Solved on Jan 24, 2024

Find the area of the region between the lines $y=x$ and $y=2\sqrt{x}$ . The area is $\frac{4}{3}$ .

STEP 1

Assumptions
1. The equations given are $y = x$ and $y = 2\sqrt{x}$ .
2. We are looking for the area of the region where these two graphs intersect and bound a region.
3. The graphs intersect where $x = y = 2\sqrt{x}$ .
4. The area can be found by integrating the difference between the two functions over the interval where they bound the region.

STEP 2

First, we need to find the points of intersection of the two graphs. To do this, we set the two equations equal to each other and solve for $x$ .
$x = 2\sqrt{x}$

STEP 3

To solve the equation from Step 2, we square both sides to eliminate the square root.
$(x)^2 = (2\sqrt{x})^2$

STEP 4

Simplify the equation by squaring the terms.
$x^2 = 4x$

STEP 5

To find the solutions for $x$ , we rearrange the equation to set it to zero.
$x^2 - 4x = 0$

STEP 6

Factor the quadratic equation.
$x(x - 4) = 0$

STEP 7

Set each factor equal to zero and solve for $x$ .
$x = 0 \quad \text{or} \quad x - 4 = 0$

STEP 8

Solve for $x$ in the second equation.
$x = 4$

STEP 9

Now we have the points of intersection, which are $(0, 0)$ and $(4, 4)$ .

STEP 10

The area of the region bounded by the graphs can be found by integrating the difference between the two functions, $y = 2\sqrt{x}$ and $y = x$ , from $x = 0$ to $x = 4$ .
$Area = \int_{0}^{4} (2\sqrt{x} - x) \, dx$

STEP 11

Split the integral into two separate integrals.
$Area = \int_{0}^{4} 2\sqrt{x} \, dx - \int_{0}^{4} x \, dx$

STEP 12

Integrate the first part $\int_{0}^{4} 2\sqrt{x} \, dx$ .
$\int 2\sqrt{x} \, dx = \int 2x^{\frac{1}{2}} \, dx$

STEP 13

Use the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ , to integrate $2x^{\frac{1}{2}}$ .
$\int 2x^{\frac{1}{2}} \, dx = 2 \cdot \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} + C$

STEP 14

Simplify the expression.
$\int 2x^{\frac{1}{2}} \, dx = \frac{4}{3}x^{\frac{3}{2}} + C$

STEP 15

Integrate the second part $\int_{0}^{4} x \, dx$ .
$\int x \, dx = \frac{x^2}{2} + C$

STEP 16

Combine the two integrals and evaluate from $0$ to $4$ .
$Area = \left[\frac{4}{3}x^{\frac{3}{2}}\right]_{0}^{4} - \left[\frac{x^2}{2}\right]_{0}^{4}$

STEP 17

Evaluate the first expression at the upper and lower limits.
$\left[\frac{4}{3}x^{\frac{3}{2}}\right]_{0}^{4} = \frac{4}{3}(4)^{\frac{3}{2}} - \frac{4}{3}(0)^{\frac{3}{2}}$

STEP 18

Simplify the expression.
$\frac{4}{3}(4)^{\frac{3}{2}} = \frac{4}{3}(8) = \frac{32}{3}$

STEP 19

Evaluate the second expression at the upper and lower limits.
$\left[\frac{x^2}{2}\right]_{0}^{4} = \frac{(4)^2}{2} - \frac{(0)^2}{2}$

STEP 20

Simplify the expression.
$\frac{(4)^2}{2} = \frac{16}{2} = 8$

STEP 21

Subtract the second result from the first to find the area.
$Area = \frac{32}{3} - 8$

STEP 22

Convert the integer to a fraction with a denominator of 3 to combine the terms.
$8 = \frac{24}{3}$

STEP 23

Subtract the fractions.
$Area = \frac{32}{3} - \frac{24}{3}$

STEP 24

Simplify the subtraction.
$Area = \frac{32 - 24}{3}$

STEP 25

Calculate the result.
$Area = \frac{8}{3}$
The area of the region bounded by the graphs of the given equations is $\frac{8}{3}$ .

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Find the area of the region between the lines y=xy=xy=x and y=2xy=2\sqrt{x}y=2x​. The area is 43\frac{4}{3}34​.

STEP 1

STEP 2

First, we need to find the points of intersection of the two graphs. To do this, we set the two equations equal to each other and solve for xxx.x=2xx = 2\sqrt{x}x=2x​

STEP 3

To solve the equation from Step 2, we square both sides to eliminate the square root.(x)2=(2x)2(x)^2 = (2\sqrt{x})^2(x)2=(2x​)2

STEP 4

Simplify the equation by squaring the terms.x2=4xx^2 = 4xx2=4x

STEP 5

To find the solutions for xxx, we rearrange the equation to set it to zero.x2−4x=0x^2 - 4x = 0x2−4x=0

STEP 6

Factor the quadratic equation.x(x−4)=0x(x - 4) = 0x(x−4)=0

STEP 7

Set each factor equal to zero and solve for xxx.x=0orx−4=0x = 0 \quad \text{or} \quad x - 4 = 0x=0orx−4=0

STEP 8

Solve for xxx in the second equation.x=4x = 4x=4

STEP 9

Now we have the points of intersection, which are (0,0)(0, 0)(0,0) and (4,4)(4, 4)(4,4).

STEP 10

The area of the region bounded by the graphs can be found by integrating the difference between the two functions, y=2xy = 2\sqrt{x}y=2x​ and y=xy = xy=x, from x=0x = 0x=0 to x=4x = 4x=4.Area=∫04(2x−x) dxArea = \int_{0}^{4} (2\sqrt{x} - x) \, dxArea=∫04​(2x​−x)dx

STEP 11

Split the integral into two separate integrals.Area=∫042x dx−∫04x dxArea = \int_{0}^{4} 2\sqrt{x} \, dx - \int_{0}^{4} x \, dxArea=∫04​2x​dx−∫04​xdx

STEP 12

Integrate the first part ∫042x dx\int_{0}^{4} 2\sqrt{x} \, dx∫04​2x​dx.∫2x dx=∫2x12 dx\int 2\sqrt{x} \, dx = \int 2x^{\frac{1}{2}} \, dx∫2x​dx=∫2x21​dx

STEP 13

STEP 14

Simplify the expression.∫2x12 dx=43x32+C\int 2x^{\frac{1}{2}} \, dx = \frac{4}{3}x^{\frac{3}{2}} + C∫2x21​dx=34​x23​+C

STEP 15

Integrate the second part ∫04x dx\int_{0}^{4} x \, dx∫04​xdx.∫x dx=x22+C\int x \, dx = \frac{x^2}{2} + C∫xdx=2x2​+C

STEP 16

Combine the two integrals and evaluate from 000 to 444.Area=[43x32]04−[x22]04Area = \left[\frac{4}{3}x^{\frac{3}{2}}\right]_{0}^{4} - \left[\frac{x^2}{2}\right]_{0}^{4}Area=[34​x23​]04​−[2x2​]04​

STEP 17

Evaluate the first expression at the upper and lower limits.[43x32]04=43(4)32−43(0)32\left[\frac{4}{3}x^{\frac{3}{2}}\right]_{0}^{4} = \frac{4}{3}(4)^{\frac{3}{2}} - \frac{4}{3}(0)^{\frac{3}{2}}[34​x23​]04​=34​(4)23​−34​(0)23​

STEP 18

Simplify the expression.43(4)32=43(8)=323\frac{4}{3}(4)^{\frac{3}{2}} = \frac{4}{3}(8) = \frac{32}{3}34​(4)23​=34​(8)=332​

STEP 19

Evaluate the second expression at the upper and lower limits.[x22]04=(4)22−(0)22\left[\frac{x^2}{2}\right]_{0}^{4} = \frac{(4)^2}{2} - \frac{(0)^2}{2}[2x2​]04​=2(4)2​−2(0)2​

STEP 20

Simplify the expression.(4)22=162=8\frac{(4)^2}{2} = \frac{16}{2} = 82(4)2​=216​=8

STEP 21

Subtract the second result from the first to find the area.Area=323−8Area = \frac{32}{3} - 8Area=332​−8

STEP 22

Convert the integer to a fraction with a denominator of 3 to combine the terms.8=2438 = \frac{24}{3}8=324​

STEP 23

Subtract the fractions.Area=323−243Area = \frac{32}{3} - \frac{24}{3}Area=332​−324​

STEP 24

Simplify the subtraction.Area=32−243Area = \frac{32 - 24}{3}Area=332−24​

STEP 25

Calculate the result.Area=83Area = \frac{8}{3}Area=38​The area of the region bounded by the graphs of the given equations is 83\frac{8}{3}38​.

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Find the area of the region between the lines $y=x$ and $y=2\sqrt{x}$ . The area is $\frac{4}{3}$ .

First, we need to find the points of intersection of the two graphs. To do this, we set the two equations equal to each other and solve for $x$ .
$x = 2\sqrt{x}$

To solve the equation from Step 2, we square both sides to eliminate the square root.
$(x)^2 = (2\sqrt{x})^2$

Simplify the equation by squaring the terms.
$x^2 = 4x$

To find the solutions for $x$ , we rearrange the equation to set it to zero.
$x^2 - 4x = 0$

Factor the quadratic equation.
$x(x - 4) = 0$

Set each factor equal to zero and solve for $x$ .
$x = 0 \quad \text{or} \quad x - 4 = 0$

Solve for $x$ in the second equation.
$x = 4$

Now we have the points of intersection, which are $(0, 0)$ and $(4, 4)$ .

The area of the region bounded by the graphs can be found by integrating the difference between the two functions, $y = 2\sqrt{x}$ and $y = x$ , from $x = 0$ to $x = 4$ .
$Area = \int_{0}^{4} (2\sqrt{x} - x) \, dx$

Split the integral into two separate integrals.
$Area = \int_{0}^{4} 2\sqrt{x} \, dx - \int_{0}^{4} x \, dx$

Integrate the first part $\int_{0}^{4} 2\sqrt{x} \, dx$ .
$\int 2\sqrt{x} \, dx = \int 2x^{\frac{1}{2}} \, dx$

Simplify the expression.
$\int 2x^{\frac{1}{2}} \, dx = \frac{4}{3}x^{\frac{3}{2}} + C$

Integrate the second part $\int_{0}^{4} x \, dx$ .
$\int x \, dx = \frac{x^2}{2} + C$

Combine the two integrals and evaluate from $0$ to $4$ .
$Area = \left[\frac{4}{3}x^{\frac{3}{2}}\right]_{0}^{4} - \left[\frac{x^2}{2}\right]_{0}^{4}$

Evaluate the first expression at the upper and lower limits.
$\left[\frac{4}{3}x^{\frac{3}{2}}\right]_{0}^{4} = \frac{4}{3}(4)^{\frac{3}{2}} - \frac{4}{3}(0)^{\frac{3}{2}}$

Simplify the expression.
$\frac{4}{3}(4)^{\frac{3}{2}} = \frac{4}{3}(8) = \frac{32}{3}$

Evaluate the second expression at the upper and lower limits.
$\left[\frac{x^2}{2}\right]_{0}^{4} = \frac{(4)^2}{2} - \frac{(0)^2}{2}$

Simplify the expression.
$\frac{(4)^2}{2} = \frac{16}{2} = 8$

Subtract the second result from the first to find the area.
$Area = \frac{32}{3} - 8$

Convert the integer to a fraction with a denominator of 3 to combine the terms.
$8 = \frac{24}{3}$

Subtract the fractions.
$Area = \frac{32}{3} - \frac{24}{3}$

Simplify the subtraction.
$Area = \frac{32 - 24}{3}$

Calculate the result.
$Area = \frac{8}{3}$
The area of the region bounded by the graphs of the given equations is $\frac{8}{3}$ .