The area under the curve y = sqrt{3x + 4} bet | vedclass

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(D) The area $A$ under the curve $y = f(x)$ from $x = a$ to $x = b$ is given by $A = \int_{a}^{b} f(x) \, dx$.Here,$f(x) = \sqrt{3x + 4}$,$a = 0$,and

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None of these

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(D) The area $A$ under the curve $y = f(x)$ from $x = a$ to $x = b$ is given by $A = \int_{a}^{b} f(x) \, dx$.Here,$f(x) = \sqrt{3x + 4}$,$a = 0$,and $b = 4$.$A = \int_{0}^{4} \sqrt{3x + 4} \, dx$Let $u = 3x + 4$,then $du = 3 \, dx$,or $dx = \frac{du}{3}$.When $x = 0$,$u = 4$. When $x = 4$,$u = 16$.$A = \int_{4}^{16} \sqrt{u} \cdot \frac{du}{3} = \frac{1}{3} \int_{4}^{16} u^{1/2} \, du$$A = \frac{1}{3} \left[ \frac{u^{3/2}}{3/2} ight]_{4}^{16} = \frac{1}{3} \cdot \frac{2}{3} \left[ u^{3/2} ight]_{4}^{16}$$A = \frac{2}{9} [16^{3/2} - 4^{3/2}]$$A = \frac{2}{9} [64 - 8] = \frac{2}{9} 	imes 56 = \frac{112}{9} 	ext{ sq. unit}$.Since $\frac{112}{9}$ is not among the options,the correct option is $(d)$.

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

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