Let F of x equals the integral from 1 to 3 times x of the natural logarithm of t squared. Use your calculator to find F″(1).

A. 12
B. 6
C. 4
D. 1/9

Find the range of the function f of x equals the integral from 0 to x of the square root of the quantity 16 minus t squared.

A. [−4, 4]
B. [0, 4]
C. [0, 4π]
D. [0, 8π]

To find the second derivative of F(x), we need to differentiate F'(x) first, and then differentiate F''(x).

Step 1: Differentiate F(x) to find F'(x)
To differentiate F(x), we need to apply the Fundamental Theorem of Calculus and use the Chain Rule.

F(x) = ∫(1 to 3x) ln(t^2) dt

Applying the Chain Rule, we get:
F'(x) = (d/dx)(∫(1 to 3x) ln(t^2) dt)

Using the Fundamental Theorem of Calculus, F'(x) is simply equal to the integrand evaluated at the upper limit of integration (3x) times the derivative of the upper limit (d/dx [3x]).

F'(x) = ln((3x)^2) * d/dx [3x] = ln(9x^2) * 3 = 3ln(9x^2)

Step 2: Differentiate F'(x) to find F''(x)
To find F''(x), we differentiate F'(x) with respect to x.

F''(x) = (d/dx)(3ln(9x^2))
= 3 * d/dx (ln(9x^2))

Applying the Chain Rule, we get:
F''(x) = 3 * (1/(9x^2)) * d/dx (9x^2)
= 3 * (1/(9x^2)) * 18x
= 2/x

Now, to find F''(1), we simply substitute x = 1 into the expression we found for F''(x).

F''(1) = 2/1
= 2

Therefore, the answer is 2, which is not among the options given.

To find F″(1), we need to find the second derivative of F(x) with respect to x and then evaluate it at x = 1.

Step 1: Find the first derivative of F(x).
F'(x) = d/dx [∫(1 to 3x) ln(t^2) dt]
= ln((3x)^2) * d/dx(3x) (using the chain rule)
= 2ln(3x) * 3
= 6ln(3x)

Step 2: Find the second derivative of F(x).
F''(x) = d/dx [6ln(3x)]
= 6 * d/dx(ln(3x))
= 6 * (1/(3x)) * 3
= 6/x

Step 3: Evaluate F''(1).
F''(1) = 6/1
= 6

Therefore, the value of F″(1) is 6.

The answer is:
B. 6

For the second question:

To find the range of the function f(x) = ∫(0 to x) √(16 - t^2) dt, we need to determine the values that f(x) can take for different values of x.

Step 1: Determine the range of the integrand.
The function inside the square root, 16 - t^2, represents a circle centered at the origin with a radius of 4. It ranges from 0 to 16, inclusive.

Step 2: Determine the range of the integral.
The integral is evaluated from 0 to x, so the range of the integral depends on the value of x. As x increases, the integration interval extends further along the positive x-axis.

Step 3: Find the maximum value in the range.
The maximum value occurs when x equals the length of the diameter of the circle. In this case, the diameter is 8, so the maximum value in the range is 8.

Step 4: Find the minimum value in the range.
The minimum value occurs at x = 0, where the integration interval is zero. Therefore, the minimum value in the range is 0.

Therefore, the range of the function f(x) is [0, 8].

The answer is:
D. [0, 8π]

Not sure whether you mean f(t) = ln(t^2) or f(t) = (ln t)^2

In either case, F'(x) = f(3x) * 3
then take the next derivative

#2. You can do the math, recalling arcsines, etc, or just realize that the integral represents the area under the circle of radius 2, in the first quadrant. As you increase x from 0 to 4, the area increases from 0 to 1/4 of a circle, or 4π.