sqrt{1+u^2}-\ln\left|\frac{1+\sqrt{1+u^2}}{u}\right|+C\]

Step 5: Evaluate at the limits

At $u=e^2$: $\sqrt{1+e^4}-\ln\dfrac{1+\sqrt{1+e^4}}{e^2}$.

At $u=1$: $\sqrt2-\ln(1+\sqrt2)$.

Step 6: Combine

\[L=\frac12\left[\sqrt{1+e^4}-\sqrt2-\ln\frac{1+\sqrt{1+e^4}}{e^2}+\ln(1+\sqrt2)\right]\]

\[\boxed{L=\frac{\sqrt{1+e^4}-\sqrt2}{2}+\frac12\ln\frac{e^2(1+\sqrt2)}{1+\sqrt{1+e^4}}}\]

Step 7: Numerical value

With $e^4\approx54.598$: $\sqrt{1+e^4}\approx7.456$, giving $L\approx3.021+0.373\approx3.394$.