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Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R is the base of a solid. For the solid, each cross section
perpendicular to the y-axis is a square. What is the volume of the solid?
A) 0.404
B)0.533
С)1.333
D)2.667

Respuesta :

jacobEwing

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

[tex]a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333[/tex]

So the correct answer is C, the area between the two functions is 4/3 units.

Ver imagen jacobEwing
shubhamchouhanvrVT

The area between the two functions is approximately 1.333 units.

What will be the area?

Area is defined as the region bounded by the sides on a plane. The areas of different shapes can be calculated by different formulas.

If I understand your question correctly, you're looking for the area surrounded by the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

[tex]a=\int\limits^2_0 {2x} \, dx -\int\limits^2_0 {x^2} \, dx[/tex]

[tex]a=(2^2-0^2)-(\dfrac{2^3}{3}-\dfrac{0^3}{3})[/tex]

[tex]a=4-\dfrac{8}{3}[/tex]

[tex]a=1.333[/tex]

Hence the area between the two functions is approximately 1.333 units.

To know more about area follow

https://brainly.com/question/25292087

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