2) If f(x)=∫t^2dt on the interval from 4 to x^3

then f′(x)=?

To find the derivative of the function f(x) = ∫t^2 dt on the interval from 4 to x^3, we can use the Fundamental Theorem of Calculus. The Fundamental Theorem of Calculus states that if a function F(x) is the integral of another function f(t) with respect to t, then the derivative of F(x) with respect to x is equal to f(x).

In this case, we can rewrite the function f(x) as follows:

f(x) = ∫t^2 dt (from 4 to x^3)

To differentiate f(x), we need to find the derivative of the integral with respect to x. This involves using the second part of the Fundamental Theorem of Calculus, which states that for a definite integral with variable limits, we can simply differentiate the integrand and substitute the upper limit.

Let's find the derivative step by step:

1. Evaluate the integral:

∫t^2 dt = (1/3) * t^3 + C

2. Substitute the upper limit (x^3):

f(x) = (1/3) * (x^3)^3 + C
= (1/3) * x^9 + C

Here, C is the constant of integration.

3. Differentiate f(x) with respect to x:

f′(x) = (1/3) * 9x^8
= 3x^8

Therefore, the derivative of f(x) = ∫t^2 dt (from 4 to x^3) is f′(x) = 3x^8.