Answer:
A) 12
Step-by-step explanation:
Notice that [tex]\int\limits^5_1 {f(x)} \, dx=-\int\limits^1_5 {f(x)} \, dx[/tex] by the exchange-of-limits property of integrals, so [tex]-\int\limits^1_5 {f(x)} \, dx=-1[/tex]
We also know that [tex]\int\limits^1_{-1} {f(x)} \, dx+\int\limits^5_1 {f(x)} \, dx=\int\limits^5_{-1} {f(x)} \, dx[/tex] by the additivity property of integrals, which means that:
[tex]\int\limits^1_{-1} {f(x)} \, dx+\int\limits^5_1 {f(x)} \, dx=\int\limits^5_{-1} {f(x)} \, dx\\\\\int\limits^1_{-1} {f(x)} \, dx-\int\limits^1_5 {f(x)} \, dx=\int\limits^5_{-1} {f(x)} \, dx\\\\5-1=\int\limits^5_{-1} {f(x)} \, dx\\\\4=\int\limits^5_{-1} {f(x)} \, dx[/tex]
Therefore, [tex]3\int\limits^5_{-1} {f(x)} \, dx=3(4)=12[/tex]
