Question
Evaluate the following integrals:
$\int^\limits4_{-4}|\text{x}+2|\text{dx}$
$\int^\limits4_{-4}|\text{x}+2|\text{dx}$
✓
Answer
We have,
$\int^\limits4_{-4}|\text{x}+2|\text{dx}$
$=\int^\limits{-2}_{-4}-(\text{x}+2)\text{dx}+\int^\limits{4}_{-2}(\text{x}+2)\text{dx}$
$=-\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^{-2}_{-4}+\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^4_{-2}$
$=-\bigg[\Big(\frac{4}{2}-4-\Big(\frac{16}{2}-8\Big)\bigg]+\bigg[\Big(\frac{16}{2}+8\Big)-\Big(\frac{4}{2}-4\Big)\bigg]$
$=-\big[(-2)-(0)\big]+\big[(16)-(-2)\big]$
$=-\big[-2\big]+\big[16+2\big]$
$=2-18$
$=20$
$\int^\limits4_{-4}|\text{x}+2|\text{dx}$
$=\int^\limits{-2}_{-4}-(\text{x}+2)\text{dx}+\int^\limits{4}_{-2}(\text{x}+2)\text{dx}$
$=-\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^{-2}_{-4}+\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^4_{-2}$
$=-\bigg[\Big(\frac{4}{2}-4-\Big(\frac{16}{2}-8\Big)\bigg]+\bigg[\Big(\frac{16}{2}+8\Big)-\Big(\frac{4}{2}-4\Big)\bigg]$
$=-\big[(-2)-(0)\big]+\big[(16)-(-2)\big]$
$=-\big[-2\big]+\big[16+2\big]$
$=2-18$
$=20$
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