How do you use part I of the Fundamental Theorem of Calculus to find the derivative of: f(x) = ∫(((1/3)t²) - 1)⁷dt from x to 3?

Solution:

Given, f(x) = \(\int (((\frac{1}{3}t^{2})-1))^{7}\) from x to 3.

We have to find the derivative using part 1 of the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus states that

\(\frac{d}{dx}\int_{a}^{x}f(t)\, dt=f(x)for any constant a\)

Now, f’(x) where f(x) = \(\int_{x}^{3}(\frac{1}{3}t^{2}-1)^{7}dt\)

\(f'(x)=\frac{d}{dx}\int_{x}^{3}(\frac{1}{3}t^{2}-1)^{7}dt\)

We have to change the limits of integration to get the function into the correct form.

\(f'(x)=\frac{d}{dx}(-\int_{x}^{3}(\frac{1}{3}t^{2}-1)^{7}dt)\)

\(f'(x)=-\frac{d}{dx}\int_{x}^{3}(\frac{1}{3}t^{2}-1)^{7}dt\)

Now, we can directly apply the Fundamental Theorem of Calculus,

\(f'(x)=-(\frac{1}{3}(x)^{2}-1)^{7}\)

Therefore, the derivative is \(f'(x)=-(\frac{1}{3}(x)^{2}-1)^{7}\).