Prove that f(x)=x2 is integrable using superior and inferior sums

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lalilulelo2k3eq · Accepted Answer

Step 1  The P1 is a partion on [−1,0] and your P2 is a partition on [0,2]. So your partition P on [−1,2] should be P=P1+P2 aka:  P={t−n=−1, t−n+1=−1+1n,⋯,t−1=−1n, t0=0, t1=2n},⋯,tn=2}  You can then start from here to show  S(f,P)−s(f,P)&amp;lt;ϵ, or  [S(f,p1)+S(f,p2)]−[s(f,p1)+s(f,p2)]&amp;lt;ϵ

Annie Gonzalez · Accepted Answer

Step 1
The key idea is, on an interval where f is monotone, the difference between the upper and lower sums with respect to an equal-length partition telescopes to

Δ f Δ t

.
Suppose f is real-valued on

[a, b]

, and without loss of generality assume f is non-decreasing. For each positive integer n, consider the equal-length partition

t i = a + i \frac{b - a}{n}

for

0 \leq i \leq n

,
whose subintervals all have length

Δ t = \frac{b - a}{n}

.
On the ith subinterval

I_{i} = [t_{i - 1}, t_{i}]

we have

m_{i} = f (t_{i - 1})

M_{i} = f (t_{i})

Since the subintervals all have the same length,

S (f, P) - s (f, P) = \sum_{i = 1}^{n} f (t_{i}) Δ t - \sum_{i = 1}^{n} f (t_{i - 1}) Δ t

= Δ t \sum_{i = 1}^{n} [f (t_{i}) - f (t_{i = 1})]

= Δ t [f (b) - f (a)] = [f (b) - f (a)] \frac{b - a}{n}

This is essentially what you have with your lengthy sum, but the way the terms are expanded may be obscuring the cancellation.

Prove that f(x)=x^{2} is integrable using superior and inferior sums

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