Answer:
[tex]5(e^3-e) + 8[/tex]
Step-by-step explanation:
Given the integral function [tex]\int\limits^3_1 {e^x} \, dx = e^3 - e[/tex], we are to use the result to evaluate [tex]\int\limits^3_1 ({5e^x+4}) \, dx[/tex], to do this the following steps must be followed.
Step 1: Write the integral sum [tex]\int\limits^3_1 ({5e^x+4}) \, dx[/tex] as sum of its individual integral:
[tex]\int\limits^3_1 ({5e^x+4}) \, dx = \int\limits^3_1 {5e^x} \, dx + \int\limits^3_1 {4} \, dx\\\int\limits^3_1 ({5e^x+4}) \, dx = 5\int\limits^3_1 {e^x} \, dx + 4\int\limits^3_1 {} \, dx\\[/tex]
Step 2: integrate the resulting individual integrals
[tex]\int\limits^3_1 ({5e^x+4}) \, dx = 5\int\limits^3_1 {e^x} \, dx + 4\int\limits^3_1 {} \, dx\\\\since \int\limits^3_1 {e^x}dx = e^3-e\ hence;\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + 4\int\limits^3_1 {} \, dx\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + [4x]\limits^3_1\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + [4(3)-4(1)]\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + [12-4]\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + 8 \\\\[/tex]
