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 . (4 points)
Question 1 options:
1)
the integral from 1 to 3 of the quantity x plus 1, dx
2)
the integral from 1 to 4 of x, dx
3)
the integral from 1 to 4 of the quantity 3 times x plus 1, dx
4)
the integral from 1 to 3 of x, dx
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 squared of 3 plus 5 times i over n and 5 over n.
the integral from 3 to 8 of x squared, dx
the integral from 0 to 3 of x plus 3 quantity squared, dx
the integral from 0 to 5 of the quantity 3 plus 5 times x, dx
the integral from 3 to 5 of x squared, dx
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 is:
3 times the integral from 1 to 4 of the quantity x plus 1, dx.
To find the integral that produces the same value as the given limit, we need to rewrite the given limit in integral form and then evaluate it.
The given limit is:
lim(n→∞) Σ(i=1 to n) [(1 + 3i/n) * 3/n]
To convert this into integral form, we rewrite it as:
lim(n→∞) Σ(i=1 to n) [3(1 + 3i/n)/n]
= lim(n→∞) (3/n) Σ(i=1 to n) (1 + 3i/n)
Now, let's focus on the expression inside the summation:
1 + 3i/n
We need to express this in terms of x, where x will be the variable of integration.
Since x ranges from 1 to 4 (as given in the options), we map i to x using the relationship:
x = a + iΔx
Here, a = 1 and Δx = (b - a)/n = (4 - 1)/n = 3/n
Substituting these values and simplifying, we get:
x = 1 + i(3/n)
=> i = (x - 1)(n/3)
Now, let's substitute this expression for i back into the summation:
lim(n→∞) (3/n) Σ(i=1 to n) (1 + 3i/n)
= lim(n→∞) (3/n) Σ(x=1 to 4) [(1 + 3(x - 1)(n/3))/n]
Simplifying further, we have:
= lim(n→∞) (3/n) Σ(x=1 to 4) [(1 + (x - 1)(n/3))/n]
= lim(n→∞) (1/n) Σ(x=1 to 4) [1 + (x - 1)(n/3)]
As n approaches infinity, the term (1/n) approaches 0, so we can remove it. The summation becomes a definite integral:
= Σ(x=1 to 4) [1 + (x - 1)(0/3)]
= Σ(x=1 to 4) 1
= ∫(1 to 4) 1 dx
Therefore, the equivalent integral is:
the integral from 1 to 4 of 1, dx
Considering the available options, the correct answer is:
4) the integral from 1 to 3 of x, dx
All those words! It appears you have
∑ (1 + 3i/n)(3/n)
This seems to be a right-hand sum on an interval of width 3, since it is
∑ (1+(3/n)i)(3/n) = ∑f(xi)∆x
That makes me think that it is
∫[1,4] x dx
If you have learned to evaluate definite integrals, you should verify that the sum and the integral are equal.