Taking number of intervals n = 3, the value of the integral
\[
\int_0^{0.3} e^{-x^2}\, dx
\]
using Trapezoidal method, is ______ (rounded off to two decimal places).
Show Hint
Use h = 0.1, evaluate e^(-x^2) at x = 0, 0.1, 0.2, 0.3, and apply the trapezoidal rule I is about (h/2)[f0 + 2f1 + 2f2 + f3].
Step 1: Split the interval into three equal strips.
We are told to use $n=3$ over $[0, 0.3]$, so each strip has width $h = 0.1$. This gives three trapezoids: one over $[0, 0.1]$, one over $[0.1, 0.2]$, and one over $[0.2, 0.3]$.
Step 2: Get the function values needed.
$f(x) = e^{-x^2}$, so
$$ f(0) = 1, \quad f(0.1) = e^{-0.01} \approx 0.99005, \quad f(0.2) = e^{-0.04} \approx 0.96079, \quad f(0.3) = e^{-0.09} \approx 0.91393 $$
Step 3: Find the area of each trapezoid separately.
Each trapezoid area is $\dfrac{h}{2}(\text{left height} + \text{right height})$.
First strip, $[0, 0.1]$:
$$ A_1 = \frac{0.1}{2}(1 + 0.99005) = 0.05 \times 1.99005 = 0.099503 $$
Second strip, $[0.1, 0.2]$:
$$ A_2 = \frac{0.1}{2}(0.99005 + 0.96079) = 0.05 \times 1.95084 = 0.097542 $$
Third strip, $[0.2, 0.3]$:
$$ A_3 = \frac{0.1}{2}(0.96079 + 0.91393) = 0.05 \times 1.87472 = 0.093736 $$
Step 4: Add the three strip areas.
$$ I \approx A_1 + A_2 + A_3 = 0.099503 + 0.097542 + 0.093736 = 0.290781 $$
This matches the composite formula result, since adding strip by strip is the same sum written a different way.
Step 5: Round to two decimals.
$$ I \approx 0.29 $$
$$ \boxed{0.29} $$
