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).
it's 6 just did the test.
what test??? please help
To find F″(1), we need to take the second derivative of the given function. Let's break it down step by step:
1. Let's start with the original function F(x). We are given that F(x) equals the integral from 1 to 3 times x of ln(t^2). We can express it as:
F(x) = ∫(1 to 3x) ln(t^2) dt
2. To proceed, we need to find the first derivative, F'(x), of F(x) with respect to x.
F'(x) = d/dx [ ∫(1 to 3x) ln(t^2) dt ]
3. Since the integral is a function of x, we will apply the Fundamental Theorem of Calculus. According to the theorem, if we have a function F(x) defined as an integral, we can differentiate it by evaluating the integrand at the upper limit, multiplied by the derivative of the upper limit with respect to x.
F'(x) = [ln((3x)^2) * d/dx (3x)] - [ln((1)^2) * d/dx (1)]
4. Simplifying the expression, we obtain:
F'(x) = [ln(9x^2) * 3] - [ln(1) * 0]
F'(x) = 3ln(9x^2)
5. Finally, to find F″(1), we need to differentiate F'(x) once more with respect to x:
F″(x) = d/dx [3ln(9x^2)]
F″(x) = 3 * (1/(9x^2)) * 18x
F″(x) = 6/x
6. Now that we have the second derivative, we can calculate F″(1) by substituting x with 1:
F″(1) = 6/1
F″(1) = 6
Therefore, F″(1) equals 6.
Please note that the above steps provide the analytical approach to solving the problem. To use a calculator, you would need to enter the original function and its derivatives into the calculator's mathematical software or programming capabilities to obtain the answer directly.
so, is
F(x) = ∫[1,3x] ln(t^2) dt
or
F(x) = ∫[1,3x] (ln t)^2 dt
?
Either way,
f(x) = F'(x) = ln(3x)^2 * 3
or
f(x) = F'(x) = (ln 3x)^2 * 3
And F" = f'
Recall that the 2nd Fundamental Theorem of Calculus says that if
F(x) = ∫[a,g(x)] f(t) dt
then
F'(x) = f(g(x)) * g'(x)