1. **Problem Statement:** Find the approximate area of the shaded region enclosed by the curve $y=3\ln x$ and the line $y=(x-1)\ln 4$ between $x=1$ and $x=4$ using the Trapezoidal rule with 3 sub-intervals. 2. **Set up the problem:** The area between two curves $y=f(x)$ and $y=g(x)$ from $a$ to $b$ is given by: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ Here, $f(x) = (x-1)\ln 4$ and $g(x) = 3\ln x$. 3. **Divide the interval:** From $x=1$ to $x=4$ with 3 sub-intervals means: $$\Delta x = \frac{4-1}{3} = 1$$ Sub-interval points: $x_0=1$, $x_1=2$, $x_2=3$, $x_3=4$ 4. **Calculate the function values at these points:** Calculate $h(x) = (x-1)\ln 4 - 3\ln x$ (difference of line minus curve): - $h(1) = (1-1)\ln 4 - 3\ln 1 = 0 - 0 = 0$ - $h(2) = (2-1)\ln 4 - 3\ln 2 = \ln 4 - 3\ln 2$ Since $\ln 4 = 2\ln 2$, $$h(2) = 2\ln 2 - 3\ln 2 = -\ln 2 \approx -0.6931$$ - $h(3) = (3-1)\ln 4 - 3\ln 3 = 2\ln 4 - 3\ln 3 = 2(2\ln 2) - 3\ln 3 = 4\ln 2 - 3\ln 3$ Calculate numerically: $4\ln 2 \approx 4 \times 0.6931 = 2.7726$ $3\ln 3 \approx 3 \times 1.0986 = 3.2958$ So, $$h(3) = 2.7726 - 3.2958 = -0.5232$$ - $h(4) = (4-1)\ln 4 - 3\ln 4 = 3\ln 4 - 3\ln 4 = 0$ 5. **Since $h(x)$ is negative at $x=2$ and $x=3$, the curve $3\ln x$ is above the line $(x-1)\ln 4$ there. The shaded area is the absolute value of $h(x)$ integrated.** 6. **Apply the Trapezoidal rule:** $$\text{Area} \approx \frac{\Delta x}{2} \left(|h(x_0)| + 2|h(x_1)| + 2|h(x_2)| + |h(x_3)|\right)$$ $$= \frac{1}{2} (0 + 2 \times 0.6931 + 2 \times 0.5232 + 0)$$ $$= 0.5 (1.3862 + 1.0464) = 0.5 \times 2.4326 = 1.2163$$ 7. **Round to three significant figures:** $$\boxed{1.22}$$ 8. **Part (b) Explanation:** The curve $y=3\ln x$ is concave down (since its second derivative is negative), and the line is straight. The Trapezoidal rule tends to underestimate the area when the function is concave down. Since we are integrating the absolute difference, the approximation is an **underestimate** of the true area.