Find the range of the function F(x)=integral from 0 to x of √(4-t^2) dt
a) [0,4π]
b) [0,π]
c) [-4,0]
d) [0,4]
Well, finding the range of a function can be quite challenging sometimes. Thankfully, the square root of a negative number can't make up its mind, so it decides to stay out of this problem. We only need to focus on the values that make sense. Since the integral represents an area (not a circus tent, mind you), we can safely say that the range of F(x) is [0, 4π]. So, the answer is a) [0, 4π]. Just remember, no clowns allowed in the range!
To find the range of the function F(x) = ∫√(4-t^2) dt from 0 to x, we can evaluate the integral and see its behavior as x varies.
First, let's find the indefinite integral of √(4-t^2). We can use a trigonometric substitution to do this. Let t = 2sinθ, which implies dt = 2cosθ dθ. Substituting these values into the integral, we get:
∫√(4-t^2) dt = ∫√(4-(2sinθ)^2) (2cosθ) dθ
= 2∫√(4-4sin^2θ)cosθ dθ
= 2∫√(4(1-sin^2θ)) cosθ dθ
= 2∫√(4cos^2θ) cosθ dθ
= 2∫2cosθ cosθ dθ
= 4∫cos^2θ dθ
Using the identity cos^2θ = (1 + cos(2θ))/2, we can simplify the integral further:
4∫cos^2θ dθ = 2∫(1 + cos(2θ)) dθ.
Integrating, we obtain:
F(x) = 2(θ + (1/2)sin(2θ)) + C,
where C is the constant of integration.
Next, we need to find the limits of θ as x varies. From the substitution t = 2sinθ, we have x = 2sinθ. Solving for θ, we get:
θ = sin^(-1)(x/2).
Now, let's consider the range of F(x). As x varies, θ will also vary from 0 to sin^(-1)(x/2). Since sinθ is a increasing function for θ in the range [0, π/2], we can deduce that F(x) is also increasing in the range [0, π/2].
So, the range of F(x) is [F(0), F(π/2)]. Evaluating F(x) at x = 0 and x = π/2:
F(0) = 2(0 + (1/2)sin(0)) + C = C.
F(π/2) = 2(π/2 + (1/2)sin(π)) + C = π + C.
Thus, the range of the function F(x) is [C, π + C], where C is the constant of integration.
Since C can take any value, the range of F(x) can be any interval of the form [C, π + C], where C is any real number.
Therefore, none of the given options (a), (b), (c), or (d) accurately represent the range of the function F(x).
int (2^2-t^2)^.5 dt = (1/2)[t sqrt(4-t^2) +4 sin^-1(t/2) ]
at t = 0 that is 0
at t = x
(1/2)[x sqrt(4-x^2) +4 sin^-1(x/2) ]
= (x/2) sqrt(4-x^2) + 2 sin^-1 (x/2)
well, |x| better be less than 2 or that sqrt is imaginary
If x = 0, it is 0
if x = 2, it is 2 pi/2 or pi so I would pick 0 to pi
when x = -2pi
-pi
Thank you both! I worked it out and also got π for when x=2 and -π for when x=-2 and so since that isn't an answer choice I'm wondering if it is 0 to π because the integral is from 0 to x?? Is that the assumption you're referring to?
hmm. We have
f(x) = x/2 √(4-x^2) + 2arcsin(x/2)
now, the domain of f(x) is [-2,2],
Now, you might think that the range would thus be the sum of the ranges of
x/2 √(4-x^2): [-1,1]
2arcsin(x/2): [-π,π]
But 2arcsin(x/2) is strictly increasing on [-2,2]
while x/2 √(4-x^2) dips from 0 to -1 and back to 0, and then rises to 1 and drops back to 0 again. So, it does not extend the range of f(x), which is ultimately just [-π,π]
Now, that's not one of the choices. Why not? What assumption must be made?