Answer is: option1
\(\frac{1}{4} \left[1 + 2\sqrt{e} + 2e + 2e\sqrt{e} + e^2\right]\)Solution:
Function: \( f(x) = e^x \)
Interval: \([0, 2]\)
Subdivisions: \( n = 4 \)
So, \( h = \frac{2 - 0}{4} = 0.5 \)
Points:
- \( x_0 = 0 \)
- \( x_1 = 0.5 \)
- \( x_2 = 1 \)
- \( x_3 = 1.5 \)
- \( x_4 = 2 \)
We approximate the integral as the sum of areas of 4 trapezoids:
\[ \int_0^2 e^x dx \approx \sum_{i=1}^4 \frac{h}{2} \left[ f(x_{i-1}) + f(x_i) \right] \]
With \( h = 0.5 \), we get:
\[ \frac{1}{2}(0.5)[e^0 + e^{0.5}] + \frac{1}{2}(0.5)[e^{0.5} + e^1] + \frac{1}{2}(0.5)[e^1 + e^{1.5}] + \frac{1}{2}(0.5)[e^{1.5} + e^2] \]
Factor out \( \frac{1}{4} \):
\[ \frac{1}{4}\left[1 + 2\sqrt{e} + 2e + 2e\sqrt{e} + e^2 \right] \]
Option A