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musickatia [10]
3 years ago
14

ts^3_t{|x+2|} \, dx " alt=" \int\limits^3_t{|x+2|} \, dx " align="absmiddle" class="latex-formula">

where t = - 3
Mathematics
1 answer:
photoshop1234 [79]3 years ago
7 0
\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

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
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