Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 10
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Sketch the region enclosed by the curves $y = 1 + \sqrt{x}$, $\displaystyle y = \frac{(3+x)}{3}$. Then find the area of the region.
By using vertical strip,
$\displaystyle A = \int^{x_2}_{x_1} \left(y_{\text{upper}} - y_{\text{lower}} \right) dx$
In order to get the values of the upper and lower limits, we equate the two functions to get its points of intersection. Thus,
$ \begin{equation} \begin{aligned} 1 + \sqrt{x} &= \frac{3+x}{3}\\ \\ 3 + 3 \sqrt{x} &= 3+x\\ \\ 3\sqrt{x} &= x \\ \\ 9x &= x^2\\ \\ x^2 - 9x &= 0\\ \\ x(x-9) &= 0 \end{aligned} \end{equation} $
We have, $x = 0$ and $x =9$
Therefore,
$ \begin{equation} \begin{aligned} A &= \int^{9}_{0} \left[ (1+\sqrt{x}) - \left( \frac{3+x}{3} \right)\right] dx \\ \\ A &= \int^9_0 \left( \sqrt{x} - \frac{x}{3} \right) dx\\ \\ A &= \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^2}{2(3)} \right]^9_0\\ \\ A &= 4.5 \text{ square units} \end{aligned} \end{equation} $
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