3 years ago

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

Mathematics

1 answer:

koban [17]3 years ago

4 0

Answer:

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

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