Let F(x)= x^2
∫ sqrt(1+t^3)dt
x
Find F'(4).
I applied the FTC and thought that F'(4) = f(4) which in this case would be sqrt(1+4^3), giving me an answer of sqrt(65), however this is incorrect. I'm not sure how to proceed with the question.
That x^2 should be the upper limit on the integral and x is the lower limit on the integral
You forgot the chain rule.
F'(x) = √(1+(x^2)^3)*(2x) - √(1+x^3)*(1)
F'(4) = √(1+4096)(8)-√(1+64)
visit this url and scroll down about halfway -- there is a good example with variable limits of integration.
https://en.wikipedia.org/wiki/Leibniz_integral_rule
To find F'(4), we need to differentiate the function F(x) = x^2. The derivative of F(x) with respect to x is given by dF/dx.
In this case, F(x) = x^2, and we want to find F'(4), so we need to compute dF/dx at x = 4.
To differentiate F(x) = x^2, we can use the power rule of differentiation. According to the power rule, if we have a function g(x) = x^n, then the derivative of g(x) with respect to x is given by dg/dx = nx^(n-1).
In our case, we have F(x) = x^2, so we will use the power rule to differentiate it. Applying the power rule, we get:
dF/dx = 2x^(2-1) = 2x
Now, to find F'(4), we substitute x = 4 into the derivative, getting:
F'(4) = 2(4) = 8
Therefore, the value of F'(4) is 8.
It seems like you have applied the Fundamental Theorem of Calculus (FTC) incorrectly in your approach. The FTC states that if F(x) is an antiderivative of a function f(x), then ∫f(x)dx from a to b equals F(b) - F(a). However, it does not imply that the derivative of F(x) at a particular point x is equal to f(x).
In this case, F'(4) is not equal to f(4). Instead, we need to differentiate F(x) to find its derivative at x = 4.