Question

Some food is placed in a freezer. After t hours, the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where r(t) = 12 + 4/(t+3)^2. (a) Compute the area under the graph

Answer 1

SOLUTION

GIVEN

Some food is placed in a freezer. After t hours, the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where

$\displaystyle \sf{r(t) = 12 + \frac{4}{ {(t + 3)}^{2} } }$

TO DETERMINE

(a) Compute the area under the graph y = r(t) for the interval 0 ≤ t ≤ 2

EVALUATION

Here it is given that Some food is placed in a freezer.

After t hours, the temperature of the food is dropping at the rate of r(t) degrees Fahrenheit per hour, where

$\displaystyle \sf{r(t) = 12 + \frac{4}{ {(t + 3)}^{2} } }$

Hence the area under the graph y = r(t) for the interval 0 ≤ t ≤ 2

$\displaystyle \sf{ = \int\limits_{0}^{2} r(t) \, dt }$

$\displaystyle \sf{ = \int\limits_{0}^{2} \bigg[12 + \frac{4}{ {(t + 3)}^{2} } \bigg ] \, dt }$

$\displaystyle \sf{ = \bigg[12t - \frac{4}{ {(t + 3)}^{} } \bigg ]_{0}^{2} }$

$\displaystyle \sf{ = \bigg[24 - \frac{4}{ {(2 + 3)}^{} } \bigg ] - \bigg[0 - \frac{4}{ {(0+ 3)}^{} } \bigg ] }$

$\displaystyle \sf{ = 24 - \frac{4}{5} + \frac{4}{3} }$

$\displaystyle \sf{ = 24 + \frac{20 - 12}{15} }$

$\displaystyle \sf{ = 24 + \frac{8}{15} }$

$\displaystyle \sf{ = 24 + 0.533}$

$\displaystyle \sf{ = 24 .533}$

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