Write the limit as n goes to infinity of the summation from k equals 1 of the product of the 10th power of the quantity 5 plus 2 times k over n and 2 over n as a definite integral.
∫[0..2] (5+x)^10 dx
Well, well, well, it seems like we're diving into the world of limits and integrals, huh? Don't worry, I'm here to put a smile on your face!
Let's break this down step by step, shall we?
First, we want to express the summation as a definite integral. So, let's rewrite the sum using a Riemann sum approach:
∑(10th power of (5 + 2k/n) * (2/n)), as n goes to infinity.
Now, let's make a few observations. As n approaches infinity, the differences between the terms inside the sum become infinitesimally small. In other words, we can think of each term as a small chunk, or width, of a continuous function. And what do we get when we add up an infinite number of tiny chunks? That's right, an integral!
So, our definite integral will look like this:
∫(10th power of (5 + 2x) * 2) dx, where x goes from 0 to 1.
We've changed k/n to x because x takes on values between 0 and 1. And remember, when we switch from a summation to an integral, we replace the "k" with "x" and drop the division by "n."
Voilà! We have transformed the summation into the definite integral. Now, go ahead and evaluate that integral to find your answer. Keep in mind that since we're dealing with a definite integral, the answer will be a specific value and not just a general expression.
Now, wasn't that an entertaining way to approach limits and integrals? I sure think so! If you have any more questions, feel free to ask, and I'll keep the jokes coming!
To write the given limit as a definite integral, we can make use of the definition of a Riemann sum.
Step 1: Let's start by finding the width of each subinterval. Since we are summing from k = 1 to n, there are n subintervals in total. Therefore, the width of each subinterval is (1/n).
Step 2: Next, we need to determine the function that represents the values we are summing. In this case, it is the product of the 10th power of the quantity (5 + 2k/n) and 2/n. So, the function f(k/n) is given by:
f(k/n) = (5 + 2k/n)^10 * 2/n
Step 3: Now we can express the given limit as a definite integral:
lim(n->∞) ∑[k=1]^{n} [(5 + 2k/n)^10 * 2/n]
can be written as
lim(n->∞) [1/n] * ∑[k=1]^{n} [(5 + 2k/n)^10 * 2]
Step 4: To convert the sum into an integral, we can rewrite it in terms of integration over the interval [0,1]. We do this by making the substitution k/n = x, where x ranges from 0 to 1.
Using the substitutions k = nx and dk = ndx, the integral becomes:
lim(n->∞) [1/n] * ∑[k=1]^{n} [(5 + 2k/n)^10 * 2]
= lim(n->∞) [1/n] * ∑[x=0]^{1} [(5 + 2nx)^10 * 2n]
= lim(n->∞) 2 * ∑[x=0]^{1} [(5 + 2nx)^10]
≈ ∫[0]^{1} (5 + 2x)^10 dx
So, the limit as n approaches infinity of the given summation can be approximated by the definite integral:
∫[0]^{1} (5 + 2x)^10 dx
To convert the given limit of a summation into a definite integral, we need to rewrite the expression in terms of the integration variable.
Looking at the given expression, we have the sum from k equals 1 to n. Let's rewrite it as a sum from k equals 0 to n-1 for easier manipulation.
Rewriting the expression with the sum from k equals 0 to n-1:
∑(k=0 to n-1) [(5 + 2k/n)^10 * (2/n)]
Now, let's consider the width of each subinterval within the summation. In this case, the width is Δx = 1/n since we have n subintervals from k = 0 to n-1.
To convert the summation into a definite integral, we consider the following steps:
1. Recognize that this is a Riemann sum, where the summand represents the height of each rectangle.
2. Express the summation in terms of the integration variable by replacing k/n with the variable x.
Let x = k/n, then k = nx.
The limits of the summation will change accordingly:
When k = 0, x = 0, and when k = n-1, x = (n-1)/n = 1 - 1/n.
Therefore, the limits of the integral will be from x = 0 to x = 1 - 1/n.
Now, let's rewrite the expression with the integration variable x:
∑(k=0 to n-1) [(5 + 2k/n)^10 * (2/n)] = ∑(x=0 to 1 - 1/n) [(5 + 2nx)^10 * (2/n)]
Taking the limit as n goes to infinity, we can convert the summation into a definite integral.
Now, let's express the limit of the summation as a definite integral using the variable x:
lim(n→∞) ∑(k=0 to n-1) [(5 + 2k/n)^10 * (2/n)] = ∫(0 to 1) [(5 + 2x)^10 * 2] dx
Therefore, the limit of the given summation as n goes to infinity can be expressed as the definite integral from 0 to 1 of [(5 + 2x)^10 * 2] dx.