To approximate the integral of $e^{-x}$ from $-1$ to $2$ using three equal subintervals, we can apply the Trapezoidal Rule or Simpson's Rule. For simplicity, I will use the Trapezoidal Rule in this case.

Steps for the Trapezoidal Rule:

1. Determine the subintervals: The interval is from $-1$ to $2$, so the total width of the interval is: $b - a = 2 - (-1) = 3$ We divide this interval into 3 equal subintervals: $\Delta x = \frac{b - a}{n} = \frac{3}{3} = 1$ Therefore, the subintervals will be: [ [-1, 0], [0, 1], [1, 2] ]

2. Evaluate the function $f(x) = e^{-x}$ at the endpoints:

   • $f(-1) = e^{-(-1)} = e^1 = 2.7183$
   • $f(0) = e^{0} = 1$
   • $f(1) = e^{-1} = 0.3679$
   • $f(2) = e^{-2} = 0.1353$

3. Apply the Trapezoidal Rule: The Trapezoidal Rule formula is: $\int_a^b f(x) \, dx \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3) \right]$ Substituting the values we calculated: $\int_{-1}^{2} e^{-x} \, dx \approx \frac{1}{2} \left[ 2.7183 + 2(1) + 2(0.3679) + 0.1353 \right]$ Simplifying: $\frac{1}{2} \left[ 2.7183 + 2 + 0.7358 + 0.1353 \right] = \frac{1}{2} \times 5.5894 = 2.7947$

Approximation of the integral:

$\int_{-1}^{2} e^{-x} \, dx \approx 2.7947$

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How does Simpson's Rule compare to the Trapezoidal Rule for approximating integrals?
2. What happens to the approximation if more subintervals are used?
3. Can you derive the formula for the Trapezoidal Rule?
4. How would you use numerical methods like the Midpoint Rule for this integral?
5. How can you approximate improper integrals using the Trapezoidal Rule?

Tip: Increasing the number of subintervals in numerical integration methods often improves the accuracy of the approximation.