Question

Hello! Could someone help me out with part d? Please show your work!

Answer 1

Answer:

1

Step-by-step 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