Use the graph of f(t) = 2t +2 on the interval [−1, 4] to write the function F(x), where
f(x)= integral from 1 to x f(t) dt
f(x)=x^2+3x
f(x)=x^2+2x-12 My Answer
f(x)x^2+2x-3
f(x)=x^2+4x-8
Find the range of the function f(x)=integral from 0 to x (sqrt 4-t^2) dt
(0, 4pi)
(0, 2pi) My Answer
(-4,0)
(0,4)
?[1,x] 2t+2 dt
= t^2+2t [1,x]
= (x^2+2x)-(1^2+2*1)
= x^2+2x-3
?[0,x] ?(4-t^2) dt
= 1/2 ?(4-t^2) + 2 arcsin(t/2) [0,x]
= 1/2 ?(4-x^2) + 2arcsin(x/2) - (1/2 ?(4-0) + 2 arcsin(0))
= 1/2 ?(4-x^2) + 2arcsin(x/2) - 1
The domain is clearly [-2,2]
The range is [-?-1,?-1]
http://www.wolframalpha.com/input/?i=range+1%2F2+%E2%88%9A(4-x%5E2)+%2B+2arcsin(x%2F2)+-+1
Well that was my issue with the second one is I originally got that answer but it isn't an option.
To write the function F(x) using the given graph of f(t) = 2t + 2 on the interval [−1, 4], we need to find the integral of f(t) from 1 to x. Let's go step by step:
1. First, let's find the integral of f(t) = 2t + 2 with respect to t. The indefinite integral of 2t + 2 is t^2 + 2t + C, where C is the constant of integration.
2. Next, let's evaluate this integral from 1 to x. To do this, we subtract the value of the integral at 1 from the value of the integral at x.
F(x) = ∫[1 to x] (2t + 2) dt
= [t^2 + 2t] evaluated from 1 to x
= (x^2 + 2x) - (1^2 + 2(1))
= (x^2 + 2x) - (1 + 2)
= x^2 + 2x - 3
Therefore, the correct function F(x) is F(x) = x^2 + 2x - 3.
For the second question, to find the range of the function f(x) = ∫[0 to x] (√(4 - t^2)) dt, where x is in the range [0, 2π], we need to evaluate the integral and find the values of f(x) over this range.
1. First, let's find the integral of √(4 - t^2) with respect to t. We can use a trigonometric substitution to solve this integral.
Let t = 2sin(θ), then dt = 2cos(θ)dθ, and when t = 0, θ = 0, and when t = x, θ = arcsin(x/2).
Therefore, the integral becomes:
∫[0 to x] (√(4 - t^2)) dt = ∫[0 to arcsin(x/2)] (√(4 - (2sin(θ))^2)) (2cos(θ)) dθ
= 2 ∫[0 to arcsin(x/2)] (√(4 - 4sin^2(θ))) cos(θ) dθ
2. Simplify the integrand. Using the trigonometric identity sin^2(θ) + cos^2(θ) = 1, we can rewrite the integrand as:
√(4 - 4sin^2(θ)) cos(θ) = √(4cos^2(θ)) cos(θ) = 2cos^2(θ)
3. Evaluate the integral:
∫[0 to arcsin(x/2)] 2cos^2(θ) dθ = ∫[0 to arcsin(x/2)] (1 + cos(2θ))/2 dθ
Using the trigonometric identity cos(2θ) = 2cos^2(θ) - 1, we can rewrite the integral as:
∫[0 to arcsin(x/2)] (1 + cos(2θ))/2 dθ = ∫[0 to arcsin(x/2)] (1/2) dθ + ∫[0 to arcsin(x/2)] (cos(2θ)/2) dθ
= (1/2)(θ) + (1/4)sin(2θ) evaluated from 0 to arcsin(x/2)
= (1/2)(arcsin(x/2)) + (1/4)sin(2(arcsin(x/2)))
4. Simplify further using trigonometric identities.
(1/2)(arcsin(x/2)) + (1/4)sin(2(arcsin(x/2)))
= (1/2)(arcsin(x/2)) + (1/4)(2sin(arcsin(x/2))cos(arcsin(x/2)))
= (1/2)(arcsin(x/2)) + (1/2)sin(arcsin(x/2))cos(arcsin(x/2))
= (1/2)(arcsin(x/2)) + (1/2)(x/2)sqrt(1 - x^2/4)
= (1/2)(arcsin(x/2)) + (x/2)sqrt(1 - x^2/4)
5. Now we have the function f(x) as:
f(x) = (1/2)(arcsin(x/2)) + (x/2)sqrt(1 - x^2/4)
To find the range of f(x), we need to find the output values of f(x) for inputs in the given range. By evaluating f(x) for different values of x in the range [0, 2π], we can find the range of f(x).
Based on your options, (0, 2π) is correct because the range of f(x) is indeed (0, 2π).