Write the limit as n goes to infinity of the summation from k equals 1 of the product of the 4th power of the quantity negative 1 plus 3 times k over n and 3 over n as a definite integral.
To rewrite the given limit as a definite integral, we first need to express the summation as a Riemann sum.
The Riemann sum for a definite integral is defined as ∑(f(xi)Δx), where f(xi) represents the function evaluated at xi, and Δx represents the width of each subinterval.
In this case, we have the following summation:
∑[(−1 + (3k/n))^4 * (3/n)]
To express this as a Riemann sum, we need to express the terms in terms of xi and Δx.
Let's consider the interval [a, b] which is divided into n subintervals. The width of each subinterval is given by Δx = (b - a)/n.
Now, let xi represent the midpoint of each subinterval. Then, xi can be given by:
xi = (a + (i - 1/2)Δx) = (a + (i - 1/2)(b - a)/n)
Next, we need to express the function evaluated at xi, f(xi), in terms of k and n. Let's rewrite the given function using xi instead of k:
f(xi) = [−1 + 3((a + (i - 1/2)(b - a)/n)/n)]^4 * (3/n)
Now, we can rewrite the given summation as a Riemann sum:
∑[(−1 + (3k/n))^4 * (3/n)] = ∑[f(xi)Δx]
where f(xi) = [−1 + 3((a + (i - 1/2)(b - a)/n)/n)]^4 * (3/n) and Δx = (b - a)/n.
Finally, we can write the given limit as a definite integral:
lim(n→∞) ∑[(−1 + (3k/n))^4 * (3/n)] = ∫[a, b] [−1 + 3((a + (x - 1/2)(b - a)/n)/n)]^4 * (3/n) dx
Keep in mind that this is an expression in terms of a general interval [a, b]. If you have specific values for a and b, you can substitute them into the integral.
To find the limit as n goes to infinity of the given summation, we can re-write it as a definite integral.
First, let's express the summation in terms of x instead of k:
x = k/n
Notice that as n approaches infinity, x approaches 0.
Now, let's express the product inside the summation:
P = (-1 + 3x)^4 * (3/n)
Substituting k/n with x, we have:
P = (-1 + 3x)^4 * (3/n)
Next, we'll find the limits of the sum:
As n goes to infinity, x ranges from 0 to 1.
Now, let's convert the summation into a definite integral using the limit of integration 0 to 1:
∫[0,1] (-1 + 3x)^4 * (3/n) dx
Since we are taking the limit as n goes to infinity, we can treat the (3/n) as a constant and move it outside the integral:
(3/n) ∫[0,1] (-1 + 3x)^4 dx
Now, let's simplify the integral:
(3/n) ∫[0,1] (81x^4 - 108x^3 + 54x^2 - 12x + 1) dx
Using the power rule for integration, we can integrate each term of the polynomial:
(3/n) * (81/5)x^5 - (108/4)x^4 + (54/3)x^3 - (12/2)x^2 + x | from 0 to 1
Now, let's evaluate the integral at the limits of integration:
(3/n) * [(81/5)(1^5) - (108/4)(1^4) + (54/3)(1^3) - (12/2)(1^2) + 1] - (3/n) * [(81/5)(0^5) - (108/4)(0^4) + (54/3)(0^3) - (12/2)(0^2) + 0]
Simplifying further:
(3/n) * [(81/5) - (108/4) + (54/3) - (12/2) + 1] - 0
This expression simplifies to:
(3/n) * [81/5 - 27] = (243/n) * (4/5)
As n approaches infinity, the expression (243/n) approaches 0. Therefore, the final limit is:
lim (243/n) * (4/5) = 0
Therefore, the limit as n goes to infinity of the given summation can be expressed as a definite integral as:
∫[0,1] (-1 + 3x)^4 dx, which evaluates to 0.
so, it appears you have the sum
n
∑ (-1+3/n k)(3/n)
k=1
It appears we have divided the interval of width 3 into n parts. That would make the limit
∫[-1,2] x dx