Question

$\huge{ \bold \red{\int_{1}^{a}(2x - 1) dx = 6}}$



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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_1^a\rm (2x - 1) \: dx \: = \: 6$

We know,

$\boxed{ \tt{ \: \displaystyle \int {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

So, using this, we get

$\rm :\longmapsto\:\bigg[\dfrac{2 {x}^{2} }{2} - x\bigg]_1^a = 6$

$\rm :\longmapsto\:\bigg[ {x}^{2} - x\bigg]_1^a = 6$

$\rm :\longmapsto\:( {a}^{2} - a) - (1 - 1) = 6$

$\rm :\longmapsto\:{a}^{2} - a= 6$

$\rm :\longmapsto\:{a}^{2} - a - 6 = 0$

$\rm :\longmapsto\:{a}^{2} - 3a + 2a- 6 = 0$

$\rm :\longmapsto\:a(a - 3) + 2(a - 3) = 0$

$\rm :\longmapsto\:(a - 3)(a + 2) = 0$

$\bf\implies \:a = 3 \: \: \: or \: \: \: a \: = \: - \: 2$

Verification :-

When a = - 2

$\rm :\longmapsto\:\displaystyle\int_1^{ - 2}\rm (2x - 1) \: dx \: = \: 6$

$\rm :\longmapsto\:\bigg[\dfrac{2 {x}^{2} }{2} - x\bigg]_1^{ - 2} = 6$

$\rm :\longmapsto\:\bigg[ {x}^{2} - x\bigg]_1^{ - 2} = 6$

$\rm :\longmapsto\:[4 - ( - 2)] - (1 - 1) = 6$

$\bf\implies \:6 = 6$

Hence, Verified

When a = 3

$\rm :\longmapsto\:\displaystyle\int_1^{3}\rm (2x - 1) \: dx \: = \: 6$

$\rm :\longmapsto\:\bigg[\dfrac{2 {x}^{2} }{2} - x\bigg]_1^{3} = 6$

$\rm :\longmapsto\:\bigg[ {x}^{2} - x\bigg]_1^{3} = 6$

$\rm :\longmapsto\:[9 - 3] - [ 1 - 1] = 6$

$\bf\implies \:6 = 6$

Hence, Verified

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