David Troyer

2021-12-30

Calculate the integral.
${\int }_{0}^{6}|3-x|dx$

Archie Jones

Step 1
Given: $I={\int }_{0}^{6}|3-x|dx$
for evaluating given integral, we break in two parts and integrate it
Step 2
so,
$I={\int }_{0}^{6}|3-x|dx$
$={\int }_{0}^{3}\left(3-x\right)dx+{\int }_{3}^{6}\left(x-3\right)dx$
$\left(\because \int \left(x-a\right)dx=\frac{{\left(x-a\right)}^{2}}{2}+c\right)$
$=-\frac{1}{2}{\left[{\left(x-3\right)}^{2}\right]}_{0}^{3}+\frac{1}{2}{\left[{\left(x-3\right)}^{2}\right]}_{3}^{6}$
$=-\frac{1}{2}\left[{\left(3-3\right)}^{2}-{\left(0-3\right)}^{2}\right]+\frac{1}{2}\left[{\left(6-3\right)}^{2}-{\left(3-3\right)}^{2}\right]$
$=-\frac{1}{2}\left[0-9\right]+\frac{1}{2}\left[9-0\right]$
$=\frac{9}{2}+\frac{9}{2}$
=9
hence, given integral is equal to 9.

Ana Robertson

Step 1
Given:
${\int }_{0}^{6}|3-x|dx$
Step 2
Solution
${\int }_{0}^{3}3-xdx+{\int }_{3}^{6}-\left(3-x\right)dx$
Evaluate the integrals
$\frac{9}{2}+\frac{9}{2}$
Calculate the sum
Step 3
9

Vasquez

$\int \left(3-x\right)dx$
$\int \left(3-x\right)dx=-\frac{{x}^{2}}{2}+3x$
Let's calculate a definite integral:
${\int }_{0}^{6}3-xdx=\left(-\frac{{x}^{2}}{2}+3x\right){|}_{0}^{6}$
F(6)=0
F(0)=0
I=6-(0)=6