Find the range of the function F(x)= definite integral from [-6,x] of sqrt(36-t^2)dt.
[0, 36π]
[0, 18π]***
[-6, 6]
[-6, 0]
The region is a semi-circle of radius 6, so the area ranges from 0 to 18π
The integral is the area moving from the left side to some point x in [-6,6].
i know this question has something to do with semicircles and radians. I got the starred answer as my own but I'm not sure if it's correct or how I got there. Help thanks!!
To find the range of the function F(x), we first need to evaluate the definite integral and see how it behaves as x varies.
The integral is given as:
F(x) = ∫[-6,x]√(36-t^2)dt
To evaluate the integral, we can use the substitution u = t/6. This gives us du = dt/6, or dt = 6 du.
Converting the limits of integration, when t = -6, u = -1 and when t = x, u = x/6.
The integral becomes:
F(x) = ∫[-6,x]√(36-t^2)dt = ∫[-1, x/6]√(36-(36u^2))(6du)
Simplifying the integral:
F(x) = ∫[-1, x/6]√(36(1-u^2))(6du) = 6∫[-1, x/6]√(1-u^2)du
Recognizing that the integral represents the area under the curve of a semicircle with radius 6, we can see that the function F(x) is equal to a constant multiple of the area of a semicircle.
The area of a semicircle with radius 6 is πr^2/2 = 36π/2 = 18π.
Therefore, the range of the function F(x) is [0, 18π].
To find the range of the function F(x), which is defined as the definite integral from -6 to x of sqrt(36 - t^2) dt, we need to evaluate the integral and determine the values that F(x) can take.
First, let's find the antiderivative of sqrt(36 - t^2) with respect to t. Using the substitution u = 36 - t^2, we have du = -2t dt. Rearranging, we get dt = -du / (2t). Substituting these values into the integral, we have:
F(x) = ∫(sqrt(36 - t^2)) dt
= ∫ (sqrt(u)) * (-du / (2t))
= -1/2 * ∫((sqrt(u)) / t) du
Next, we need to determine the limits of the integral. The given function is the definite integral from -6 to x, so the limits of integration are -6 and x.
F(x) = -1/2 * ∫((sqrt(u)) / t) du, from -6 to x
Evaluating the integral, we have:
F(x) = -1/2 * [∫((sqrt(u)) / t) du] evaluated from -6 to x
To find the values that F(x) can take, we substitute the limits of integration into the integral expression:
F(x) = -1/2 * [∫((sqrt(u)) / t) du] evaluated from -6 to x
= -1/2 * (∫ ((sqrt(u)) / t) du) evaluated at x minus (∫ ((sqrt(u)) / t) du) evaluated at -6
Since the antiderivative of sqrt(u) / t is (2/3) * (u^(3/2) / sqrt(36 - x^2)), we have:
F(x) = -1/2 * [(2/3) * (u^(3/2) / sqrt(36 - t^2))] evaluated from -6 to x
= -1/2 * [(2/3) * ((36 - x^2)^(3/2) / sqrt(36 - x^2))] - (-1/2 * [(2/3) * (36^(3/2) / sqrt(36 - 36^2))]
Simplifying further, we have:
F(x) = -1/2 * (2/3) * ((36 - x^2)^(3/2) / sqrt(36 - x^2)) + (1/2 * (2/3) * (36^(3/2) / sqrt(36)))
= -1/3 * ((36 - x^2)^(3/2) / sqrt(36 - x^2)) + (1/3 * (2 * 36^(3/2)) / 6)
Simplifying the expression, we get:
F(x) = -1/3 * ((36 - x^2)^(3/2) / sqrt(36 - x^2)) + (2/9 * 36^(3/2))
Now, let's find the range of F(x). The range is the set of all possible values that F(x) can take as x varies.
Since the term ((36 - x^2)^(3/2) / sqrt(36 - x^2)) always results in a positive value for -6 ≤ x ≤ 6 (since the square root and exponentiation are always positive), and 36^(3/2) = 36 * (sqrt(36)) = 36 * 6 = 216, the range of F(x) is:
[0, (2/9) * 216]
= [0, 48]
Therefore, the correct answer is none of the given options. The range of the function F(x) is [0, 48].