Let's break down the problem step by step: ### **Step 1: Write the Area Formula** We want the area **between** \( f(x) = \ln(x) \) and \( g(x) = -\frac{2x}{e} + 3 \) over the interval \([1, 7]\): \[ A = \int_{1}^{7} \left[ f(x) - g(x) \right] \, dx \] \[ A = \int_{1}^{7} \left[ \ln(x) - \left(-\frac{2x}{e} + 3\right) \right] dx \] \[ A = \int_{1}^{7} \left[ \ln(x) + \frac{2x}{e} - 3 \right] dx \] --- ### **Step 2: Integrate Each Term Separately** **a.** \(\int \ln(x) dx = x\ln(x) - x\) **b.** \(\int \frac{2x}{e} dx = \frac{2}{e} \cdot \int x dx = \frac{2}{e} \cdot \frac{x^2}{2} = \frac{x^2}{e}\) **c.** \(\int 3 dx = 3x\) So, the antiderivative is: \[ F(x) = x\ln(x) - x + \frac{x^2}{e} - 3x \] --- ### **Step 3: Evaluate the Definite Integral** \[ A = F(7) - F(1) \] Calculate each term: #### **At \( x = 7 \):** - \( 7\ln(7) \) - \( -7 \) - \( \frac{49}{e} \) - \( -21 \) So, \[ F(7) = 7\ln(7) - 7 + \frac{49}{e} - 21 = 7\ln(7) + \frac{49}{e} - 28 \] #### **At \( x = 1 \):** - \( 1\ln(1) = 0 \) - \( -1 \) - \( \frac{1}{e} \) - \( -3 \) So, \[ F(1) = 0 - 1 + \frac{1}{e} - 3 = \frac{1}{e} - 4 \] --- ### **Step 4: Subtract** \[ A = F(7) - F(1) = \left[ 7\ln(7) + \frac{49}{e} - 28 \right] - \left[ \frac{1}{e} - 4 \right] \] \[ A = 7\ln(7) + \frac{49}{e} - 28 - \frac{1}{e} + 4 \] \[ A = 7\ln(7) + \frac{48}{e} - 24 \] --- ### **Final Answer** \[ \boxed{A = 7\ln(7) + \frac{48}{e} - 24 \text{ units}^2} \]

Let's carefully re-evaluate the problem step-by-step to ensure accuracy. --- ### **Given:** - \( f(x) = \ln(x) \) - \( g(x) = -\frac{2x}{e} + 3 \) - Interval: \([1, 7]\) --- ### **Step 1: Write the expression for the area \(A\)** Since we want the region **bounded between** the two curves, the area \(A\) is: \[ A = \int_{x=1}^{x=7} \left[ f(x) - g(x) \right] dx \] \[ A = \int_{1}^{7} \left[ \ln(x) - \left( -\frac{2x}{e} + 3 \right) \right] dx \] Simplify the integrand: \[ A = \int_{1}^{7} \left[ \ln(x) + \frac{2x}{e} - 3 \right] dx \] --- ### **Step 2: Find the antiderivative** Break the integral into parts: \[ A = \int_{1}^{7} \ln(x) dx + \int_{1}^{7} \frac{2x}{e} dx - \int_{1}^{7} 3 dx \] **a.** \(\int \ln(x) dx\): Using integration by parts: Let: - \( u = \ln(x) \Rightarrow du = \frac{1}{x} dx \) - \( dv = dx \Rightarrow v = x \) Then: \[ \int \ln(x) dx = x \ln(x) - \int x \cdot \frac{1}{x} dx = x \ln(x) - \int 1 dx = x \ln(x) - x \] **b.** \(\int \frac{2x}{e} dx\): Constants can be factored out: \[ \frac{2}{e} \int x dx = \frac{2}{e} \cdot \frac{x^2}{2} = \frac{x^2}{e} \] **c.** \(\int 3 dx = 3x\) --- ### **Step 3: Write the combined antiderivative** \[ F(x) = x \ln(x) - x + \frac{x^2}{e} - 3x \] --- ### **Step 4: Evaluate \(F(x)\) at the bounds** Compute \(F(7)\): \[ F(7) = 7 \ln(7) - 7 + \frac{49}{e} - 21 \] Compute \(F(1)\): \[ F(1) = 1 \cdot 0 - 1 + \frac{1}{e} - 3 = -1 + \frac{1}{e} - 3 = -4 + \frac{1}{e} \] --- ### **Step 5: Find the area \(A = F(7) - F(1)\)** \[ A = \left( 7 \ln(7) - 7 + \frac{49}{e} - 21 \right) - \left( -4 + \frac{1}{e} \right) \] Simplify: \[ A = 7 \ln(7) - 7 + \frac{49}{e} - 21 + 4 - \frac{1}{e} \] Combine like terms: - Constants: \(-7 - 21 + 4 = -24\) - \( \frac{49}{e} - \frac{1}{e} = \frac{48}{e} \) Thus: \[ A = 7 \ln(7) - 24 + \frac{48}{e} \] --- ### **### **Final answer:** \[ \boxed{ A = 7 \ln(7) + \frac{48}{e} - 24 } \] This matches the previous calculation but confirms the correctness. --- ### **Summary:** - The area between the given curves over \([1,7]\) is: \[ \boxed{ A = 7 \ln(7) + \frac{48}{e} - 24 } \] **This is the exact form of the area.**