(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Calculate the integral:

from 0 to pi/2

I=[itex]\int[/itex]sin x {2f′(cos x) − 1} dx

When f(0) = 2 and f(1) = 4

2. Relevant equations

3. The attempt at a solution

integral from zero to pi/2 of 2[itex]\int[/itex]sin(x){f'(Cos (x) -1}dx

u = cos (x)

du = -sinx dx

(Here is where I run into problems....i do not know know to do with the -1)

1. We insert the limits of integration into the U = cos (x) and get

0 = cos (pi/2)

1 = cos (1)

so now write

-2[itex]\int[/itex]f'(u)-1du (limits of integration are from 1 to zero

flip limits of integration

2[itex]\int[/itex] f'u-1du (limits of integration are zero to 1)

integrate

2(F(1)-1) - (2(F(0)-0)

2(4-1) - 2(2-0)

6 - 4 = 2.

I tried 2 in my HM system and it is incorrect. What am I doing wrong?

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# Question regarding integration

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