The Fundamental Theorem of Calculus Part 1 states that if a function f is continuous on the interval [a, b] and F is an antiderivative of f on [a, b], then the definite integral from a to b of f(x) dx equals F(b) - F(a).

In this case, we are given the function g(s) = ∫ from 2 to s of (t - t^3)^6 dt. We want to find the derivative of g(s), which is g'(s).

According to the Fundamental Theorem of Calculus Part 1, the derivative of the integral of a function from a constant to a variable is just the original function evaluated at that variable.

So, g'(s) = (s - s^3)^6.

Question

The Fundamental Theorem of Calculus Part 1 states that if a function f is continuous on the interval [a, b] and F is an antiderivative of f on [a, b], then the definite integral from a to b of f(x) dx equals F(b) - F(a).

In this case, we are given the function g(s) = ∫ from 2 to s of (t - t^3)^6 dt. We want to find the derivative of g(s), which is g'(s).

According to the Fundamental Theorem of Calculus Part 1, the derivative of the integral of a function from a constant to a variable is just the original function evaluated at that variable.

So, g'(s) = (s - s^3)^6.

Knowee AI · Accepted Answer

The Fundamental Theorem of Calculus Part 1 states that if a function f is continuous on the interval [a, b] and F is an antiderivative of f on [a, b], then the definite integral from a to b of f(x) dx equals F(b) - F(a).

In this case, we are given the function g(s) = ∫ from 2 to s of (t - t^3)^6 dt. We want to find the derivative of g(s), which is g'(s).

According to the Fundamental Theorem of Calculus Part 1, the derivative of the integral of a function from a constant to a variable is just the original function evaluated at that variable.

So, g'(s) = (s - s^3)^6.

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.g(s) = s(t − t3)6dt2

Question

Solution

Similar Questions

Upgrade your grade with Knowee