Answer is: option4
\( \left(e - \dfrac{6}{5} \right)\pi \)Solution:
The volume when rotating about the x-axis is:
\[ V = \pi \int_0^1 \left[ \left(e^{x/2}\right)^2 - \left((x - 1)^2\right)^2 \right] dx \]
\[ V = \pi \int_0^1 \left[ e^x - (x - 1)^4 \right] dx \]
First term:
\[ \int_0^1 e^x \, dx = e^1 - e^0 = e - 1 \]
Second term:
Let \( u = x - 1 \)
- \( du = dx \)
- When \( x = 0 \), \( u = -1 \)
- When \( x = 1 \), \( u = 0 \)
\[ \int_0^1 (x - 1)^4 dx = \int_{-1}^0 u^4 \, du \]
\[ \int_{-1}^0 u^4 \, du = \left[ \frac{u^5}{5} \right]_{-1}^0 = \frac{0^5}{5} - \frac{(-1)^5}{5} = 0 - \left(-\frac{1}{5}\right) = \frac{1}{5} \]
Final Volume:
\[ V = \pi \left[ (e - 1) - \frac{1}{5} \right] = \pi \left( e - \frac{6}{5} \right) \]
Option D