Question

Given that 3 ex dx 1 = e3 − e, use this result to evaluate 3 5ex + 4 dx. 1

Answer 1

Answer:

$5(e^3-e) + 8$

Step-by-step explanation:

Given the integral function $\int\limits^3_1 {e^x} \, dx = e^3 - e$, we are to use the result to evaluate $\int\limits^3_1 ({5e^x+4}) \, dx$, to do this the following steps must be followed.

Step 1: Write the integral sum $\int\limits^3_1 ({5e^x+4}) \, dx$ as sum of its individual integral:

$\int\limits^3_1 ({5e^x+4}) \, dx = \int\limits^3_1 {5e^x} \, dx + \int\limits^3_1 {4} \, dx\\\int\limits^3_1 ({5e^x+4}) \, dx = 5\int\limits^3_1 {e^x} \, dx + 4\int\limits^3_1 {} \, dx$

Step 2: integrate the resulting individual integrals

$\int\limits^3_1 ({5e^x+4}) \, dx = 5\int\limits^3_1 {e^x} \, dx + 4\int\limits^3_1 {} \, dx\\\\since \int\limits^3_1 {e^x}dx = e^3-e\ hence;\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + 4\int\limits^3_1 {} \, dx\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + [4x]\limits^3_1\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + [4(3)-4(1)]\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + [12-4]\\\\\int\limits^3_1 ({5e^x+4}) \, dx = 5(e^3-e) + 8$