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

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:

t² / 3 -  1 = u   ------> (3)

(2/3)tdt = du

dt = (3/2t )du       ------> (4)

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

⇒  t²  =  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)t²) - 1)⁷dt?

Summary:

 By using  part I of the Fundamental Theorem of Calculus the derivative of:f(x)=_x ∫³(((1/3)t²) - 1)⁷dt is given by  - \(\frac{\sqrt{3}}{2}\frac{x^{7}}{\sqrt{x+1}}\)