Question

$\int\limits^3_t{|x+2|} \, dx$

where t = - 3

Answer 1

$\displaystyle\int_{-3}^3|x+2|\,\mathrm dx$

Recall the definition of absolute value:

$|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}$

So we can split up the integration interval at $x=-2$ and apply this definition to rewrite the integral as

$\displaystyle\int_{-3}^{-2}(-(x+2))\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx$
$=\displaystyle-\int_{-3}^{-2}(x+2)\,\mathrm dx+\int_{-2}^3(x+2)\,\mathrm dx$
$=-\left(\dfrac12x^2+2x\right)\bigg|_{x=-3}^{x=-2}+\left(\dfrac12x^2+2x\right)\bigg|_{x=-2}^{x=3}$
$=\dfrac12+\dfrac{25}2=13$