In a chemical processing plant, a liquid substance flows through a pipe, and the buildup of sediment in the pipe is causing a reduction in flow efficiency. The rate at which the pipe's cross-sectional area is being blocked by sediment over time 𝑡 (in hours) is modeled by the function 𝑏(𝑡)= 6𝑡+5 (𝑡+1)(𝑡+2) , where 𝑏(𝑡) is the blockage rate in square centimeters per hour. Determine the total area of the pipe blocked by sediment over the first 4 hours of operation. Solution To find the total area of the pipe blocked by sediment over the first 4 hours, we need to integrate the blockage rate function over the interval from 0 to 4 hours. 1. Set up the integral: The total blocked area 𝐴 is given by: 𝐴=∫ 𝑏(𝑡) 4 0  𝑑𝑡 Substitute the given blockage rate function: � �=∫ 6𝑡+5 (𝑡+1)(𝑡+2) 4 0  𝑑𝑡 2. Partial Fractions Decomposition We decompose the function 6𝑡+5 (𝑡+1)(𝑡+2) into partial fractions: 6𝑡+5 (𝑡+1)(𝑡+2) = 𝐴 𝑡+1+ 𝐵 𝑡+2 Multiply both sides by (𝑡+1)(𝑡+2) to eliminate the denominator: 6𝑡+5=𝐴(𝑡+2)+𝐵(𝑡+1) (…) Complete the calculations