Question

use part one of the fundamental theorem of calculus to find the derivative of the function.
g(w)=\int_{0}^{w}\sin(8 + t^{9})dt
g(w)=\square

Explanation:

Step1: Recall the Fundamental Theorem of Calculus Part 1

If \(g(w)=\int_{a}^{w}f(t)dt\), then \(g^{\prime}(w) = f(w)\) by the Fundamental Theorem of Calculus Part 1. Here \(a = 0\) and \(f(t)=\sin(8 + t^{9})\).

Step2: Apply the theorem

Substitute \(t = w\) into \(f(t)\). So \(g^{\prime}(w)=\sin(8+w^{9})\).

Answer:

\(\sin(8 + w^{9})\)