Answer:
1
Explanation:
Part d
∫f′(t)dt as long as the function is continuous
Let u = 6-2x
du = -2dx so -1/2 du = dx
upper limit = 6- 2(4) = 6-8 = -2
lower limit = 6 - 2(2) = 6 - 4 = 2
∫f′(u)(-1/2)du limits -2 to 2 =-1/2( f(-2) - f(2))
f(-2) is 1
f(2) is 3
-1/2(f(-2) - f(2)) = -1/2( 1-3) = -1/2(-2) = 1