Compute ∫√(25 - x^2) dx from 0 to 5 using an infinite Riemann Sum

[s3a]

Homework Statement

Integrate √(25 - x^2) dx from 0 to 5 using an infinite Riemann Sum

Homework Equations

lim n→∞ Σ_(i=1)^n i = n(n+1)/2
lim n→∞ Σ_(i=1)^n i^2 = n(n+1)(2n+1)/6

The Attempt at a Solution

Δx = (b - a)/n
Δx = (5 - 0)/n
Δx = 5/n

f(x_i) = √(25 - [a + iΔx]^2)
f(x_i) = √(25 - [0 + 5i/n]^2)
f(x_i) = √(25 - [5i/n]^2)
f(x_i) = √(25 - 25 i^2/n^2)
f(x_i) = √(25) √(1 - i^2/n^2)
f(x_i) = 5 √(1 - i^2/n^2)

lim n→∞ Σ_(i=1)^n [f(x_i) Δx]
lim n→∞ Σ_(i=1)^n [ [5 √(1 - i^2/n^2)] [5/n] ]
lim n→∞ 5/n Σ_(i=1)^n [5 √(1 - i^2/n^2)] (This is where I'm stuck.)

Is it impossible to compute the definite integral of √(25 - x^2) dx from 0 to 5 using an infinite Riemann sum (such that I have to use the regular integral method of trigonometric substitution instead)?

If it is possible, how do I proceed from where I am stuck?

Any help in getting unstuck would be GREATLY appreciated!

 

[gopher_p]

Homework Statement

Integrate √(25 - x^2) dx from 0 to 5 using an infinite Riemann Sum

Homework Equations

lim n→∞ Σ_(i=1)^n i = n(n+1)/2
lim n→∞ Σ_(i=1)^n i^2 = n(n+1)(2n+1)/6

The Attempt at a Solution

Δx = (b - a)/n
Δx = (5 - 0)/n
Δx = 5/n

f(x_i) = √(25 - [a + iΔx]^2)
f(x_i) = √(25 - [0 + 5i/n]^2)
f(x_i) = √(25 - [5i/n]^2)
f(x_i) = √(25 - 25 i^2/n^2)
f(x_i) = √(25) √(1 - i^2/n^2)
f(x_i) = 5 √(1 - i^2/n^2)

lim n→∞ Σ_(i=1)^n [f(x_i) Δx]
lim n→∞ Σ_(i=1)^n [ [5 √(1 - i^2/n^2)] [5/n] ]
lim n→∞ 5/n Σ_(i=1)^n [5 √(1 - i^2/n^2)] (This is where I'm stuck.)

Is it impossible to compute the definite integral of √(25 - x^2) dx from 0 to 5 using an infinite Riemann sum (such that I have to use the regular integral method of trigonometric substitution instead)?

If it is possible, how do I proceed from where I am stuck?

Any help in getting unstuck would be GREATLY appreciated!

Your work looks correct, but I am unaware of any way to compute $\lim\limits_{n\rightarrow\infty}\sum\limits_{i=1}^n5\sqrt{1-\frac{i^2}{n^2}}\frac{5}{n}$ without identifying it as a definite integral (which we already know here) and doing it the "easy" way. For the record the "easy" way involves knowing what the graph of $y=\sqrt{25-x^2}$ is along with some very basic geometry.

Also, I'm not a fan of the terminology "infinite Riemann sum". Riemann sums are finite. Integrals are defined to be limits of Riemann sums. There are no infinite Riemann sums.

 

[pasmith]

One method is, in effect, to prove a special case of the fundamental theorem by finding an antiderivative F: [0,5] \to \mathbb{R} of \sqrt{25 - x^2} and applying the mean value theorem to F on each subinterval of an arbitrary partition to conclude that <br /> \int_0^5 \sqrt{25 - x^2}\,dx = F(5) - F(0).