Question

Area under the curve $y=\sqrt{3x+4}$between x=0 and x=4, is

• A. $\frac{56}{9}sq.unit$
• B. $\frac{64}{9}sq.unit$
• C. 8 sq. unit
• D. None of these

Answer 1

The correct option is D

None of these

$Area={\int }_0^4\sqrt{3x+4dx}={\left|\frac{(3x+4{)}^{\frac{3}{2}}}{3.\left(\frac{3}{2}\right)}\right|}_0^4$

$=\frac{2}{9}\times 56=\frac{112}{9}sq.unit$