# How do you find the area between f(x)=root3(x-1), g(x)=x-1?

Apr 29, 2017

For a given domain, subtract the lesser function from the greater and integrate over that region.

#### Explanation:

Here is a graph of the two functions, $f \left(x\right) = \sqrt[3]{x - 1} , g \left(x\right) = x - 1$:

Please observe that from $x = 0$ to $x = 1$, $g \left(x\right) > f \left(x\right)$, therefore, the integral is:

$A = {\int}_{0}^{1} g \left(x\right) - f \left(x\right) \mathrm{dx}$

From $x = 1$ to $x = 2$ $f \left(x\right) > g \left(x\right)$, therefore, we subtract $g \left(x\right)$ from #f(x) in this region:

$A = {\int}_{0}^{1} g \left(x\right) - f \left(x\right) \mathrm{dx} + {\int}_{1}^{2} f \left(x\right) - g \left(x\right) \mathrm{dx}$

Substitute for $f \left(x\right) \mathmr{and} g \left(x\right)$:

$A = {\int}_{0}^{1} x - 1 - \sqrt[3]{x - 1} \mathrm{dx} + {\int}_{1}^{2} \sqrt[3]{x - 1} - x + 1 \mathrm{dx}$

$A = {\left({x}^{2} / 2 - x - \frac{3}{4} {\left(x - 1\right)}^{\frac{4}{3}}\right]}_{0}^{1} + {\int}_{1}^{2} \sqrt[3]{x - 1} - x + 1 \mathrm{dx}$

$A = \left({1}^{2} / 2 - 1 - \frac{3}{4} {\left(1 - 1\right)}^{\frac{4}{3}}\right) - \left({0}^{2} / 2 - 0 - \frac{3}{4} {\left(0 - 1\right)}^{\frac{4}{3}}\right) + {\int}_{1}^{2} \sqrt[3]{x - 1} - x + 1 \mathrm{dx}$

$A = - \frac{1}{2} + \frac{3}{4} + {\int}_{1}^{2} \sqrt[3]{x - 1} - x + 1 \mathrm{dx}$

$A = \frac{1}{4} + {\int}_{1}^{2} \sqrt[3]{x - 1} - x + 1 \mathrm{dx}$

$A = \frac{1}{4} + {\left(\frac{3}{4} {\left(x - 1\right)}^{\frac{4}{3}} - {x}^{2} / 2 + x\right]}_{1}^{2}$

$A = \frac{1}{4} + \left(\frac{3}{4} {\left(2 - 1\right)}^{\frac{4}{3}} - {2}^{2} / 2 + 2\right) - \left(\frac{3}{4} {\left(1 - 1\right)}^{\frac{4}{3}} - {1}^{2} / 2 + 1\right)$

$A = \frac{1}{4} + \frac{3}{4} - \frac{1}{2}$

$A = \frac{1}{2}$