009B Sample Midterm 3, Problem 4

The rate of reaction to a drug is given by:

$r'(t)=2t^{2}e^{-t}$

where  $t$  is the number of hours since the drug was administered.

Find the total reaction to the drug from  $t=1$  to  $t=6.$

Foundations:
If we calculate  $\int _{a}^{b}r'(t)~dt,$  what are we calculating?
We are calculating  $r(b)-r(a).$  This is the total reaction to the
drug from  $t=a$  to  $t=b.$

Solution:

Step 1:
To calculate the total reaction to the drug from  $t=1$  to  $t=6,$
we need to calculate
$\int _{1}^{6}r'(t)~dt=\int _{1}^{6}2t^{2}e^{-t}~dt.$

Step 2:
We proceed using integration by parts.
Let  $u=2t^{2}$  and  $dv=e^{-t}dt.$
Then,  $du=4t~dt$  and  $v=-e^{-t}.$
Then, we have
$\int _{1}^{6}2t^{2}e^{-t}~dt=\left.-2t^{2}e^{-t}\right|_{1}^{6}+\int _{1}^{6}4te^{-t}~dt.$

Step 3:
Now, we need to use integration by parts again.
Let  $u=4t$  and  $dv=e^{-t}dt.$
Then,  $du=4dt$  and  $v=-e^{-t}.$
Thus, we get
${\begin{array}{rcl}\displaystyle {\int _{1}^{6}2t^{2}e^{-t}~dt}&=&\displaystyle {\left.(-2t^{2}e^{-t}-4te^{-t})\right|_{1}^{6}+\int _{1}^{6}4e^{-t}}\\&&\\&=&\displaystyle {\left.(-2t^{2}e^{-t}-4te^{-t}-4e^{-t})\right|_{1}^{6}}\\&&\\&=&\displaystyle {(-2(6)^{2}e^{-6}-4(6)e^{-6}-4e^{-6})-(-2(1)^{2}e^{-1}-4(1)e^{-1}-4e^{-1})}\\&&\\&=&\displaystyle {{\frac {-100+10e^{5}}{e^{6}}}.}\end{array}}$

Final Answer:
${\frac {-100+10e^{5}}{e^{6}}}$

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