Find the area of the shaded region.
the following curves are;
y=x^3
y=x+6
y=-1/2x
If you plot the curves, as at
http://www.wolframalpha.com/input/?i=plot+y%3Dx^3%2C+y+%3D+x%2B6%2C+y+%3D+-1%2F2+x
it is clear that the area is
∫[-4,0] (x+6)-(-x/2) dx + ∫[0,2] (x+6)-x^3 dx
= 12 + 10
= 22
To find the area of the shaded region, we first need to determine the bounds of the region. To do this, we need to find the x-values where the curves intersect.
First, let's find where the curve y = x^3 intersects the curve y = x + 6:
Setting the two equations equal to each other:
x^3 = x + 6
Rearranging the equation:
x^3 - x - 6 = 0
Since this is a cubic equation, we can use factoring or numerical methods like Newton's method to find the solutions. Let's use factoring and see if we can find any rational solutions:
By plugging in possible rational solutions, we find that x = 2 is a solution. Using synthetic division or long division, we can divide x^3 - x - 6 by x - 2:
2 | 1 0 -1 -6
---------------
1 2 3 0
This shows that (x - 2) is a factor of the equation, so we can rewrite it as:
(x - 2)(x^2 + 2x + 3) = 0
The quadratic equation x^2 + 2x + 3 does not have real solutions, so the only solution is x = 2.
Now, let's find where the curve y = x^3 intersects the curve y = -1/2x:
Setting the two equations equal to each other:
x^3 = -1/2x
Multiplying both sides by 2:
2x^3 = -x
Rearranging the equation:
2x^3 + x = 0
Again, this is a cubic equation, so we can use factoring or numerical methods to find the solutions. By factoring, we can see that x = 0 is a solution. Dividing 2x^3 + x by x, we find:
0 | 2 0 1
------------
2 0 1
This shows that x is a factor of the equation, so we can rewrite it as:
x(x^2 + 1) = 0
The quadratic equation x^2 + 1 does not have real solutions, so the only solution is x = 0.
Now that we have the x-values where the curves intersect, we can find the area of the shaded region by integrating the curves between these bounds. The shaded region is a bit difficult to determine exactly without a visual reference, so for this explanation, let's assume we are looking at the region between x = 0 and x = 2.
The area of the shaded region is given by the integral of the curve y = x^3 minus the curve y = -1/2x:
Area = ∫ (x^3 - (-1/2x)) dx
Evaluating this integral between x = 0 and x = 2 will give us the area of the shaded region. To simplify the calculation, you can also rearrange the equation and subtract the two integrals:
Area = ∫ (x^3) dx - ∫(-1/2x) dx
Area = ∫x^3 dx + ∫(1/2x) dx
Integrating each term separately:
Area = 1/4 x^4 + 1/2 ln(abs(x))
To evaluate the definite integral between x = 0 and x = 2, we substitute the limits into the equation and subtract the value of the integral at the lower limit from the value at the upper limit:
Area = (1/4 * 2^4 + 1/2 ln(abs(2))) - (1/4 * 0^4 + 1/2 ln(abs(0)))
Area = (1/4 * 16 + 1/2 ln(2)) - (0 + undefined)
Area = 4 + 1/2 ln(2)
So, the area of the shaded region bounded by the curves y = x^3, y = x + 6, and y = -1/2x is 4 + 1/2 ln(2).