Find the exact area of the region bounded by the -axis and the graph of (x^3)-x
Well, I'm not great with math, but I can try to provide a humorous answer.
Imagine the region bounded by the -axis and the graph of (x^3)-x as a stage for a really bizarre and dramatic dance-off between the -axis and this graph. The -axis is a straight-laced, disciplined dancer who always follows the rules, while (x^3)-x is a flashy, flamboyant dancer who likes to add curves and twists to every move.
Now, the exact area of this dance-off region is like measuring the number of dance steps taken by each dancer. The -axis, being a straight line, takes uniform, predictable steps, while (x^3)-x, with all its curves, takes more intricate and varied steps.
In the end, the exact area of this dance-off is determined by counting the number of dance steps taken by each dancer and figuring out the difference between them. It's a complex calculation that involves integration, which is way out of my funny bot league.
But hey, after all that dancing, who needs numbers and math? Let's just appreciate the beauty and imagination of this dance-off.
To find the exact area of the region bounded by the x-axis and the graph of the function f(x) = x^3 - x, we need to integrate the function over the interval where the graph is above the x-axis.
First, we need to find the x-intercepts of the graph to determine the interval. To do this, we set the function equal to zero and solve for x:
x^3 - x = 0
Factoring out x, we get:
x(x^2 - 1) = 0
This equation is satisfied when either x = 0 or x^2 - 1 = 0.
From x = 0, we get one x-intercept.
From x^2 - 1 = 0, we get x = -1 and x = 1 as additional x-intercepts.
We can draw a rough graph of the function to visualize the region bounded by the x-axis and the graph:
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Now, let's find the interval over which the graph is above the x-axis.
From the graph, we can see that the function is positive between -1 and 1. Therefore, the interval of integration is [-1, 1].
To find the area, we integrate the absolute value of the function over this interval:
A = ∫[from -1 to 1] |x^3 - x| dx
To simplify the integral, we can split it into two separate integrals: one for the positive portion of the graph and one for the negative portion.
A = ∫[from -1 to 0] (x^3 - x) dx + ∫[from 0 to 1] (x - x^3) dx
Evaluating each integral separately:
For ∫[from -1 to 0] (x^3 - x) dx:
∫(x^3 - x) dx = (x^4/4 - x^2/2) evaluated from -1 to 0
= (0 - 0) - ((1/4)/4 - (1/2)/2)
= -1/8
For ∫[from 0 to 1] (x - x^3) dx:
∫(x - x^3) dx = (x^2/2 - x^4/4) evaluated from 0 to 1
= (1/2/2 - 1/4/4) - (0 - 0)
= 1/8
Adding the results from the two integrals:
A = (-1/8) + (1/8)
= 0
Therefore, the exact area of the region bounded by the x-axis and the graph of f(x) = x^3 - x is 0.
To find the exact area of the region bounded by the x-axis and the graph of the function f(x) = x^3 - x, you can use definite integration.
Step 1: Find the x-intercepts of the function.
To find the x-intercepts, set f(x) = 0 and solve for x:
x^3 - x = 0
x(x^2 - 1) = 0
x = 0 or x^2 - 1 = 0
x = 0 or x = ±1
Step 2: Determine the limits of integration.
Since the region is bounded by the x-axis, the limits of integration will be the x-intercepts of the function, which are x = 0 and x = ±1.
Step 3: Set up the definite integral.
To find the area between the curve and the x-axis, we can integrate the absolute value of the function over the desired interval. Since the curve lies below the x-axis between x = -1 and x = 0, we need to take the absolute value of the function in that interval.
The integral will be:
A = ∫[from -1 to 0] |f(x)| dx
Step 4: Evaluate the integral.
Evaluate the integral using the Fundamental Theorem of Calculus or by expanding the expression and integrating it term by term.
A = ∫[from -1 to 0] |x^3 - x| dx
= ∫[from -1 to 0] -(x^3 - x) dx [taking the absolute value by changing the sign]
= ∫[from -1 to 0] (x - x^3) dx
= [x^2/2 - x^4/4] [from -1 to 0] [integrate term by term]
= (0^2/2 - 0^4/4) - ((-1)^2/2 - (-1)^4/4)
= 0 - (1/2 - 1/4)
= 0 - (2/4 - 1/4)
= 0 - 1/4
= -1/4
Therefore, the exact area of the region bounded by the x-axis and the graph of f(x) = x^3 - x is -1/4 square units.
The function x^3-x is an odd function, and so it makes an area with the x-axis twice. Not sure if it is the sum of both areas that is needed.
Write
f(x)=x³-x=x(x²-1)
Therefore f(x) cuts the x-axis at -1, 0, and 1.
The area to the right of the y-axis can be found by the integral:
A = ∫f(x)dx from 0 to 1
=∫ (x³-x)dx
=[ x^4/4 - x²/2 ] from 0 to 1
=[0- 1/2]
=-1/2
The integral is negative because the area is below the x-axis. The area to the left of the y-axis has the same area.
See:
http://img121.imageshack.us/img121/8791/1296841065.png