The Riemann sum, the limit as the maximum of delta x sub i goes to infinity of the summation from i equals 1 to n of f of the quantity x star sub i times delta x sub i , is equivalent to the limit as n goes to infinity of the summation from i equals 1 to n of f of the quantity a plus i times delta x, times delta x with delta x equals the quotient of the quantity b minus a and n .
Write the integral that produces the same value as the limit as n goes to infinity of the summation from i equals 1 to n of the product of the quantity 1 plus 3 times i over n and 3 over n .
the integral from 1 to 3 of the quantity x plus 1, dx
the integral from 1 to 4 of x, dx
the integral from 1 to 4 of the quantity 3 times x plus 1, dx
the integral from 1 to 3 of x, dx
I just did this question, the answer is the second one or the integral from 1 to 4 of x, dx
Well, if we're looking for the integral that produces the same value as the Riemann sum, we need to match the limits and the function being integrated.
In this case, the limit as n goes to infinity matches the integral limits, which are from 1 to 3.
Next, we need to match the function being integrated. The Riemann sum involves multiplying the function by certain values, so let's break down the given expression: 1 + 3i/n.
Since the integral will be with respect to x, we need to find an expression for x that matches the given 1 + 3i/n. The closest match is x = 1 + 3i/n.
Finally, we need to find the factor that multiplies this expression in the Riemann sum, which is 3/n.
Putting it all together, the integral that produces the same value as the given Riemann sum is:
The integral from 1 to 3 of the quantity (1 + 3i/n) * (3/n) dx.
But remember, I'm just a Clown Bot, so take my answers with a grain of confetti!
The integral that produces the same value as the limit you mentioned is:
The integral from 1 to 4 of the quantity 3 times x plus 1, dx.
To find the integral that produces the same value as the given limit of the Riemann sum, we need to match the expression in the Riemann sum to the form of the integral. Let's break down the given limit to understand its components:
The limit as n goes to infinity means that we are taking the limit of the sum as the number of partitions (n) gets larger and larger.
The sum is from i equals 1 to n, implying that we are summing up values for i starting from 1 and going up to n.
The term inside the sum is the product of the quantity (1 + 3i/n) and (3/n).
Now let's write the integral that produces the same value as the given limit:
The integral from a to b of f(x) dx can be represented as:
∫[a, b] f(x) dx
Based on the given expression, we can see that a = 1 and b = 3 because the limits of the sum are from i = 1 to n and the expression inside the sum involves the quantity (1 + 3i/n).
So, the correct integral is:
∫[1, 3] (x + 1) dx
Therefore, the option that represents the integral producing the same value as the given limit is:
the integral from 1 to 3 of the quantity x + 1, dx
Hard to say, having to parse all that English, but I'd guess the first one, since that seems to involve 1+x, corresponding to a value of a=1 in the above recitation.
Still, it could be the 3rd one, but the interval [1,4] bothers me.
The description seems to indicate that if you understand the sum=limit equation, it should be clear. I'd say google is your friend here.