In this question, you will find the exact value of the equation using Riemann sums.

Question

In this question, you will find the exact value of ∫⁵₀ x²/2 + 2xdx using Riemann sums.

1. simplify ∑ⁿ_i=1 (i²/2) and ∑ⁿ_i=1 2i
2. approximate ∫⁵₀ x²/2 + 2xdx via theRiemann sum, using the partition of [0, 5] into n equal intervals, and using the right endpoint of each interval as the sample point. Your answer should depend on n and use (a) to simplify. Let this Riemann sum be denoted S_n
3. Compute the limit Lim n−>∞ S_n

Answer 1

Answer 2

For part (1), it may be useful to visit the following

website:

https://courses.cs.washington.edu/courses/cse373/19sp/resources/math/summation/

(1) To simplify the first summation, we first use the rule

that allows us to factor out the constant. So,

∑(i^2 / 2) = (1 / 2) * ∑(i^2).

Then, by the sum of squares rule,

(1 / 2) * ∑(i^2) = (1 / 2) * [ (n * (n+1) * (2n + 1)) / 6 ]

= (n * (n+1) * (2n + 1)) / 12.

Hence, ∑(i^2 / 2) = (n * (n+1) * (2n + 1)) / 12.

Now moving on to the second summation, we once

again begin with the rule of factoring out a

constant:

∑(2i) = 2 * ∑(i).

Then, by Gauss's Identity,

2 * ∑(i) = 2 * [ (n * (n+1)) / 2 ]

= n * (n+1).

In conclusion, ∑(2i) = n * (n+1).

(2) The interval [0,5] is divided into n equal

intervals of width

Δx = (5 − 0) / n = 5 / n.

Using the right endpoint for each interval, the

sample points are x_i = 5i / n. The Riemann sum,

S, is given by

S = ∑[f(x_i) * ​Δx].

Recall, f(x) = (x^2 / 2) + 2x, so f(x_i) = (x_i^2 / 2) + 2x_i,

and, since x_i = 5i / n,

f(5i / n) = ((5i / n)^2 / 2) + 2(5i / n)

= ((25i^2 / n^2) / 2) + (10i / n)

= (25 / 2) * (i^2 / n^2) + (10i / n)

= 12.5(i^2 / n^2) + (10i / n).

Also, Δx = 5 / n. So,

S = ∑[f(x_i) * ​Δx]

= ∑{[(x_i^2 / 2) + 2x_i] * (5 / n)}

= ∑{[((5i / n)^2 / 2) + 2(5i / n)] * (5 / n)}

= ∑{[12.5(i^2 / n^2) + (10i / n)] * (5 / n)}

= (5 / n) * {∑[12.5(i^2 / n^2) + (10i / n)]}

by factoring out the constant

= (5 / n) * {∑[12.5(i^2 / n^2)] + ∑(10i / n)]}

by splitting the sum

= (5 / n) * {[(12.5 / n^2) * ∑(i^2)] + [(10 / n) * ∑(i)]}

by factoring out the constants

= (5 / n) * {[(12.5 / n^2) * ({n * (n+1) * (2n + 1)} / 6 )]

+ [(10 / n) * ∑(i)]}

by the sum of squares

= (5 / n) * {[(12.5 / 6) * ({(n+1) * (2n + 1)} / n )]

+ [(10 / n) * ({n * (n+1)} / 2)]}

by Gauss's identity

= (5 / n) * {[(12.5 / 6) * ({(n+1) * (2n + 1)} / n )]

+ [5 * (n+1)]}

= (5 / n^2) * {[(12.5 / 6) * (n+1) * (2n + 1)] + [5 * (n+1) * n]}

= (5 / n^2) * {[(12.5 / 6) * (2n^2 + 3n + 1)] + [5 * (n^2 + n)]}

= (5 / n^2) * {[(25/6)n^2 + (25/4)n + (25/12)] + [5n^2 + 5n]}

= (5 / n^2) * [(25/6)n^2 + (25/4)n + (25/12) + 5n^2 + 5n]

= (5 / n^2) * [(55/6)n^2 + (45/4)n + (25/12)]

= (1 / n^2) * [(275/6)n^2 + (225/4)n + (125/12)]

= [(275/6)n^2 + (225/4)n + (125/12)] / n^2.

So, our final solution is

S = [(275/6)n^2 + (225/4)n + (125/12)] / n^2.

(3) The final part asks us to compute the limit as n

tends to infinity. Well, for very large n values the n^2

terms will dominate the behavior, so we essentially ignore

all other terms,

[(275/6)n^2 + (225/4)n + (125/12)] / n^2 --> [(275/6)n^2] / n^2

then the n^2 cancels out from the numerator and denominator,

leaving us with the coefficient of 275/6.

Thus, the limit of S as n tends to infinity is 275/6.