Question

The value of \( \int_0^1 e^x \, dx \) using the trapezoidal rule with four equal subintervals is 
 

• A. 1.718
• B. 1.727
• C. 2.192
• D. 2.718

Hint:

The trapezoidal rule is a simple method for numerical integration. Divide the interval into equal subintervals, calculate function values at those points, and apply the formula for an approximation.

Answer 1

The Correct Option is B

To apply the trapezoidal rule, we divide the interval \( [0, 1] \) into 4 equal subintervals. The trapezoidal rule for numerical integration is given by: \[ \int_a^b f(x) \, dx \approx \frac{b - a}{2n} \left[ f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right] \] Where:
- \( a = 0 \), \( b = 1 \), and \( n = 4 \), - \( f(x) = e^x \),
- The points are \( x_0 = 0, x_1 = 0.25, x_2 = 0.5, x_3 = 0.75, x_4 = 1 \).
Now, calculate the values of \( f(x) \):
- \( f(0) = e^0 = 1 \),
- \( f(0.25) = e^{0.25} \approx 1.284 \),
- \( f(0.5) = e^{0.5} \approx 1.649 \),
- \( f(0.75) = e^{0.75} \approx 2.117 \),
- \( f(1) = e^1 = 2.718 \).
Using the trapezoidal rule: \[ \int_0^1 e^x \, dx \approx \frac{1 - 0}{2 \times 4} \left[ 1 + 2(1.284 + 1.649 + 2.117) + 2.718 \right] \] \[ \approx \frac{1}{8} \left[ 1 + 2(5.05) + 2.718 \right] \] \[ \approx \frac{1}{8} \left[ 1 + 10.1 + 2.718 \right] = \frac{1}{8} \times 13.818 \approx 1.727 \] Thus, the value of the integral is approximately 1.727, which corresponds to option (B).
Final Answer: (B) 1.727