Question

simplify the following expression.

$$\frac{d}{dx}int_{6}^{x}(3t^{2}+t + 8)dt$$

$$\frac{d}{dx}int_{6}^{x}(3t^{2}+t + 8)dt = square$$

Explanation:

Step1: Apply Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus, Part 1 states that if \( F(x)=\int_{a}^{x}f(t)dt \), then \( F^\prime(x) = f(x) \), where \( a \) is a constant and \( f \) is continuous on the interval \([a,x]\). Here, \( f(t)=3t^{2}+t + 8 \), \( a = 6 \), and we are taking the derivative with respect to \( x \) of the integral from 6 to \( x \) of \( f(t)dt \).

Step2: Substitute \( x \) into \( f(t) \)

Using the theorem, we substitute \( x \) for \( t \) in the function \( f(t) \). So \( f(x)=3x^{2}+x + 8 \).

Answer:

\( 3x^{2}+x + 8 \)