Question

For a certain continuous function f, the right Riemann sum approximation of\int_{0}^{2} f(x) dx with n subintervals of equal length is\frac{2(n+1)(3n+2)}{n^2}for all n. What is the value of\int_{0}^{2} f(x) dx?

A. 2
B. 6
C. 12
D. 20

Answer 1

The value of the integral ∫₀² f(x) dx is 6.

Step-by-step explanation:

The right Riemann sum approximation of the integral ∫02 f(x) dx with n subintervals of equal length is given by the formula 2(n+1)(3n+2)/n²for all n.

To find the value of the integral, we need to take the limit of the right Riemann sum as n approaches infinity, since the right Riemann sum is an approximation.

Taking the limit, we have:

limn→∞ 2(n+1)(3n+2)/n² = 2 × 3 = 6

Therefore, the value of the integral ∫02 f(x) dx is 6.

Answer 2

The value of the integral ∫02 f(x) dx with the given right Riemann sum approximation is 12.

Step-by-step explanation:

The right Riemann sum approximation of an integral is given by the sum of the areas of rectangles, where the height of each rectangle is the value of the function at the right endpoint of each subinterval. In this case, the right Riemann sum approximation of ∫02 f(x) dx with n subintervals of equal length is ∫02 f(x) dx ∼ ∫02 (2(n+1)(3n+2)/n^2) dx. To find the value of the integral, we need to evaluate this expression.

First, let's rewrite the expression as a sum:

∫02 (2(n+1)(3n+2)/n^2) dx = 2∫02 [(3n+2)(n+1)/n^2] dx.

Next, we can evaluate the integral by integrating each term:

2∫02 [(3n+2)(n+1)/n^2] dx = 2[(3n+2)/n^2 * ∫02 (n+1) dx + 2/n^2 * ∫02 (3n+2) dx].

Finally, we can substitute the limits of integration:

2[(3n+2)/n^2 * (n+1)(2-0) + 2/n^2 * (3n+2)(2-0)] = 2[(3n+2)(n+1)/n^2 * 2 + (3n+2)/n^2 * 2].

Simplifying the expression gives:

2[(3n+2)(n+1)/n^2 * 2 + (3n+2)/n^2 * 2] = 2[(6n^2 + 10n + 4)/n^2] = 12.

Therefore, the value of the integral ∫02 f(x) dx is 12.