Find the volume of the solid of revolution formed by rotating the region bounded by \( y=2 x+1, y=0, x=1 \) and \( x=2 \) is rotated about the \( x \)-axis. A. None of the optians B. \( \pi\left(\frac{11}{3}+7\right) \) units \( ^{3} \) C. \( \pi\left(\frac{23}{3}+7\right) \) units \( ^{3} \) D. \( \pi\left(\frac{21}{2}-1\right) \) units \( ^{3} \) E. \( -\pi\left(\frac{23}{3}-7\right) \) units \( ^{3} \)

To find the volume of the solid of revolution formed by rotating the region bounded by the lines \( y = 2x + 1 \), \( y = 0 \), \( x = 1 \), and \( x = 2 \) about the \( x \)-axis, we can use the disk method. The volume \( V \) of the solid of revolution is given by the integral: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where \( f(x) \) is the function being rotated, and \( [a, b] \) are the bounds of integration. In this case, the function is \( f(x) = 2x + 1 \), and the bounds are \( x = 1 \) and \( x = 2 \). Now we can set up the integral: \[ V = \pi \int_{1}^{2} (2x + 1)^2 \, dx \] First, we need to expand \( (2x + 1)^2 \): \[ (2x + 1)^2 = 4x^2 + 4x + 1 \] Now we can substitute this back into the integral: \[ V = \pi \int_{1}^{2} (4x^2 + 4x + 1) \, dx \] Next, we can compute the integral: \[ V = \pi \left[ \frac{4}{3}x^3 + 2x^2 + x \right]_{1}^{2} \] Now we evaluate this expression at the bounds: 1. Evaluate at \( x = 2 \): \[ \frac{4}{3}(2^3) + 2(2^2) + 2 = \frac{4}{3}(8) + 2(4) + 2 = \frac{32}{3} + 8 + 2 = \frac{32}{3} + \frac{24}{3} + \frac{6}{3} = \frac{62}{3} \] 2. Evaluate at \( x = 1 \): \[ \frac{4}{3}(1^3) + 2(1^2) + 1 = \frac{4}{3} + 2 + 1 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \] Now we subtract the two results: \[ V = \pi \left( \frac{62}{3} - \frac{10}{3} \right) = \pi \left( \frac{52}{3} \right) \] Thus, the volume of the solid of revolution is: \[ V = \frac{52\pi}{3} \text{ cubic units} \] Now, we need to check which option matches this result. None of the options provided directly match \( \frac{52\pi}{3} \). Thus, the answer is: **A. None of the options**.