It seems like you're asking for the volume of a solid of revolution, specifically the volume of the solid created by rotating the region under the curve y = 4e^(-x^2) and above the x-axis between x = 0 and x = 1 around the y-axis.

The formula for the volume of a solid of revolution when rotating around the y-axis is given by V = ∫[a,b] 2πxh dx, where h is the height of the cylindrical shell, in this case given by the function y = 4e^(-x^2), and x is the distance from the y-axis to the shell.

Here are the steps to solve this:

1. Set up the integral: V = ∫[0,1] 2πx(4e^(-x^2)) dx.

2. Simplify the integral: V = 8π ∫[0,1] xe^(-x^2) dx.

3. Now, we need to solve this integral. We can do this by using the method of integration by substitution. Let u = x^2. Then, du = 2x dx.

4. Rewrite the integral in terms of u: V = 8π ∫[0,1] 0.5e^(-u) du.

5. Now, this is a simple integral to solve. The integral of e^(-u) is -e^(-u), so V = -8π [0.5e^(-u)] evaluated from 0 to 1.

6. Evaluate the integral at the limits of integration: V = -8π [0.5e^(-1) - 0.5e^(0)].

7. Simplify to get the final answer: V = -8π [0.5/e - 0.5] = 8π [0.5 - 0.5/e].

So, the volume of the solid created by rotating the region under the curve y = 4e^(-x^2) and above the x-axis between x = 0 and x = 1 around the y-axis is V = 8π [0.5 - 0.5/e].

Question

It seems like you're asking for the volume of a solid of revolution, specifically the volume of the solid created by rotating the region under the curve y = 4e^(-x^2) and above the x-axis between x = 0 and x = 1 around the y-axis.

The formula for the volume of a solid of revolution when rotating around the y-axis is given by V = ∫[a,b] 2πxh dx, where h is the height of the cylindrical shell, in this case given by the function y = 4e^(-x^2), and x is the distance from the y-axis to the shell.

Here are the steps to solve this:

1. Set up the integral: V = ∫[0,1] 2πx(4e^(-x^2)) dx.

2. Simplify the integral: V = 8π ∫[0,1] xe^(-x^2) dx.

3. Now, we need to solve this integral. We can do this by using the method of integration by substitution. Let u = x^2. Then, du = 2x dx.

4. Rewrite the integral in terms of u: V = 8π ∫[0,1] 0.5e^(-u) du.

5. Now, this is a simple integral to solve. The integral of e^(-u) is -e^(-u), so V = -8π [0.5e^(-u)] evaluated from 0 to 1.

6. Evaluate the integral at the limits of integration: V = -8π [0.5e^(-1) - 0.5e^(0)].

7. Simplify to get the final answer: V = -8π [0.5/e - 0.5] = 8π [0.5 - 0.5/e].

So, the volume of the solid created by rotating the region under the curve y = 4e^(-x^2) and above the x-axis between x = 0 and x = 1 around the y-axis is V = 8π [0.5 - 0.5/e].

Knowee AI · Accepted Answer

It seems like you're asking for the volume of a solid of revolution, specifically the volume of the solid created by rotating the region under the curve y = 4e^(-x^2) and above the x-axis between x = 0 and x = 1 around the y-axis.

The formula for the volume of a solid of revolution when rotating around the y-axis is given by V = ∫[a,b] 2πxh dx, where h is the height of the cylindrical shell, in this case given by the function y = 4e^(-x^2), and x is the distance from the y-axis to the shell.

Here are the steps to solve this:

1. Set up the integral: V = ∫[0,1] 2πx(4e^(-x^2)) dx.

2. Simplify the integral: V = 8π ∫[0,1] xe^(-x^2) dx.

3. Now, we need to solve this integral. We can do this by using the method of integration by substitution. Let u = x^2. Then, du = 2x dx.

4. Rewrite the integral in terms of u: V = 8π ∫[0,1] 0.5e^(-u) du.

5. Now, this is a simple integral to solve. The integral of e^(-u) is -e^(-u), so V = -8π [0.5e^(-u)] evaluated from 0 to 1.

6. Evaluate the integral at the limits of integration: V = -8π [0.5e^(-1) - 0.5e^(0)].

7. Simplify to get the final answer: V = -8π [0.5/e - 0.5] = 8π [0.5 - 0.5/e].

So, the volume of the solid created by rotating the region under the curve y = 4e^(-x^2) and above the x-axis between x = 0 and x = 1 around the y-axis is V = 8π [0.5 - 0.5/e].

Now we can say that the volume of the solid created by rotating the region under y = 4e−x2and above the x-axis between x = 0 and x = 1 around the y-axis isV = b2𝜋rh dxa= 2𝜋x dx.0

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