Find the derivative of F(x)= integral from -2 to x of sqrt(t^3+8)dt?
My answer: sqrt(x^3+8)? Is this right?
Yes, the value of the function at the upper limit where t = x.
To find the derivative of the function F(x) = ∫ from -2 to x of sqrt(t^3 + 8) dt, we can apply the Fundamental Theorem of Calculus.
According to the Fundamental Theorem of Calculus, if a function F(x) is equal to the integral of another function f(t) from a constant a to x, then the derivative of F(x) with respect to x is equal to f(x).
First, let's find the integrand. The integrand is sqrt(t^3 + 8). Now, let's find its derivative with respect to t.
d/dt (sqrt(t^3 + 8)) = (1/2) * (t^3 + 8)^(-1/2) * 3t^2.
Now, we can apply this result to find the derivative of F(x).
F'(x) = sqrt(x^3 + 8).
Therefore, your answer is correct. The derivative of F(x) is sqrt(x^3 + 8).