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The difference between the greatest and he least value of the function

F(x)=0x(t+1)dt on [2, 3] is

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a
3
b
2
c
7/2
d
3/2

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detailed solution

Correct option is C

Differentiating the given function, we getF′(x)=t+1t=xdxdx−[t+1]t=0d0dx=x+1This is positive for all x ∈ [2, 3], so F is an increasing function in this interval. Therefore its greatest value isF(3)=∫03 (t+1)dt and its least value is F(2)=∫02 (t+1)dt,so that the required difference between these values is∫03 (t+1)dt−∫02 (t+1)dt=∫23 (t+1)dt=72.


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