AP Calculus AB 8.1 Finding the Average Value of a Function on an Interval - FRQs - Exam Style Questions

(A) Average value of \(C\) on \([0,4]\)
The average number of acres is the average value of \(C\) over \([0,4]\):
\(\displaystyle \text{Avg}=\frac{1}{4-0}\int_{0}^{4} C(t)\,dt=\frac{1}{4}\int_{0}^{4} 7.6\arctan(0.2t)\,dt.\)
Numerically (radians), \(\displaystyle \int_{0}^{4} 7.6\arctan(0.2t)\,dt\approx 11.112896.\)
\(\displaystyle \text{Avg}=\frac{1}{4}\times 11.112896\approx \boxed{2.778\ \text{acres}}\) (to three decimals).

(B) Time when instantaneous rate equals average rate
Average rate of change of \(C\) on \([0,4]\):
\(\displaystyle \frac{C(4)-C(0)}{4-0}=\frac{7.6\arctan(0.8)-0}{4}\approx \frac{5.128031}{4}=1.282008.\)
Set \(C'(t)\) equal to this value and solve:
\(\displaystyle \frac{38}{25+t^{2}}=1.282008\ \Rightarrow\ 25+t^{2}=\frac{38}{1.282008}\approx 29.641006\)
\(\displaystyle t^{2}\approx 29.641006-25=4.641006\ \Rightarrow\ t\approx \sqrt{4.641006}\approx \boxed{2.154\ \text{weeks}}.\)

(C) End behavior of the rate of change
The rate of change is \(C'(t)=\dfrac{38}{25+t^{2}}\). The required limit is
\(\displaystyle \lim_{t\to\infty} C'(t)=\lim_{t\to\infty}\frac{38}{25+t^{2}}=0,\)
since the denominator grows without bound while the numerator is constant.
\(\boxed{0}\)

(D) Time when \(A(t)\) is maximized on \([4,36]\)
By the Fundamental Theorem of Calculus, for \(t>0\):
\(\displaystyle A'(t)=C'(t)-0.1\times \ln t=\frac{38}{25+t^{2}}-0.1\ln t.\)
Critical points occur when \(A'(t)=0\):
\(\displaystyle \frac{38}{25+t^{2}}=0.1\times \ln t.\) (Solve numerically.)
This gives \(t\approx \boxed{11.441700}.\)
To apply the candidates test, evaluate \(A(t)\) at the endpoints and the critical point. First compute the antiderivative needed for the integral:
\(\displaystyle \int 0.1\times \ln x\,dx=0.1\big(x\ln x – x\big)+C.\)
Hence \(\displaystyle \int_{4}^{t} 0.1\times \ln x\,dx=0.1\big(t\ln t – t\big)-0.1\big(4\ln 4-4\big).\)
Now compute the three candidate values (calculator, radians):
\(\displaystyle A(4)=C(4)-\int_{4}^{4}\cdots = 7.6\arctan(0.8)\approx 5.128031.\)
\(\displaystyle A(11.441700)=C(11.441700)-\int_{4}^{11.441700}\cdots \approx 7.316978.\)
\(\displaystyle A(36)=C(36)-\int_{4}^{36}\cdots \approx 1.743056.\)
Therefore the maximum value of \(A\) on \(4\le t\le 36\) occurs at \(\boxed{t\approx 11.442\ \text{weeks}}\) (to three decimals).
Justification: We evaluated \(A\) at both endpoints and the only critical point in the interval and chose the greatest value; additionally, \(A'(t)\) changes sign from positive to negative at \(t\approx 11.442\), confirming an absolute maximum there.