Consider two functions fi: (0, 1]R and fa: 1,2R defined by : h(2) = 4z -1 and fa(x) = 6 – 2r. 15150n () Then f is Riemann integrable since Let f(r) Ua(z) , if 1

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Consider two functions fi : [0,1] → R and fa : [1,2) R defined by:
fi(2) = 42" - 1 and fa(x) = 6 – 2r.
Let f(z) = h(x) if 0szs1
Then f is Riemann integrable since
La(x) , if 1<r< 2
1. f is bounded on (0, 2).
2. f is a bounded piecewise continuous function over (0, 2).
3. lim S,(P) ER and lim s/(P) ER.
4. f is monotonic on (0, 2).
O 1
O 2
Оз
O 4
Transcribed Image Text:Consider two functions fi : [0,1] → R and fa : [1,2) R defined by: fi(2) = 42" - 1 and fa(x) = 6 – 2r. Let f(z) = h(x) if 0szs1 Then f is Riemann integrable since La(x) , if 1<r< 2 1. f is bounded on (0, 2). 2. f is a bounded piecewise continuous function over (0, 2). 3. lim S,(P) ER and lim s/(P) ER. 4. f is monotonic on (0, 2). O 1 O 2 Оз O 4
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