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]?

Aug 31, 2015

For $f$ as stated, the derivative is: $f ' \left(x\right) = - \sqrt{4 + \sec \left(x\right)}$

Explanation:

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

$f \left(x\right) = {\int}_{a}^{x} g \left(t\right) \mathrm{dt}$ for $x$ in some interval $\left[a , b\right]$ on which $g$ is continuous.

$f ' \left(x\right) = g \left(x\right)$

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

$f \left(x\right) = {\int}_{x}^{\pi} \sqrt{4 + \sec \left(t\right)} \mathrm{dt}$

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.

$f \left(x\right) = - {\int}_{\pi}^{x} \sqrt{4 + \sec \left(t\right)} \mathrm{dt}$,

so $f ' \left(x\right) = - \sqrt{4 + \sec \left(x\right)}$

If the intended function is

$f \left(x\right) = {\int}_{\pi}^{x} \sqrt{4 + \sec \left(t\right)} \mathrm{dt}$, the omit the reversing of limits of integration and we get $f ' \left(x\right) = \sqrt{4 + \sec \left(x\right)}$