# How do you find the area between f(x)=sqrt(3x)+1, g(x)=x+1?

Mar 5, 2017

The area is $\frac{3}{2}$ square units.

#### Explanation:

Start by finding the intersection points.

$\sqrt{3 x} + 1 = x + 1$

$\sqrt{3 x} = x$

${\left(\sqrt{3 x}\right)}^{2} = {x}^{2}$

$3 x = {x}^{2}$

${x}^{2} - 3 x = 0$

$x \left(x - 3\right) = 0$

$x = 0 \mathmr{and} 3$

These will be our bounds of integration. Next, we use a test point in $\left[0 , 3\right]$ to determine which function is higher up.

Let the test point be $x = 2$.

$f \left(2\right) = \sqrt{6} + 1 \approx 3.449$

$g \left(2\right) = 2 + 1 = 3$

Therefore, $f \left(x\right)$ will be above $g \left(x\right)$. We can now find an expression for the area:

$A = {\int}_{0}^{3} f \left(x\right) - g \left(x\right) \mathrm{dx}$

$A = {\int}_{0}^{3} \sqrt{3 x} + 1 - \left(x + 1\right) \mathrm{dx}$

$A = {\int}_{0}^{3} \sqrt{3 x} + 1 - x - 1 \mathrm{dx}$

$A = {\int}_{0}^{3} \sqrt{3 x} - x \mathrm{dx}$

$A = {\int}_{0}^{3} \sqrt{3} \sqrt{x} - x \mathrm{dx}$

$A = {\left[\frac{2 \sqrt{3}}{3} {x}^{\frac{3}{2}} - \frac{1}{2} {x}^{2}\right]}_{0}^{3}$

$A = \frac{2}{3} \sqrt{3} {\left(3\right)}^{\frac{3}{2}} - \frac{1}{2} {\left(3\right)}^{2} - \left(\frac{2}{3} \sqrt{3} \left({0}^{\frac{3}{2}} - \frac{1}{2} {\left(0\right)}^{2}\right)\right)$

$A = 6 - \frac{9}{2}$

$A = \frac{3}{2} \text{ } {u}^{2}$

Hopefully this helps!