Question
Asked Dec 11, 2019
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Let f(x)=2x2-1 and consider the partition {-1,0,2} of the interval [-1,2].

Find Lf(P) and Uf(P). 

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Expert Answer

Step 1

Formula used:

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Lower sum L(P,f)defines as L; (P)=Emin f(x)Ar on(x,,x) n+1 n=-1 Upper sum U (P,f)defines as U;(P)=E Max f (x)Ax on(x,xm1) n+1 n=-1

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Step 2

Given:

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f(x)= 2x² -1.[-1, 2] Partition, P = {-1, 0, 2} To find L, (P)and U, (P):

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Step 3

To begin with the first partition:

...
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Ar = 0+1=1 Now, Substitute the end points and check the value, f(-1)= 2(-1)* –1=2-1=1 f(0)=2(0) –1=-1 Thus, Min value is -1 Max value is 1 on the given interval.

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