Can someone help me check my answers
1. Find the range of the function f(x) = x∫-6 √36-t^2 dt
- [-6, 0]
- [0, 6]
- [0, 9π] (my answer)
- [0, 18π]
2. Use the graph of f(t) = 2t + 2 on the interval [-1, 4] to write the function F(x), where f(x) = x∫-1 f(t) dt
a. F(x) = x^2 + 3x
b. F(x) = x^2 + 2x - 12 (my answer)
c. F(x) = x^2 + 2x - 3
d. F(x) = x^2 + 4x - 8
I guess you mean
f(x) = x∫-6 [√(36-t^2) ] dt
in general ∫ (p^2-u^2)^.5 du = (1/2)[ u sqrt(p^2-u^2)+p^2(sin^-1(u/p)
here p = 6 , u = t
(1/2) [ t sqrt (36-t^2) + 36 sin^-1 (t/6) ] at t = x - at t = -6
(1/2) [ x sqrt(36-x^2) +36 sin^-1 (x/6)]
- (1/2) [-6sqrt(36-36) + 36 sin^-1(-1)]
= (1/2) [ x sqrt (36-x^2) +36 sin^-1(x/6) - 36 sin^-1 (-1)]
= (x/2) sqrt (36-x^2) + 18 sin^-1(x/6) - 18 sin^-1(-1)
well we can not take sqrt of a negative so x is between -6 and+ 6
when x = 0 I get
0 + 18(0 or pi) -18 (3 pi/4)
I am not sure I am going to get any of those answers, not sure I understood you question.
To check your answers for these questions, we can walk through the process of how to find the range of the function and how to write the function using the given graph.
1. Finding the range of the function f(x) = x∫-6 √36-t^2 dt:
To find the range of this function, we need to evaluate the definite integral. In this case, the integral of √36-t^2 dt represents the area between the curve and the x-axis from t = -6 to t = x. When we integrate, we find the antiderivative of the function inside the integral.
Since the integral is with respect to t, we can treat x as a constant when evaluating the integral. The integral of √36-t^2 dt is equal to (1/2) * (t * √(36 - t^2) + 18 * arcsin(t/6)) + C, where C is the constant of integration.
To find the range, we need to evaluate the integral at the upper limit of integration, which is x. When we plug in x to the expression above and simplify, we get:
f(x) = (1/2) * (x * √(36 - x^2) + 18 * arcsin(x/6)) + C
Now, to find the range of the function, we can evaluate f(x) at its endpoints, -6 and 0:
Range = [f(-6), f(0)]
= [(1/2) * (-6 * √(36 - (-6)^2) + 18 * arcsin(-6/6)) + C, (1/2) * (0 * √(36 - 0^2) + 18 * arcsin(0/6)) + C]
Simplifying this expression, we can obtain the range of the function.
2. Writing the function F(x) using the graph of f(t) = 2t + 2 on the interval [-1, 4]:
To find the function F(x), we need to evaluate the integral of f(t) with respect to t and then replace t with x. In this case, the integral of 2t + 2 dt represents the area between the curve and the t-axis from t = -1 to t = x. When we integrate, we find the antiderivative of the function inside the integral.
The integral of 2t + 2 dt is equal to t^2 + 2t + C, where C is the constant of integration.
To find the function F(x), we can replace t with x in the expression above:
F(x) = x^2 + 2x + C
Now, to determine the value of the constant C, we can use the fact that F(x) is equal to the definite integral of f(t) from -1 to x.
Since the graph of f(t) = 2t + 2 on the interval [-1, 4] is a line, the area under the curve represents a trapezoid. We can calculate the area of this trapezoid using the formula (1/2) * (base1 + base2) * height.
The area of the trapezoid is equal to the definite integral of f(t) from -1 to 4. This means we need to evaluate ∫[-1, 4] (2t + 2) dt.
Evaluating this integral, we get:
∫[-1, 4] (2t + 2) dt = [(t^2 + 2t) from -1 to 4]
= [(4^2 + 2 * 4) - (-1^2 + 2 * (-1))]
= [18]
Now, since F(x) is equal to the definite integral of f(t), F(x) = ∫[-1, x] (2t + 2) dt.
So, F(x) = x^2 + 2x + 18
Based on this explanation, it seems that your answer for question 1 was correct, but your answer for question 2 was incorrect. The correct answer for question 2 should be:
b. F(x) = x^2 + 2x + 18