Find the derivative of F(x)= integral from -2 to x of sqrt(t^3+8)dt?
My answer: sqrt(x^3+8)?
To find the derivative of the function F(x) = ∫(-2 to x)√(t^3 + 8)dt, we can use the Fundamental Theorem of Calculus which states that if f(x) is continuous on the interval [a, b] and F(x) is the integral of f(x) from a to x, then the derivative of F(x) is equal to f(x).
In this case, we have F(x) = ∫(-2 to x)√(t^3 + 8)dt.
To differentiate F(x), we need to evaluate the integrand function inside the integral with respect to x. However, since the limits of integration also depend on x, we need to apply the Chain Rule.
Let's say that G(x) is the function inside the integral, i.e., G(x) = √(x^3 + 8).
Then, we have F(x) = ∫(-2 to x)G(t)dt.
Now, to differentiate F(x), we need to differentiate G(x) and multiply by the derivative of the upper limit of integration, in this case, x.
So, G'(x) = (3x^2)/(2√(x^3 + 8)), which is found by differentiating the function G(x) = √(x^3 + 8).
Finally, we can write the derivative of F(x) as:
F'(x) = G(x) * d/dx(x) + ∫(-2 to x)G(t)dt * d/dx(x)
Since the derivative of x with respect to x is 1, we can simplify this as:
F'(x) = G(x) + ∫(-2 to x)G(t)dt
Substituting G(x) = √(x^3 + 8) into this equation, we get:
F'(x) = √(x^3 + 8) + ∫(-2 to x)√(t^3 + 8)dt
Therefore, the derivative of F(x) is indeed √(x^3 + 8) + ∫(-2 to x)√(t^3 + 8)dt.