# How do you use the Fundamental Theorem of Calculus to find the derivative of int {1} / {1+t^{2}} dt from x to 5?

Oct 27, 2017

#### Answer:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{5} \frac{1}{1 + {t}^{2}} \setminus \mathrm{dt} = - \frac{1}{1 + {x}^{2}}$

#### Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{x} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(x\right)$ for any constant $a$

(ie the derivative of an integral gives us the original function back).

We are asked to find:

$E = \frac{d}{\mathrm{dx}} \setminus {\int}_{x}^{5} \frac{1}{1 + {t}^{2}} \setminus \mathrm{dt}$

(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using integral properties:

$E = \frac{d}{\mathrm{dx}} \setminus - {\int}_{5}^{x} \frac{1}{1 + {t}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = - \frac{d}{\mathrm{dx}} \setminus {\int}_{5}^{x} \frac{1}{1 + {t}^{2}} \setminus \mathrm{dt}$

We can now apply the FTOC to get;

$E = - \frac{1}{1 + {x}^{2}}$