Find a good numerical approximation to F(5) for the function with the properties that F′(x)=e^(−x^2/5) and F(0)=3.
To approximate F(5), we can use the numerical integration method called the trapezoidal rule. This method approximates the area under the curve of a function by dividing it into trapezoids and adding up their areas.
We can first use integration by substitution to find an antiderivative of the given function F'(x) = e^(-x^2/5). Let u = -x^2/5, then du/dx = -2x/5 and dx = -5/2udu. We can rewrite the integral as follows:
∫ e^(-x^2/5) dx = -∫ e^u (5/2) du = -5/2 ∫ e^u du = -5/2 e^u + C, where C is the constant of integration.
Since F(0) = 3, we have C = F(0) + 5/2 = 3 + 5/2 = 13/2.
Therefore, F(x) = -5/2 e^(-x^2/5) + 13/2.
To use the trapezoidal rule to approximate F(5), we need to divide the interval [0, 5] into n subintervals of equal width h = (5-0)/n. Then we can approximate the area under the curve by adding up the areas of the trapezoids formed by adjacent subintervals, as follows:
T_n = h/2 [f(0) + 2f(h) + 2f(2h) + ... + 2f((n-1)h) + f(nh)], where f(x) = e^(-x^2/5).
Using n = 10, we get h = 1/2 and the following values for f(x):
f(0) = e^(0) = 1,
f(1/2) = e^(-1/20) ≈ 0.975,
f(1) = e^(-1/5) ≈ 0.904,
f(3/2) = e^(-9/20) ≈ 0.824,
f(2) = e^(-1/2) ≈ 0.607,
f(5/2) = e^(-25/20) ≈ 0.449,
f(3) = e^(-9/5) ≈ 0.222,
f(7/2) = e^(-49/20) ≈ 0.110,
f(4) = e^(-4/1) ≈ 0.0183, and
f(9/2) = e^(-225/20) ≈ 0.00226.
Substituting these values into the trapezoidal rule formula, we get:
T_10 = (1/2)(1 + 2(0.975) + 2(0.904) + 2(0.824) + 2(0.607) + 0.449 + 0.222 + 0.110 + 0.0183 + 0.00226) ≈ 3.054.
Therefore, a good numerical approximation to F(5) is T_10 ≈ 3.054.