I apologize in advance for posting a homework problem. Here it is anyway.
An 18-wheeler traveling at speed v mph gets about $4+0.01v$ mpg (miles per gallon) of diesel fuel. If its speed is $80 \dfrac{t+1}{t+2}$ mph at time t, then the amount, in gallons, of diesel fuel used during the first 2 hr is approximately: Answer $23.1$.
I can only think of one approach (which doesn't feel right tbh). Find the total number of miles traveled during first 2 hours of trip:$$\int_0^2 80 \dfrac{t+1}{t+2}\,dt \approx104.548$$ Then, compute for the total gallons per mile, whatever that means. $$\int_0^2\dfrac{1}{4+0.01(80\frac{t+1}{t+2})}\,dt \approx 0.442$$
Multiplying these two values results in $\sim46.239$ gallons, which is twice the actual answer. What am I missing here?
$4+0.01v$ is miles per gallon, $\frac1{4+0.01v}$ is gallons per mile, so $v\cdot\frac1{4+0.01v}$ will be gallons per hour, and that's your integrand. You want to calculate $$\int_0^2\dfrac{80\frac{t+1}{t+2}}{4+0.01\left(80\frac{t+1}{t+2}\right)}\,dt=80\int_0^2\dfrac{t+1}{4.8 t+8.8}\,dt\approx 23.0889.$$