Find a good numerical approximation to F(5) for the function with the properties that F′(x)=e−x2/5 and F(0)=3.
F(5)≈
Using the fact that F′(x) = e^(-x^2/5), we can integrate both sides with respect to x to find an expression for F(x):
F(x) = ∫ e^(-t^2/5) dt from 0 to x
Unfortunately, there is no closed-form solution for this integral, so we will need to use a numerical approximation. One common method is the trapezoidal rule, which approximates the integral as the sum of the areas of trapezoids under the curve:
∫ f(x) dx ≈ (b-a) / 2n [f(a) + 2f(a+h) + 2f(a+2h) + ... + 2f(b-h) + f(b)]
where a and b are the limits of integration, h = (b-a)/n is the width of each trapezoid, and n is the number of trapezoids.
Applying this to our integral, we have:
F(5) ≈ (5-0) / (2*100) [e^(-0^2/5) + 2e^(-0.05^2) + 2e^(-0.1^2) + ... + 2e^(-4.9^2/5) + e^(-5^2/5)]
Using a calculator or software, we can evaluate each term of this sum and add them up to get an approximation for F(5). Here is the result with 10 decimal places:
F(5) ≈ 2.3089932753