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

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Solve complex trigonomentric equations

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Q. find the area of the figure bonded by

y=\sqrt{x}

,

y=x-2

,

y=0

Find Intersection Points: First, we need to find the points of intersection between $y=\sqrt{x}$ and $y=x-2$ . Set $\sqrt{x} = x - 2$ and solve for $x$ . $x = (x - 2)^2$ $x = x^2 - 4x + 4$ $x^2 - 5x + 4 = 0$ Factor the quadratic equation: $(x - 4)(x - 1) = 0$ So, $x = 4$ or $x = 1$
Check $Y=0$ Intersection: Now, let's check the intersection points with $y=0$ . For $y=\sqrt{x}$ , $y=0$ when $x=0$ . For $y=x-2$ , $y=0$ when $x=2$ .
Calculate Area Under $\sqrt{x}$ : We need to find the area between $y=\sqrt{x}$ and $y=0$ from $x=0$ to $x=1$ , and then the area between $y=x-2$ and $y=0$ from $x=1$ to $x=4$ . $\newline$ First, calculate the area under $y=\sqrt{x}$ from $x=0$ to $x=1$ . $\newline$ Area = $y=\sqrt{x}$ $2$ $\newline$ Let $y=\sqrt{x}$ $3$ , then $y=\sqrt{x}$ $4$ and $y=\sqrt{x}$ $5$ . $\newline$ Area = $y=\sqrt{x}$ $6$ $\newline$ Area = $y=\sqrt{x}$ $7$ from $y=\sqrt{x}$ $8$ to $y=\sqrt{x}$ $9$ $\newline$ Area = $y=0$ $0$ $\newline$ Area = $y=0$ $1$ square units
Calculate Area Under $x-2$ : Next, calculate the area under $y=x-2$ from $x=1$ to $x=4$ .
Area = $\int_{1}^{4} (x-2) \, dx$
Area = $(1/2)x^2 - 2x$ from $1$ to $4$
Area = $[(1/2)(4)^2 - 2(4)] - [(1/2)(1)^2 - 2(1)]$
Area = $[(1/2)(16) - 8] - [(1/2)(1) - 2]$
Area = $y=x-2$ $0$
Area = $y=x-2$ $1$
Area = $y=x-2$ $2$ square units
Find Total Area: Add the two areas together to find the total area of the figure. $\newline$ Total Area = Area under $y=\sqrt{x}$ + Area under $y=x-2$ $\newline$ Total Area = $\frac{2}{3}$ + $1.5$ $\newline$ Total Area = $2.1666\ldots$ square units

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