Suppose your annual income is $I ( t ) = 60,000 + 1,800 t ( 0 \leq t \leq 3 )$ dollars per year, where t represents the number of years since you began your job, while your annual expenses are $E ( t ) = 50,000 + 1,400 t ( 0 \leq t \leq 3 )$ dollars per year. Find the area between the graphs of $I ( t )$ and $E ( t )$ for $( 0 \leq t \leq 3 )$ . Round your answer to the nearest dollar.
A)Area = $31,800
B) Area = $33,600
C) Area = $188,100
D) Area = $344,400
E) Area = $156,300

Question

Suppose your annual income is $I ( t ) = 60,000 + 1,800 t ( 0 \leq t \leq 3 )$ dollars per year, where t represents the number of years since you began your job, while your annual expenses are $E ( t ) = 50,000 + 1,400 t ( 0 \leq t \leq 3 )$ dollars per year. Find the area between the graphs of $I ( t )$ and $E ( t )$ for $( 0 \leq t \leq 3 )$ . Round your answer to the nearest dollar.  
A)Area = $31,800
B) Area = $33,600
C) Area = $188,100
D) Area = $344,400
E) Area = $156,300

Quizplus · Accepted Answer

The area between the graphs of $I(t)$ and $E(t)$ represents the total savings over the 3 years. To find this, we calculate the integral of the difference between $I(t)$ and $E(t)$ from $t = 0$ to $t = 3$. The difference $I(t) - E(t) = (60,000 + 1,800t) - (50,000 + 1,400t) = 10,000 + 400t$. Integrating this from 0 to 3 gives $\int_{0}^{3} (10,000 + 400t) dt = [10,000t + 200t^2]_{0}^{3} = (10,000(3) + 200(9)) - (10,000(0) + 200(0)^2) = 30,000 + 1,800 = 31,800$.

Suppose Your Annual Income Is I(t)=60,000+1,800t(0≤t≤3)I ( t ) = 60,000 + 1,800 t ( 0 \leq t \leq 3 )I(t)=60,000+1,800t(0≤t≤3)

Suppose Your Annual Income Is $I ( t ) = 60,000 + 1,800 t ( 0 \leq t \leq 3 )$