Mathematics

# If $\int _{ 1 }^{ k }{ \left( 2x-3 \right) } dx=12$, then $k=....$

$-2$ and $5$

##### SOLUTION
$\int _{ 1 }^{ k }{ \left( 2x-3 \right) } dx$

$={ \left[ { x }^{ 2 }-3x \right] }_{ 1 }^{ k }$

$=\left( { k }^{ 2 }-3k \right) -\left( 1-3 \right)$

$={ k }^{ 2 }-3k+2$

this should be equal to 12

$k^2-3k-10=0$

$k^2-5k+2k-10=0$

$k(k-5)+2(k-5)$

$(k+2)(k-5)$

$k=-2,5$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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