Find the exact area of the region that lies beneath the curve y = 3(x^[1/3]) 0 ≤ x ≤ 27
find the exact area of the region that lies beneath the curve y=6cubesqrtx 0≤x≤27
Well, well, well, it looks like we've got ourselves a mathematical conundrum here. To find the exact area under the curve, we need to integrate. But fear not, my intrepid friend, for I am here to guide you through this hairy situation with a dash of humor.
So, let's start by setting up the integral. We want to find the area under the curve y = 3(x^(1/3)) from x = 0 to x = 27. To do this, we'll integrate with respect to x:
∫[0 to 27] 3(x^(1/3)) dx
Now, let's tackle this integral with enthusiasm! First, we can pull out the constant 3:
3 ∫[0 to 27] (x^(1/3)) dx
The integral of x raised to the power of 1/3 is x^(4/3) divided by 4/3 (because who doesn't love fractions?):
3 * (3/4) * (x^(4/3) evaluated from 0 to 27
Now, let's evaluate this expression at the upper limit (27) and lower limit (0):
3 * (3/4) * (27^(4/3) - 0^(4/3))
Wait a minute, anything raised to the power of 0 is 1, so the second term is simply 0. This simplifies things quite a bit:
3 * (3/4) * (27^(4/3))
Now let's calculate this beastly expression:
3 * (3/4) * (81)
Multiply, add a sprinkle of division, and we get:
81/4 * 81
Take a deep breath, my friend, for we're almost done:
6561/4
And there you have it! The exact area under the curve y = 3(x^(1/3)) from x = 0 to x = 27 is a glorious 6561/4. So let's celebrate this mathematical triumph with a grand round of applause... or maybe a few laughs!
To find the exact area of the region that lies beneath the curve y = 3(x^(1/3)) where 0 ≤ x ≤ 27, you can use the definite integral.
The area can be calculated by integrating the function from the lower limit of 0 to the upper limit of 27, as follows:
∫[0 to 27] 3(x^(1/3)) dx
First, we need to find the integral of the function 3(x^(1/3)):
∫ 3(x^(1/3)) dx = ∫ 3x^(1/3) dx
To integrate x^(1/3), you can add 1 to the exponent and divide by the new exponent:
= 3 * (3/2)x^(4/3) + C
Next, we evaluate the integral within the given limits:
= 3 * (3/2)(27)^(4/3) - 3 * (3/2)(0)^(4/3)
= 3 * (3/2)(27)^(4/3)
Now, we substitute the value for 27:
= 3 * (3/2)(3^3)^(4/3)
= 3 * (3/2)(3^4)
= 3 * (3/2)(81)
= 3 * (3/2) * 81
= (9/2) * 81
= 729/2
Hence, the exact area of the region that lies beneath the curve y = 3(x^(1/3)) where 0 ≤ x ≤ 27 is 729/2 square units.
To find the area of the region that lies beneath the curve y = 3(x^(1/3)) for the interval 0 ≤ x ≤ 27, we need to integrate the function over that interval.
First, we can rewrite the equation as follows: y = 3x^(1/3).
To integrate this function, we will use the definite integral. The integral of the function f(x) over the interval [a, b] is denoted as ∫(a to b) f(x) dx.
In this case, we want to find the area beneath the curve from x = 0 to x = 27, so we need to calculate the integral ∫(0 to 27) 3x^(1/3) dx.
To integrate this function, we can use the power rule for integration, which states that for any real number n not equal to -1, the integral of x^n dx is (x^(n+1))/(n+1).
Using this rule, we can now find the integral of 3x^(1/3) dx:
∫(0 to 27) 3x^(1/3) dx = (3/((1/3)+1)) * x^((1/3)+1) evaluated from 0 to 27.
Simplifying further:
(3/(4/3)) * x^((1/3)+1) evaluated from 0 to 27
= (9/4) * x^(4/3) evaluated from 0 to 27.
To find the area, we evaluate the definite integral at the upper and lower limits:
= (9/4) * (27^(4/3)) - (9/4) * (0^(4/3))
= (9/4) * (27^(4/3)) - 0
= (9/4) * (81) (since 27^(4/3) = (3^3)^(4/3) = 3^4 = 81)
= 9 * 81
= 729.
Therefore, the exact area of the region that lies beneath the curve y = 3(x^(1/3)) for 0 ≤ x ≤ 27 is 729 square units.
that would, of course, be
∫[0,27] 3x^(1/3) dx