Question

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

• A. $8$ sq. unit
• B. $\frac{112}{9}$ sq. unit
• C. $\frac{56}{9}$ sq. unit
• D. None of these

Answer 1

Answer: (B)

$\frac{112}{9}$ sq. unit

$m=3x+4$

$dm=3dx$

When $x=0$ , $m=4$ ,

$x=4$ , $m=16$

${\int }_0^4\sqrt{3x+4}dx$

$=\frac{1}{3}{\int }_4^{16}\sqrt{m}dm$

$=\frac{1}{3}[\frac{m^{\frac{3}{2}}}{\frac{3}{2}}{]}_4^{16}$

$\frac{2}{9}[4^3-2^3]=\frac{2\times 56}{9}squnits$

$=\frac{112}{9}squnits$