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find the area of the figure bonded by $y=\sqrt{x}$ , $y=x-2$ , $y=0$ . use integral

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Complex conjugate theorem

Full solution

Q. find the area of the figure bonded by

y=\sqrt{x}

,

y=x-2

,

y=0

. use integral

Find Intersection Points: First, find the points of intersection between $y=\sqrt{x}$ and $y=x-2$ by setting them equal to each other. $\sqrt{x} = x - 2$ $x = (x - 2)^2$ $x = x^2 - 4x + 4$ $x^2 - 5x + 4 = 0$
Factor Quadratic Equation: Factor the quadratic equation to find the $x$ -values of the intersection points. $(x - 4)(x - 1) = 0$ So, $x = 4$ or $x = 1$
Set Up Integral for Area: Now, set up the integral to find the area between $y=\sqrt{x}$ and $y=0$ from $x=0$ to $x=1$ , and between $y=x-2$ and $y=0$ from $x=1$ to $x=4$ . $\newline$ Area = $\int_{0}^{1} \sqrt{x} \, dx + \int_{1}^{4} (x - 2) \, dx$
Calculate Integral from $0$ to $1$ : Calculate the first integral from $0$ to $1$ . $\newline$ $\int_{0}^{1} \sqrt{x} \, dx = \left[\frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{1}$ $\newline$ $= \frac{2}{3} \times (1)^{\frac{3}{2}} - \frac{2}{3} \times (0)^{\frac{3}{2}}$ $\newline$ $= \frac{2}{3}$
Calculate Integral from $1$ to $4$ : Calculate the second integral from $1$ to $4$ . $\newline$ $\int_{1}^{4} (x - 2) \, dx = \left[\frac{1}{2} x^2 - 2x\right]_{1}^{4}$ $\newline$ = $\left(\frac{1}{2} \cdot 4^2 - 2 \cdot 4\right) - \left(\frac{1}{2} \cdot 1^2 - 2 \cdot 1\right)$ $\newline$ = $(8 - 8) - (0.5 - 2)$ $\newline$ = $0 - (-1.5)$ $\newline$ = $1.5$
Find Total Area: Add the results of the two integrals to find the total area. $\newline$ Total Area = $\frac{2}{3} + 1.5$ $\newline$ = $2.166666\ldots$

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