Figure 1 shows a sketch of part of the curve with equation $y=x^{\frac{1}{2}} \ln 2 x$. The finite region $\mathrm{R}$, shown shaded in Figure 3, is bounded by the curve, the $x$ - axis and the lines $x=1$ and $x=4$.
2. Figure 1 shows a sketch of part of the curve with equation $y=x^{\frac{1}{2}} \ln 2 x$. The finite region $\mathrm{R}$, shown shaded in Figure 3, is bounded by the curve, the $\mathrm{x}$-axis and the lines $\mathrm{x}$ $=1$ and $x=4$.

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∻Solution by Steps∻ ‖ step 1 ⋮ We need to evaluate the integral of the function $π (x^{\frac{1}{2}} \ln(2x))^2$ from $x=1$ to $x=4$. ‖ ‖ step 2 ⋮ The integral is given by: $$ \int_{1}^{4} π (x^{\frac{1}{2}} \ln(2x))^2 \, dx $$ Using the result from the Asksia-LL calculator, we have: $$ \int_{1}^{4} π (x^{\frac{1}{2}} \ln(2x))^2 \, dx = \frac{1}{4} π (15 + (143 \ln(2) - 47) \ln(4)) \approx 68.529 $$ ‖ ‖ step 3 ⋮ The indefinite integral of the function is: $$ \int π (x^{\frac{1}{2}} \ln(2x))^2 \, dx = π \left( \frac{x^2}{4} + \frac{1}{2} x^2 \ln^2(2x) - \frac{1}{2} x^2 \ln(2x) \right) + C $$ where $C$ is the constant of integration. ‖ ‖ step 4 ⋮ Evaluating the definite integral from $x=1$ to $x=4$: $$ \left[ π \left( \frac{x^2}{4} + \frac{1}{2} x^2 \ln^2(2x) - \frac{1}{2} x^2 \ln(2x) \right) \right]_{1}^{4} $$ Substituting the limits, we get: $$ \frac{1}{4} π (15 + (143 \ln(2) - 47) \ln(4)) \approx 68.529 $$ ‖ ∻Answer∻ ⚹ The value of the integral is approximately 68.529. ⚹ ∻Key Concept∻ ⚹Definite Integral⚹ ∻Explanation∻ ⚹The definite integral of a function over a specified interval gives the area under the curve of the function between the given limits. In this case, we evaluated the integral of $π (x^{\frac{1}{2}} \ln(2x))^2$ from $x=1$ to $x=4$.⚹◊What is the equation of the curve in Figure 1?⍭ Generate me a similar question◊

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