Find the area enclosed by the curves y = x3 + x and y = 3x2 − x.

To find the area enclosed by the curves y = x^3 + x and y = 3x^2 - x, we need to determine the points of intersection between the two curves.

Step 1: Set the two equations equal to each other: x^3 + x = 3x^2 - x

Step 2: Rearrange the equation to get it in standard form: x^3 + x - 3x^2 + x = 0 x^3 - 3x^2 + 2x = 0

Step 3: Factor out an x from the equation: x(x^2 - 3x + 2) = 0

Step 4: Solve for x by setting each factor equal to zero: x = 0 or x^2 - 3x + 2 = 0

Step 5: Solve the quadratic equation x^2 - 3x + 2 = 0 using factoring or the quadratic formula: (x - 1)(x - 2) = 0

So, x = 0, x = 1, or x = 2.

Step 6: Now that we have the x-values of the points of intersection, we can find the corresponding y-values by substituting these x-values into either of the original equations.

For y = x^3 + x: When x = 0, y = 0^3 + 0 = 0 When x = 1, y = 1^3 + 1 = 2 When x = 2, y = 2^3 + 2 = 10

For y = 3x^2 - x: When x = 0, y = 3(0^2) - 0 = 0 When x = 1, y = 3(1^2) - 1 = 2 When x = 2, y = 3(2^2) - 2 = 10

Step 7: Now we can plot the two curves on a graph and shade the region between them.

Step 8: To find the area enclosed by the curves, we need to calculate the definite integral of the difference between the two curves over the interval where they intersect.

The area can be calculated as follows: Area = ∫[a, b] (f(x) - g(x)) dx

Where f(x) is the upper curve (y = x^3 + x) and g(x) is the lower curve (y = 3x^2 - x), and [a, b] represents the interval of intersection.

Step 9: Calculate the definite integral: Area = ∫[0, 1] ((x^3 + x) - (3x^2 - x)) dx + ∫[1, 2] ((3x^2 - x) - (x^3 + x)) dx

Step 10: Evaluate the integrals: Area = ∫[0, 1] (x^3 + x - 3x^2 + x) dx + ∫[1, 2] (3x^2 - x - x^3 - x) dx

Step 11: Simplify the integrals: Area = ∫[0, 1] (-3x^2 + 2x) dx + ∫[1, 2] (-x^3 + 2x^2 - 2x) dx

Step 12: Evaluate the integrals using the power rule: Area = [-x^3 + x^2] from 0 to 1 + [-x^4/4 + 2x^3/3 - x^2] from 1 to 2

Step 13: Substitute the limits of integration and calculate the area: Area = (-(1)^3 + (1)^2) - (-(0)^3 + (0)^2) + (-(2)^4/4 + 2(2)^3/3 - (2)^2) - (-(1)^4/4 + 2(1)^3/3 - (1)^2)

Step 14: Simplify the expression: Area = (1 - 1) + (-16/4 + 16/3 - 4) - (-1/4 + 2/3 - 1)

Step 15: Calculate the final result: Area = 0 + (-4 + 16/3 - 4) - (-1/4 + 2/3 - 1) Area = -4 + 16/3 - 4 + 1/4 - 2/3 + 1 Area = -12/3 + 16/3 - 12/3 + 1/4 - 8/12 + 12/12 Area = -8/3 + 1/4 + 4/12 Area = -32/12 + 3/12 + 4/12 Area = -32/12 + 7/12 Area = -25/12

Therefore, the area enclosed by the curves y = x^3 + x and y = 3x^2 - x is -25/12 square units.