If \(\int_{0}^{k}\displaystyle (2x+1)\,dx = 6\) and \(k \gt 0\), then the value of k is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
Evaluating the integral,
\( \displaystyle \int_{0}^{k}(2x+1)\,dx = \left[x^{2}+x\right]_{0}^{k} = k^{2}+k \)
Setting this equal to 6:
\(k^{2}+k-6 = 0\)
\((k+3)(k-2) = 0\)
So \(k = 2\) or \(k = -3\). Since \(k \gt 0\), k = 2.
Why the other options are incorrect
Option (1) 3 — Incorrect. \(3^{2}+3 = 12\), not 6.
Option (3) 1 — Incorrect. \(1^{2}+1 = 2\), not 6.
Option (4) \(\sqrt{6}\) — Incorrect. It integrates only the 2x part; with the +1 term the value is \(6+\sqrt{6}\).