Find the area bounded by the following functions:
[tex]
\begin{array}{c}
y=\sqrt{x} \\
y=x-2 \\
y=0 \\
\text { Area }=[?] \\
\end{array}
[/tex]
Round your answer to the nearest thousandth.
Answer (2)
Find the intersection points of the curves y = x , y = x − 2 , and y = 0 , which are ( 0 , 0 ) , ( 2 , 0 ) , and ( 4 , 2 ) .
Set up the integral as the sum of two integrals: ∫ 0 2 x d x + ∫ 2 4 ( x − ( x − 2 )) d x .
Evaluate the first integral: ∫ 0 2 x d x = 3 4 2 .
Evaluate the second integral: ∫ 2 4 ( x − ( x − 2 )) d x = 3 10 − 3 4 2 .
Add the two integrals to find the total area: 3 4 2 + 3 10 − 3 4 2 = 3 10 ≈ 3.333 .
Explanation
Problem Analysis We are asked to find the area bounded by the curves y = x , y = x − 2 , and y = 0 . Let's first analyze the curves and their intersections to set up the integral correctly.
Finding Intersection Points First, we find the intersection points of the curves:
y = x and y = x − 2 :
Setting x = x − 2 , we square both sides to get x = ( x − 2 ) 2 = x 2 − 4 x + 4 . This simplifies to x 2 − 5 x + 4 = 0 , which factors as ( x − 4 ) ( x − 1 ) = 0 . Thus, x = 1 or x = 4 . When x = 1 , y = 1 = 1 and y = 1 − 2 = − 1 , so ( 1 , 1 ) is not an intersection point since we are looking for the area bounded by y = 0 as well. When x = 4 , y = 4 = 2 and y = 4 − 2 = 2 , so ( 4 , 2 ) is an intersection point.
y = x and y = 0 :
Setting x = 0 , we get x = 0 . The intersection point is ( 0 , 0 ) .
y = x − 2 and y = 0 :
Setting x − 2 = 0 , we get x = 2 . The intersection point is ( 2 , 0 ) .
Setting up the Integral Now, we set up the integral. The area can be found by integrating x from x = 0 to x = 2 , and then integrating x − ( x − 2 ) from x = 2 to x = 4 . Therefore, the area is given by:
Area = ∫ 0 2 x d x + ∫ 2 4 ( x − ( x − 2 )) d x
Evaluating the First Integral Let's evaluate the first integral:
∫ 0 2 x d x = ∫ 0 2 x 1/2 d x = 3 2 x 3/2 0 2 = 3 2 ( 2 3/2 ) − 0 = 3 2 ( 2 2 ) = 3 4 2 ≈ 1.8856
Evaluating the Second Integral Now, let's evaluate the second integral:
∫ 2 4 ( x − ( x − 2 )) d x = ∫ 2 4 ( x 1/2 − x + 2 ) d x = ( 3 2 x 3/2 − 2 1 x 2 + 2 x ) 2 4
= ( 3 2 ( 4 3/2 ) − 2 1 ( 4 2 ) + 2 ( 4 ) ) − ( 3 2 ( 2 3/2 ) − 2 1 ( 2 2 ) + 2 ( 2 ) )
= ( 3 2 ( 8 ) − 8 + 8 ) − ( 3 2 ( 2 2 ) − 2 + 4 ) = 3 16 − 3 4 2 − 2 = 3 10 − 3 4 2 ≈ 1.4477
Calculating the Total Area Finally, we add the two integrals to find the total area:
Total Area = 3 4 2 + 3 10 − 3 4 2 = 3 10 ≈ 3.333
Final Answer The area bounded by the curves y = x , y = x − 2 , and y = 0 is 3 10 , which is approximately 3.333 when rounded to the nearest thousandth.
Examples
Imagine you are designing a solar panel where the sunlight capture area is defined by these curves. Knowing the precise area helps you calculate the panel's efficiency and energy output. Understanding how different functions intersect and bound areas is crucial for optimizing the design and performance of such solar energy systems.
The area bounded by the curves y = x , y = x − 2 , and y = 0 is calculated to be approximately 3.333 square units. This is determined by finding the intersection points and setting up appropriate integrals for the bounded area. Ultimately, the total area is 3 10 ≈ 3.333 after evaluating the integrals.
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