How do I prove that integral e^((x^2020)/n))dx where the lim n tends to infinity and the boundaries of the integral is from 0 to 2019
How do I do this??
They want me to show that it is equal to 2019
But I don't know how?
as n→∞, (x^2020)/n → 0
lim ∫ = ∫ lim
so, as n→∞ you have ∫[0,2019] e^0 dx = x[0,2019] = 2019
Thank you sir
To prove the value of the integral
∫ e^((x^2020)/n))dx
where the limit of n tends to infinity and the boundaries of the integral are from 0 to 2019, we will follow these steps:
Step 1: Begin by substituting u = x/n in the integral. As n approaches infinity, we can define a new variable u that tends to zero.
∫ e^((x^2020)/n))dx = ∫ e^(u^2020) n du
Step 2: Now, consider the boundaries of the integral. When x = 0, the value of u also becomes zero, and when x = 2019, u = 2019/n. As n tends to infinity, the value of u at the upper limit becomes zero.
Boundaries: 0 to 2019/n
Step 3: Now, we have the following integral:
∫ e^(u^2020) n du
Step 4: To evaluate this integral, we need to know the antiderivative of e^(u^2020). Unfortunately, there is no known closed-form antiderivative for this function.
Step 5: However, we can apply a numerical method to approximate the value of the integral. One common method is numerical integration, such as using the midpoint rule or Simpson's rule.
Step 6: Using numerical methods, you can divide the interval [0, 2019/n] into smaller subintervals and approximate the integral using the values of the function at certain points within each subinterval.
Step 7: As the limit of n approaches infinity, the numerical approximation becomes more accurate, allowing you to estimate the value of the integral with increasing precision.
In conclusion, while we can't calculate the exact value of the integral analytically, you can approximate it using numerical methods as n tends to infinity.