Question

How do you do these two questions?

Answer 1

Step-by-step explanation:

(a) ∫₋ₒₒ°° f(x) dx

We can split this into three integrals:

= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx

Since the function is even (symmetrical about the y-axis), we can further simplify this as:

= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx

The first integral is finite, so it converges.

For the second integral, we can use comparison test.

g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.

We can show that g(x) converges:

∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).

Therefore, the smaller function f(x) also converges.

(b) The width of the intervals is:

Δx = (3 − -3) / 6 = 1

Evaluating the function at the beginning and end of each interval:

f(-3) = e^(-9/2)

f(-2) = e^(-2)

f(-1) = e^(-1/2)

f(0) = 1

f(1) = e^(-1/2)

f(2) = e^(-2)

f(3) = e^(-9/2)

Apply Simpson's rule:

S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]

S ≈ 2.5103