The area bounded by $y = 3x + 1, x = 0, y = 0$ and $x = a$ is 8 Sq.units. Then value of $a$ (where $a > 0$) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
$y=3x+1,\;x=0,\;y=0,\;x=a.$
$\text{Area}=\int_{0}^{a}(3x+1)\,dx.$
$=\left[\frac{3x^2}{2}+x\right]_{0}^{a}.$
$=\frac{3a^2}{2}+a.$
$\frac{3a^2}{2}+a=8.$
$3a^2+2a-16=0.$
$a=\frac{-2\pm\sqrt{4+192}}{6}.$
$a=\frac{-2\pm14}{6}.$
$a=2\;\text{or}\;-\frac{8}{3}.$
$a>0.$
$a=2.$