# If int_1^3 \ f(x) \ dx = 5 and int_3^8 \ f(x) \ dx = 10, what is  int_1^8 \ f(x) \ dx ?

Jan 13, 2018

${\int}_{1}^{8} \setminus f \left(x\right) \setminus \mathrm{dx} = 15$

#### Explanation:

Using the properties of definite integral we have:

${\int}_{a}^{c} \setminus f \left(x\right) \setminus \mathrm{dx} = {\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx} + {\int}_{b}^{c} \setminus f \left(x\right) \setminus \mathrm{dx}$

Hence we can write:

${\int}_{1}^{8} \setminus f \left(x\right) \setminus \mathrm{dx} = {\int}_{1}^{3} \setminus f \left(x\right) \setminus \mathrm{dx} + {\int}_{3}^{8} \setminus f \left(x\right) \setminus \mathrm{dx}$
$\text{ } = 5 + 10$
$\text{ } = 15$

Jan 13, 2018

${\int}_{1}^{8} f \left(x\right) \mathrm{dx} = 15$.

#### Explanation:

We are asked to find ${\int}_{1}^{8} f \left(x\right) \mathrm{dx} ,$ which can be thought of as the total area between $f \left(x\right)$ and the $x$-axis, from $x = 1$ on the left to $x = 8$ on the right.

Just like how the area of any geometric shape can be found by breaking it into 2 pieces and adding together the two smaller areas, an integral ${\int}_{a}^{c} f \left(x\right) \mathrm{dx}$ over an interval $\left[a , c\right]$ can be found by breaking it into two smaller intervals, $\left[a , b\right]$ and $\left[b , c\right]$, and adding together their values. In other words, for any $b \in \left(a , c\right) :$

${\int}_{a}^{c} f \left(x\right) \mathrm{dx} = {\int}_{a}^{b} f \left(x\right) \mathrm{dx} + {\int}_{b}^{c} f \left(x\right) \mathrm{dx}$

Here, we are given ${\int}_{1}^{3} f \left(x\right) \mathrm{dx} = 5$ and ${\int}_{3}^{8} f \left(x\right) \mathrm{dx} = 10.$ So, we can compute:

${\int}_{1}^{8} f \left(x\right) \mathrm{dx} = {\int}_{1}^{3} f \left(x\right) \mathrm{dx} + {\int}_{3}^{8} f \left(x\right) \mathrm{dx}$

$\textcolor{w h i t e}{{\int}_{1}^{8} f \left(x\right) \mathrm{dx}} = \text{ "5" "+" } 10$
$\textcolor{w h i t e}{{\int}_{1}^{8} f \left(x\right) \mathrm{dx}} = 15$.