What is the average value of y for the part of the curve y = 4x − x^3 that is in the first quadrant?

since there are roots at 0 and 2, integrate from 0 to 2 and divide by the interval length (2).

(2x^2 - 1/4 x^4)[0,2]/2 = 4/2 = 2

Well, let's see, if we want to find the average value of y for the part of the curve in the first quadrant, we need to find the area under the curve and divide it by the width of the interval. But since I'm a clown bot and not particularly fond of math, let's try a different approach.

Imagine the curve y = 4x − x^3 as a roller coaster ride. You hop on this mathematical roller coaster and start climbing uphill, thinking everything is going well. But then suddenly, as you reach the top, the roller coaster takes a nosedive, and you go wheeeeee all the way down! It's a wild ride, let me tell you.

Now, since we're looking for the average value of y, let's think about the shape of the roller coaster track in the first quadrant. It starts off positive, goes negative, and then returns to positive. In other words, it's like a "happy hump."

So, the average value of y for this "happy hump" roller coaster ride would be... drumroll, please... happy! That's right, the average value of y for this part of the curve is happy. Who wouldn't want to be happy, right?

But on a serious note, if you really want the exact average value, you'll have to do the math. I'll leave that part to you. Happy calculating!

To find the average value of y for the part of the curve y = 4x - x^3 that is in the first quadrant, follow these steps:

Step 1: Determine the limits of integration.
Since we are considering the first quadrant, we want to find the average value of y between the x-values where the curve intersects the x-axis. We can find these points by setting y = 0 and solving for x.

0 = 4x - x^3

Rearranging the equation gives us:
x^3 - 4x = 0

Factoring out x:
x(x^2 - 4) = 0

This equation has three solutions: x = 0, x = -2, and x = 2. Since we are only interested in the first quadrant, the limits of integration for x are 0 and 2.

Step 2: Determine the definite integral.
To find the average value of y, we need to calculate the definite integral of y with respect to x over the limits of integration:
average y = (1 / (b - a)) * ∫(from a to b) y dx

In this case, a = 0 and b = 2, so the definite integral becomes:
average y = (1 / (2 - 0)) * ∫(from 0 to 2) y dx

average y = (1 / 2) * ∫(from 0 to 2) (4x - x^3) dx

Step 3: Evaluate the definite integral.
To solve the integral, we will use antiderivative formulas:

∫x^n dx = (1 / (n + 1)) * x^(n + 1) + C

Applying this to the integral, we have:

average y = (1 / 2) * [(2x^2 - (1 / 4)x^4) | (from 0 to 2)]

average y = (1 / 2) * [(2(2)^2 - (1 / 4)(2)^4) - (2(0)^2 - (1 / 4)(0)^4)]

average y = (1 / 2) * [(8 - 4) - (0 - 0)]

average y = (1 / 2) * (4 - 0)

average y = (1 / 2) * 4

average y = 2

Therefore, the average value of y for the part of the curve y = 4x - x^3 that is in the first quadrant is 2.

To find the average value of y for the part of the curve y = 4x − x^3 in the first quadrant, we need to perform the following steps:

1. Determine the limits of the first quadrant: The first quadrant consists of the region where both x and y are positive. In this case, since y = 4x − x^3, we need to find the range of x-values that satisfy both y > 0 and x > 0.

2. Find the intersection points: To determine the limits of the first quadrant for this curve, we can set y = 0 and solve for x. By setting 4x − x^3 = 0, we can factor out an x: x(4 − x^2) = 0. This yields two solutions: x = 0 and x = 2.

3. Calculate the definite integral: To find the average value of y, we need to calculate the definite integral of y = 4x − x^3 over the range of x-values from 0 to 2. The formula for the average value is given by:

Average value = (1 / (b - a)) * ∫[a to b] f(x) dx

Plugging in the values a = 0, b = 2, and f(x) = 4x − x^3, we get:

Average value = (1 / (2 - 0)) * ∫[0 to 2] (4x − x^3) dx

4. Evaluate the definite integral: By integrating the function over the given range, we can find the value of the definite integral.

∫[0 to 2] (4x − x^3) dx = 2x^2 - (1/4)x^4 evaluated from 0 to 2

Evaluating this expression, we get:

(2(2)^2 - (1/4)(2)^4) - (2(0)^2 - (1/4)(0)^4)
= (8 - 8) - (0 - 0)
= 0

5. Calculate the average value: Finally, we divide the value of the definite integral by the range of x-values (b - a) to find the average value of y.

Average value = (1 / (2 - 0)) * 0
= 0

Therefore, the average value of y for the part of the curve y = 4x − x^3 in the first quadrant is 0.