Question

14t/(6t^2 + 8) = f(t) where f (t) is in thousands of antibodies per minute, and t is measured in minutes with 0≤t≤110. Assuming that there are no antibodies present at time t =0, find the total number of antibodies in the blood at the end of 110 minutes. Round your answer to two decimal places.

Answer 1

\[=1000\int\limits_{0}^{110}\frac{14t}{6t^{2}+8}dt\]
\[u = 6t^{2} +8\]
\[du =12tdt \rightarrow dt =\frac{du}{12t}\]
\[=1000\int\limits\limits_{0}^{110}\frac{14t}{12t*u}du=1000*\frac{7}{6}\int\limits_{0}^{110}\frac{du}{u}\]
\[=\frac{7000}{6} \ln u = \frac{7000}{6}\ln (6t^{2}+8) {0 \rightarrow 110}\]
\[=\frac{7000}{6}[\ln (6*110^{2}+8) - \ln 8]\]
\[\approx 10,632.29\]