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Question

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


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