Sketch the region enclosed by the given curves. Decide whether to integrate with with respect to x or y. Draw a typical approximating rectangle and label its height and width. Then find the area of the region.
y = (x^3) - x , y = 3x
Any and all help is much appreciated! Thanks so much in advance!
y = (x^3) - x , y = 3x
Any and all help is much appreciated! Thanks so much in advance!
-
y = x^3 - x , y = 3x
Set these equal to each other.
x^3 - x = 3x
x(x^2 - 1) = 3x
x^2 - 1 = 3
x^2 = 4
x = +/- 2.
Our bounds will be -2 & 2.
By induction, our second equation is greater than our first equation in the interval [-2 , 2].
Integrating with respect to x, yields:
∫ 3x - x^3 + x dx
∫ -x^3 + 4x dx
-x^4/4 + 2x^2 [ (-2 , 2)
-4 + 8 = 4
-4 + 8 = 4
4 - 4 = 0.
This means that this graph is an odd function and is symmetric about the origin.
Whenever we've an odd function, the areas cancel.
Hope this helped.
Set these equal to each other.
x^3 - x = 3x
x(x^2 - 1) = 3x
x^2 - 1 = 3
x^2 = 4
x = +/- 2.
Our bounds will be -2 & 2.
By induction, our second equation is greater than our first equation in the interval [-2 , 2].
Integrating with respect to x, yields:
∫ 3x - x^3 + x dx
∫ -x^3 + 4x dx
-x^4/4 + 2x^2 [ (-2 , 2)
-4 + 8 = 4
-4 + 8 = 4
4 - 4 = 0.
This means that this graph is an odd function and is symmetric about the origin.
Whenever we've an odd function, the areas cancel.
Hope this helped.