$\int_{2}^{k} (2 x + 1) d x = 6$ , then $k =$

$4$

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$- 2$

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$- 3$

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$3$

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Solution

The correct option is D
$3$

Explanation for the correct option:
Finding the value for the given integral:
Consider $\int_{2}^{k} (2 x + 1) d x = 6$
$\Rightarrow$ $2 \int_{2}^{k} x d x + \int_{2}^{k} 1 d x = 6$
$\Rightarrow$ $2 {[\frac{x^{2}}{2}]}_{2}^{k} + {[x]}_{2}^{k} = 6$
$\Rightarrow$ ⁠ $2 [\frac{k^{2}}{2} - \frac{2^{2}}{2}] + [k - 2] = 6$
$\Rightarrow$ $k^{2} - 4 + k - 2 = 6$
$\Rightarrow$ $k^{2} + k - 12 = 0$
$\Rightarrow$ $k^{2} + 4 k - 3 k - 12 = 0$
$\Rightarrow$ $(k + 4) (k - 3) = 0$
$\Rightarrow$ $k = - 4, 3$
Thus, option D is the correct answer.

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