Question

a college language class was chosen for a learning experiment. using a list of 50 words, the experiment measured the rate of vocabulary memorization at different times during a continuous 5 - hour study session. the average rate of learning for the entire class was inversely proportional to the time spent studying, and was given approximately by ( v(t)=\frac{11}{t} ) for ( 1leq tleq5 ). find the area between the graph of ( v ) and the t - axis over the interval (2,4), and interpret the results. the area between the graph of ( v ) and the t - axis over the interval (2,4) is approximately (square). (round to the nearest whole number as needed.) interpret the results. choose the correct answer below. a. the area between the graph of ( v ) and the t - axis over the interval (2,4) is the average number of words learned from ( t = 2 ) hr to ( t = 4 ) hr. b. the area between the graph of ( v ) and the t - axis over the interval (2,4) is the average rate of learning for the entire class from ( t = 2 ) hr to ( t = 4 ) hr. c. the area between the graph of ( v ) and the t - axis over the interval (2,4) is the average difference between the rate of learning at ( t = 2 ) hr and the rate of learning at ( t = 4 ) hr.

Explanation:

Step1: Use the definite - integral formula for area

The area \(A\) between the graph of \(y = V^{\prime}(t)\) and the \(t\) - axis over the interval \([a,b]\) is given by \(A=\int_{a}^{b}V^{\prime}(t)dt\). Here, \(V^{\prime}(t)=\frac{11}{t}\), \(a = 2\), and \(b = 4\). So, \(A=\int_{2}^{4}\frac{11}{t}dt\).

Step2: Apply the integral formula \(\int\frac{1}{x}dx=\ln|x|+C\)

We know that \(\int\frac{11}{t}dt=11\int\frac{1}{t}dt\). Using the formula \(\int\frac{1}{t}dt=\ln(t)+C\) (\(t>0\) in our interval \([2,4]\)), we get \(11\int_{2}^{4}\frac{1}{t}dt=11[\ln(t)]_{2}^{4}\).

Step3: Evaluate the definite integral

By the fundamental theorem of calculus \(F(b)-F(a)\) where \(F(t)=11\ln(t)\), we have \(11(\ln(4)-\ln(2))\). Using the property of logarithms \(\ln(m)-\ln(n)=\ln(\frac{m}{n})\), then \(11\ln(\frac{4}{2})=11\ln(2)\).
Since \(\ln(2)\approx0.6931\), then \(11\times0.6931 = 7.6241\approx8\).

For the interpretation:
The definite integral \(\int_{a}^{b}V^{\prime}(t)dt\) gives the net change of \(V(t)\) over the interval \([a,b]\). Here, \(V(t)\) represents the number of words learned. So the area (definite integral) between the graph of \(V^{\prime}(t)\) and the \(t\) - axis over the interval \([2,4]\) is the average number of words learned from \(t = 2\) hr to \(t = 4\) hr.

Answer:

The area is approximately \(8\).
A. The area between the graph of \(V^{\prime}\) and the \(t\) - axis over the interval \([2,4]\) is the average number of words learned from \(t = 2\) hr to \(t = 4\) hr.