Question

The integral of a certain graph is given by $A=(1+3x{)}^{\frac{1}{2}}-1$; $x\ge 0$. The average value of $A$ w.r.t. $x$ as $x$ increases from $1$ to $8$ is

• A. $\frac{3}{7}$
• B. $\frac{1}{2}$
• C. $\frac{3}{8}$
• D. $\frac{4}{7}$

Answer 1

The correct option is A

$\frac{3}{7}$

Given $\int f\left(x\right)dx=\sqrt{1+3x}-1$

Average value under the curve $f\left(x\right)$ between $x_1$ and $x_2$ is given by

$\frac{{\int }_{x_1}^{x_2}f\left(x\right)dx}{x_2-x_1}$

$\therefore$ the average value of $A$ from $1$ to $8=\frac{{\int }_1^{8}f\left(x\right)dx}{8-1}$

$=\frac{[\sqrt{1+3x}-1{]}_1^8}{8-1}$

$=\frac{3}{7}$