we do
how much added-howmuch subtracted
[tex] \int\limits^5_0 {F(t)} \, dt [/tex] - [tex] \int\limits^5_0 {E(t)} \, dt [/tex]
or
[tex] \int\limits^5_0 {F(t)-E(t)} \, dt [/tex]
so
first simlify F(t)-E(t) (if you want, not necicary if use calculator)
(t+7-ln(t+4))/(t+2)=F(t)-E(t)
evaluate
[tex]\int\limits^5_0 \frac{t+7-ln(t+4)}{t+2} } \, dt[/tex]≈9.05584
about 9 pints
