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# How do you use part I of the Fundamental Theorem of Calculus to find the derivative of: f(x)=_{x }∫^{3}(((1/3)t^{2}) - 1)^{7}dt?

**Solution:**

The Fundamental Theorem of Calculus Part 1 states:

If f is continuous on [a, b] the F(x) = \(\int_{a}^{x}\) f(t)dt is continuous on [a,b] and differentiable on (a,b) and its derivative is f(x);

F’(x) = \(\frac{\mathrm{d}\int_{a}^{x}f(t)dt }{\mathrm{d} x}\) = f(x)------> (1)

f(x) = \(\int_{x}^{3}(((\frac{1}{3})t^{2})-1)^{7}dt\)

F’(x) = \(\frac{\mathrm{d} }{\mathrm{d} x}\int_{x}^{3}(((\frac{1}{3})t^{2})-1)^{7}dt\) ------>(2)

Using the u-substitution we get:

t^{2} / 3 - 1 = u ------> (3)

(2/3)tdt = du

dt = (3/2t )du ------> (4)

To find t we use equation (3) from where we get

⇒ t^{2} = 3( u+1)

t = √3√u + 1 ------> (5)

Substituting (5) in (4) we get

dt = 3 / ( 2√3√u+1) du ------>(6)

dt = √3 / ( 2√u + 1) du ------> (7)

Substituting (7) in (2) we get

= \(\frac{\mathrm{d} }{\mathrm{d} x}\int_{x}^{3}(u)^{7}(\frac{\sqrt{3}}{2\sqrt{u+1}})du\)

= \(\frac{\mathrm{d} }{\mathrm{d} x}\int_{3}^{x}-(u)^{7}(\frac{\sqrt{3}}{2\sqrt{u+1}})du\)

= - \(\frac{\sqrt{3}}{2}\frac{x^{7}}{\sqrt{x+1}}\)

## How do you use part I of the Fundamental Theorem of Calculus to find the derivative of: f(x)= _{x }∫^{3}(((1/3)t^{2}) - 1)^{7}dt?

**Summary:**

By using part I of the Fundamental Theorem of Calculus the derivative of:f(x)=_{x }∫^{3}(((1/3)t^{2}) - 1)^{7}dt is given by - \(\frac{\sqrt{3}}{2}\frac{x^{7}}{\sqrt{x+1}}\)

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