Find the area of the indicated region. Enclosed by y = −x and y = x4
intersection:
x^4 = -x
x^4 + x = 0
x(x^3 +1) = 0
x(x+1)(x^2 -x + 1)= 0
2 real solutions,
x = 0 or x = -1
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E4+%2C+y+%3D+-x
area = ∫(-x - x^4) from -1 to 0
= [- x^2/2 - x^5/5] form -1 to 0
= (0 - (-1 -(-1)/5)
= 1 - 1/5
= 4/5
To find the area of the region enclosed by the functions y = -x and y = x^4, we need to first determine the points where the two curves intersect.
Setting the two equations equal to each other:
-x = x^4
Rearranging the equation:
x^4 + x = 0
Factoring out an x:
x(x^3 + 1) = 0
Setting each factor equal to zero:
x = 0 or x^3 + 1 = 0
The solution x = 0 corresponds to the point of intersection at the origin.
To find the value of x when x^3 + 1 = 0, we can subtract 1 from both sides:
x^3 = -1
Taking the cube root of both sides:
x = -1^(1/3)
Therefore, the two curves intersect at x = 0 and x = -1^(1/3).
To find the area of the region between the curves, we integrate the difference of the two functions with respect to x over the interval where the curves intersect. Since one curve is above the other in the given interval, we subtract the function for the curve below from the function for the curve above.
The limits of integration will be x = -1^(1/3) to x = 0.
The area A can be calculated as follows:
A = ∫[x = -1^(1/3) to x = 0] (x^4 - (-x)) dx
Evaluating the integral:
A = ∫[x = -1^(1/3) to x = 0] (x^4 + x) dx
Integrating term by term:
A = (1/5)x^5 + (1/2)x^2 ] [x = -1^(1/3) to x = 0
Substituting the limits of integration:
A = (1/5)(0)^5 + (1/2)(0)^2 - [(1/5)(-1^(1/3))^5 + (1/2)(-1^(1/3))^2]
Simplifying further, we find:
A = 0 - [(-1/5) + (1/2)]
Finally,
A = 1/2 - 1/5
Therefore, the area of the region enclosed by the functions y = -x and y = x^4 is 3/10 square units.