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Question

The difference between the greatest and least values of the function F(x)=x0(t+1)dt on [2,3] is

A
3
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B
2
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C
7/2
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D
11/2
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Solution

The correct option is C 7/2
F(x)=x0(t+1)dtF(x)=x+1>0x[2,3]
So the difference of greatest and least value is
F(3)F(2)=32(t+1)dt=72

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