Question

Find the interval in which$$\displaystyle f\left( x \right) =\int _{ -1 }^{ x }{ \left( { e }^{ t }-1 \right) \left( 2-t \right) dt }$$, $$\left( x>1 \right)$$ isincreasing

A
[3,5]
B
[1,3]
C
[0,3]
D
[0,2]

Solution

The correct option is D $$\left[ 0,2 \right]$$$$f\left( x \right) =\displaystyle \int _{ -1 }^{ x }{ \left( { e }^{ t }-1 \right) \left( 2-t \right) dt }$$The function is increasing if$$f'\left( x \right) >0$$ By Lebinitz's theoram$$f'\left( x \right) =\left( { e }^{ x }-1 \right) \left( 2-x \right)$$ We see that $$f'\left( x \right) >0$$ for $$x\in [0,2]$$So, $$\displaystyle f\left( x \right) =\int _{ -1 }^{ x }{ \left( { e }^{ t }-1 \right) \left( 2-t \right) dt }$$ will be increasing in interval $$x\in [0,2]$$Mathematics

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