Answer :
Given
we have a function
[tex]f(x)=-\frac{x^2}{2}+x-\frac{5}{2};[-1,0][/tex]Required
we need to find the value of c that satisfy the mean value theorem for integrals.
Explanation
we know that
[tex]f(c)=\frac{1}{b-a}\int_a^bf(x)dx[/tex]Here a= -1 and b=0
so
[tex]\begin{gathered} f(c)=\frac{1}{0-(-1)}\int_{-1}^0-\frac{x^2}{2}+x-\frac{5}{2}dx \\ f(c)=[-\frac{x^3}{6}+\frac{x^2}{2}-\frac{5}{2}x]_{-1}^0=-\frac{19}{6} \\ i.e.\text{ }-\frac{c^2}{2}+c-\frac{5}{2}=-\frac{19}{6} \\ c^2-2c+5=\frac{19}{3} \\ 3c^2-6c+15-19=0 \\ 3c^2-6c-4=0 \\ \end{gathered}[/tex]Here c has two values 2.52 and -0.526, but 2.52 does not lie in the interval . So the correct answer is c=-0.526