# To find total cost, to the nearest cent, to cool the house for this 24-hour period

To find total cost, to the nearest cent, to cool the house for this 24-hour period
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Alannej
Given:
Rate at which the cost of cooing the house accumulates = \$0.05
Concept Used:
Applications of integral
Calculation:
Rate at which the cost of cooing the house accumulates
$=0.05{\int }_{5.231}^{18.770}\left(80-10\mathrm{cos}\left(\frac{\pi }{12}\right)-78\right)dt$
$=0.05{\int }_{5.231}^{18.770}\left(2-10\mathrm{cos}\left(\frac{\pi t}{12}\right)\right)dt$
$=0.05\left[2{\int }_{5.231}^{18.770}dt-10{\int }_{5.231}^{18.770}\mathrm{cos}\left(\frac{\pi t}{12}\right)dt\right]$
$=0.05\left[2\left(t\right){|{}_{\left(5.231\right)}^{18.770}-\frac{120}{\pi }\left(\mathrm{sin}\left(\frac{\pi t}{12}\right)\right)|}_{5.231}^{18.770}\right]$
putting the limits to get answer
$=0.05{\int }_{5.231}^{18.770}\left(80-10\mathrm{cos}\left(\frac{\pi }{12}\right)-78\right)dt=5.096$