How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #f(x)=int sqrt(4+sec(t))dt# from #[x, pi]#?

Question

William Abdalla · Accepted Answer

For #f# as stated, the derivative is: #f'(x) = -sqrt(4+sec(x))#

We can use the fundamental theorem of calculus to find the derivative of a function of the form:

#f(x) = int_a^x g(t) dt# for #x# in some interval #[a,b]# on which #g# is continuous.

#f'(x) = g(x)#

Your question asks about the function defined by integrating from #x# tp #pi#:

#f(x)=int_x^pi sqrt(4+sec(t))dt#

so we first need to reverse the order of integration to go from a constant (in this case #pi#) to the variable, #x#. Reversing the order of integration simply introduces a negative sign.

#f(x)=-int_pi^x sqrt(4+sec(t))dt#,

so #f'(x) = -sqrt(4+sec(x))#

If the intended function is

#f(x)=int_pi^x sqrt(4+sec(t))dt#, the omit the reversing of limits of integration and we get #f'(x) = sqrt(4+sec(x))#

Emma Aldridge · Answer

undefined To use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function $ f(x) = \int_{x}^{\pi} \sqrt{4 + \sec(t)} \, dt $, first, we define a new function, say $ F(x) $, as the integral of $ f(x) $ from a constant lower limit to $ x $:

$$ F(x) = \int_{a}^{x} f(t) \, dt $$

Then, according to Part 1 of the Fundamental Theorem of Calculus, the derivative of $ F(x) $ with respect to $ x $ is equal to the integrand function $ f(x) $:

$$ F'(x) = f(x) $$

So, to find the derivative of $ f(x) $, we just need to evaluate $ F'(x) $. In this case, $ F(x) = \int_{a}^{x} \sqrt{4 + \sec(t)} \, dt $, with $ a $ being a constant lower limit.

Once we have $ F(x) $, we can find its derivative $ F'(x) $ using the chain rule and then substitute $ x = \pi $ into $ F'(x) $ to obtain the derivative of $ f(x) $ at $ x = \pi $.