Answer:
a) 3
b) 3
c) 0
d) 15
Explanations:
Given the following integral functions
[tex]\int_0^3f(x)dx=6\text{ and }\int_3^6f(x)=-3[/tex]
a) We are to evaluate the following integrals:
[tex]\begin{gathered} \int_0^6f(x)=\int_0^3f(x)+\int_3^6f(x) \\ \int_0^6f(x)=6+(-3) \\ \int_0^6f(x)=3 \end{gathered}[/tex]
b)
[tex]\begin{gathered} \int_6^3f(x)dx=-\int_3^6f(x) \\ \int_6^3f(x)dx=-(-3) \\ \int_6^3f(x)dx=3 \end{gathered}[/tex]
c) For this integral since the upper and lower limit are the same, the integral of the function will always be zero.
[tex]\int_3^3f(x)dx=0[/tex]
d) Note that every constant inside an integral sign will always come out of the integral. Therefore;
[tex]\begin{gathered} \int_3^6-5f(x)dx=-5\int_3^6f(x)dx \\ \int_3^6-5f(x)dx=-5(-3) \\ \int_3^6-5f(x)dx=15 \end{gathered}[/tex]
