Certainly! Let's go through each step clearly and carefully. --- ## **Step 1:** **Set up the Integral for the Area** Given: - \( f(x) = \left(\frac{c-2}{4}\right)x + 2 \) - \( g(x) = 2 - x \) - Limits: \( x = 0 \) to \( x = 4 \) The area **between** the two curves from \( x = 0 \) to \( x = 4 \) is: \[ A = \int_{0}^{4} \left[ g(x) - f(x) \right] dx \] Substitute in the given functions: \[ A = \int_{0}^{4} \left[ (2 - x) - \left(\frac{c-2}{4}x + 2\right) \right] dx \] Simplify inside the integral: \[ A = \int_{0}^{4} \left[ 2 - x - \frac{c-2}{4}x - 2 \right] dx \] \[ A = \int_{0}^{4} \left[ - x - \frac{c-2}{4}x \right] dx \] \[ A = \int_{0}^{4} \left[ -x \left(1 + \frac{c-2}{4}\right) \right] dx \] \[ A = -\int_{0}^{4} x \left(1 + \frac{c-2}{4}\right) dx \] Factor: \[ 1 + \frac{c-2}{4} = \frac{4 + c - 2}{4} = \frac{c + 2}{4} \] So, \[ A = -\int_{0}^{4} x \cdot \frac{c + 2}{4} dx \] \[ A = -\frac{c + 2}{4} \int_{0}^{4} x \, dx \] --- ## **Step 2:** **Integrate** We know: \[ \int_{0}^{4} x \, dx = \left. \frac{x^2}{2} \right|_{0}^{4} = \frac{16}{2} - 0 = 8 \] So, \[ A = -\frac{c + 2}{4} \cdot 8 = -2(c + 2) \] --- ## **Step 3:** **Check obvious value of \( c \) for area 0** Set \( A = 0 \): \[ -2(c + 2) = 0 \implies c + 2 = 0 \implies c = -2 \] Does this make sense? If \( c = -2 \), the line \( f(x) \) goes from \( (0,2) \) to \( (4,-2) \). That is, it coincides with \( g(x) = 2 - x \). So, the area between them is zero, as expected. --- ## **Step 4:** **Find \( c \) for area = 10** Set \( A = 10 \): \[ -2(c + 2) = 10 \] \[ c + 2 = -5 \] \[ c = -7 \] --- ## **Step 5:** **Check: Plug \( c = -7 \) into \( f(x) \) and recompute area** First, write \( f(x) \) with \( c = -7 \): \[ f(x) = \left(\frac{-7 - 2}{4}\right)x + 2 = \left(\frac{-9}{4}\right)x + 2 \] Compute the area: \[ A = \int_{0}^{4} \left[ (2 - x) - \left(-\frac{9}{4}x + 2\right) \right] dx \] \[ = \int_{0}^{4} \left[ 2 - x + \frac{9}{4}x - 2 \right] dx \] \[ = \int_{0}^{4} \left( \left(\frac{9}{4} - 1\right)x \right) dx \] \[ = \int_{0}^{4} \left( \frac{5}{4}x \right) dx \] \[ = \frac{5}{4} \int_{0}^{4} x dx = \frac{5}{4} \cdot 8 = 10 \] **Checks out!** --- ## **Bonus: Geometric Explanation** The region bounded between the two lines from \( x = 0 \) to \( x = 4 \) always forms a triangle. The base is \( 4 \) units long (from \( x = 0 \) to \( x = 4 \)), and the height is the vertical distance at \( x = 4 \) between the two lines: - \( g(4) = 2 - 4 = -2 \) - \( f(4) = c \) So, the height is \( |c - (-2)| = |c + 2| \). Area of triangle: \[ A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 4 \cdot |c + 2| = 2|c + 2| \] Set area to 10: \[ 2|c + 2| = 10 \implies |c + 2| = 5 \implies c = -7 \text{ or } c = 3 \] But from the calculus approach, we only get \( c = -7 \) because we set up the area with a negative sign (the order of subtraction). If you want the area above \( g(x) \), you'd use \( c = -7 \). --- ## **Final Answers:** **Step 1:** \[ A = -\frac{c + 2}{4} \int_{0}^{4} x \, dx \] **Step 2:** \[ A = -2(c + 2) \] **Step 3:** Obvious value for area zero: \( c = -2 \) (lines coincide). **Step 4:** For area = 10: \( c = -7 \) **Step 5:** With \( c = -7 \), \( f(x) = -\frac{9}{4}x + 2 \), and the area computes to \( 10 \). **Bonus:** The region is a triangle with base 4 and height \( |c + 2| \), so area is \( 2|c + 2| \). Set \( 2|c + 2| = 10 \implies |c + 2| = 5 \implies c = -7 \) or \( c = 3 \). --- If you need screenshots for the Symbolab steps, let me know!