\int\limits^3_t \, dx where t = - 3

Question

`\int\limits^3_tx+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(\frac12x^2+2x\right)\bigg|_(x=-3)^(x=-2)+\left(\frac12x^2+2x\right)\bigg|_(x=-2)^(x=3)`

`=\frac12+\frac{25}2=13`