Question

Find the area under the curve $y=\sqrt{6x+4}$ above $x$-axis from $x=0$ to $x=2.$

Answer 1

Area under the curve

$={\int }_0^2\sqrt{6x+4}dx$

$=\frac{1}{6}\times \frac{2}{3}{\left[{\left(6x+4\right)}^{3/2}\right]}_0^2=\frac{1}{9}\left[64-8\right]$

$=\frac{56}{9}unit{s}^2$

Hence, this is the answer.