a.)
Given y=square root of (x) and y= (x) Find the area between two curves from x=0 to x=1 by integrating with respect to x (top-bottom)
I'm more concerned about this one though:
b.) Now integrate with respect to y given between y=0 and y=1
Do I just graph it on my calculator and find which one is right and which is left? Then left - right. I take the antiderivative of both the x and square root of x and then just plug in the 0 and 1. Then minus the answers from eachother? Is that wrong?
Could someone teach me step by step how to do it? I'm in intro to calc cp. I have a TI-83 graphing calculator.
Sure, I can guide you step by step on how to find the area between two curves using integration.
a.) To find the area between two curves from x = 0 to x = 1, we need to find the difference between the top curve and the bottom curve.
Let's define the top curve as y = x and the bottom curve as y = √(x).
Step 1: Determine the intersection points of the two curves.
To find the intersection points, set the two curves equal to each other and solve for x:
x = √(x)
Squaring both sides:
x^2 = x
Rearranging the equation:
x^2 - x = 0
Factoring out x:
x(x - 1) = 0
From this equation, we can see that either x = 0 or x - 1 = 0, which gives us x = 0 or x = 1. These are the intersection points.
Step 2: Determine the top and bottom curves within the given interval.
Between x = 0 and x = 1, the top curve is y = x, and the bottom curve is y = √(x).
Step 3: Set up the integral for the area.
To find the area between two curves, we need to integrate the difference in the y-values of the curves.
The integral for the area is:
A = ∫[from 0 to 1] (top curve - bottom curve) dx
In this case, it becomes:
A = ∫[from 0 to 1] (x - √(x)) dx
Step 4: Evaluate the integral.
To evaluate the integral on your calculator, follow these steps:
1. Press the "PRGM" button.
2. Choose "NEW" and create a new program.
3. Type in the integral expression: ∫(x - √(x)) dx
4. Press "2nd" and then "CALC" (or "MATH" and then "9") to access the "fnInt" function.
5. Enter the expression, the variable (x), the lower limit (0), and the upper limit (1).
6. Press "ENTER" to calculate the result.
The result will give you the area between the two curves from x = 0 to x = 1.
b.) To integrate with respect to y, we need to rearrange the equations to solve for x in terms of y.
Starting with the top curve y = x, we rearrange it to get x in terms of y: x = y.
For the bottom curve y = √(x), we square both sides to get x = y^2.
Now we have the top curve as x = y and the bottom curve as x = y^2.
To find the area between the curves with respect to y, we follow similar steps as part a:
Step 1: Determine the intersection points of the two curves.
In this case, the intersection points are the same as in part a, which are x = 0 and x = 1.
Step 2: Determine the right and left curves within the given interval.
Between x = 0 and x = 1, the right curve is x = y^2, and the left curve is x = y.
Step 3: Set up the integral for the area.
To find the area between the curves with respect to y, we integrate the difference in the x-values of the curves:
A = ∫[from 0 to 1] (right curve - left curve) dy
In this case, it becomes:
A = ∫[from 0 to 1] (y^2 - y) dy
Step 4: Evaluate the integral.
Follow the same steps as in part a to evaluate the integral on your calculator, but this time, use the "fnInt" function with respect to y instead of x.
The result will give you the area between the two curves from y = 0 to y = 1.
Remember to use parentheses and proper notation while entering the integral expression in your calculator to ensure accurate results.
To find the area between two curves, you need to integrate the difference between the top curve and the bottom curve with respect to the variable you are integrating with respect to (in this case, x or y).
For part a) where you are integrating with respect to x, here are the steps:
1. Start by finding the points of intersection between the two curves. Set the equations equal to each other and solve for x.
y = sqrt(x) and y = x
sqrt(x) = x
Square both sides: x = x^2
Rearrange the equation: x^2 - x = 0
Factor out x: x(x - 1) = 0
So, x = 0 and x = 1 are the points of intersection.
2. Determine which curve is on top and which is on the bottom on the interval [0, 1]. To do this, plug in values of x in the interval [0, 1] into both equations and compare the resulting y-values. The curve with the larger y-values on this interval will be the top curve, and the other will be the bottom curve.
For x = 0, y = 0 and y = 0, so they intersect at this point.
For x = 1, y = 1 and y = 1, so they intersect at this point as well.
Since both curves have the same y-values at the endpoints, you can choose whichever curve you prefer as the top curve.
3. Set up the integral to find the area between the curves:
A = ∫(top curve - bottom curve) dx from x = 0 to x = 1
In this case, since you want to integrate with respect to x, you will use the equation y = sqrt(x) as the top curve and y = x as the bottom curve.
The integral would be: A = ∫(sqrt(x) - x) dx from x = 0 to x = 1.
4. Evaluate the integral:
Integrate (sqrt(x) - x) with respect to x:
A = ∫(sqrt(x) - x) dx = (2/3)x^(3/2) - (1/2)x^2
Evaluate the integral from x = 0 to x = 1:
A = [(2/3)(1)^(3/2) - (1/2)(1)^2] - [(2/3)(0)^(3/2) - (1/2)(0)^2]
Simplify the expression:
A = (2/3) - (1/2) = 1/6
Therefore, the area between the curves from x = 0 to x = 1 is 1/6 square units.
Now moving on to part b) where you integrate with respect to y, the steps are slightly different:
1. Solve the equations for x in terms of y to find the x-values as functions of y.
For y = sqrt(x), square both sides to get x = y^2.
For y = x, the equation is already solved for x.
2. Determine the bounds of integration with respect to y by finding the y-values at which the curves intersect. In this case, the curves intersect at y = 0 and y = 1.
3. Set up the integral by integrating the difference between the x-values as functions of y:
A = ∫(right curve - left curve) dy from y = 0 to y = 1
In this case, since you want to integrate with respect to y, you will use the equation x = sqrt(y) as the right curve and x = y as the left curve.
The integral would be: A = ∫(sqrt(y) - y) dy from y = 0 to y = 1.
4. Evaluate the integral:
Integrate (sqrt(y) - y) with respect to y:
A = (∫sqrt(y) dy) - (∫y dy) = (2/3)y^(3/2) - (1/2)y^2
Evaluate the integral from y = 0 to y = 1:
A = [(2/3)(1)^(3/2) - (1/2)(1)^2] - [(2/3)(0)^(3/2) - (1/2)(0)^2]
Simplify the expression:
A = (2/3) - (1/2) = 1/6
Therefore, the area between the curves from y = 0 to y = 1 is also 1/6 square units.
You can verify this result by graphing the curves on your TI-83 calculator and using the "INTERSECT" function to find the points of intersection. Additionally, you can use the "fnInt" function on your calculator to directly perform the integrations for both parts a) and b) and compare the results.