When we integrate a function f(t), we generate its antiderivative F(t)
[tex]\int f(t)dt=F(t)[/tex]When we integrate on an interval, the result will be the new function evaluated on the upper boundary minus the new function evaluated on the lower boundary.
[tex]\int_a^bf(t)dt=F(b)-F(a)[/tex]item a)
Evaluating this integral, we have
[tex]\int_{11}^{11}f(t)dt=F(11)-F(11)=0[/tex]item b)
When we invert the limits of integration, we invert the sign of the integral
[tex]\begin{gathered} \int_{15}^3f(t)dt=F(3)-F(15) \\ \begin{equation*} =-(F(15)-F(3)) \end{equation*} \\ =-\int_3^{15}f(t)dt \\ =-6 \end{gathered}[/tex]item c)
The integral of a function times a constant, is the constant value times the result of the integral
[tex]\int kf(t)dt=k\int f(t)dt[/tex]Using this property in our problem, we have
[tex]\int_3^{15}7f(t)dt=7\int_3^{15}f(t)dt=7\cdot6=42[/tex]item d)
The integral of a sum is the sum of the integrals
[tex]\int f(t)+g(t)dt=\int f(t)dt+\int g(t)dt[/tex]and the antiderivative of the unitary function is the variable itself
[tex]\int1dt=t[/tex]Using those two properties in our problem, we have
[tex]\begin{gathered} \int_3^{15}f(t)+7dt \\ =\int_3^{15}f(t)dt+\int_3^{15}7dt \\ =\int_3^{15}f(t)dt+7\int_3^{15}1dt \\ =6+7(15-3) \\ =6+7(12) \\ =90 \end{gathered}[/tex]item e)
If we sum two integrals where the upper boundary of the first is equal to the lower boundary of the second, it is the same as calculating the integral from the lower boundary of the first to the upper boundary of the second
[tex]\int_a^bf(t)dt+\int_b^cf(t)dt=\int_a^cf(t)dt[/tex]Using this property in our problem, we have
[tex]\int_3^{11}f(t)dt+\int_{11}^{15}f(t)dt=\int_3^{15}f(t)dt=6[/tex]item f)
The average value of a function on an interval [a, b] is given by
[tex]\frac{\int_a^bf(t)dt}{b-a}[/tex]Then, the average value of the function on the interval [3, 15] is
[tex]\frac{\int_3^{15}f(t)dt}{15-3}=\frac{6}{12}=\frac{1}{2}[/tex]