# calculus

integrate 2/((2x-7)^2) from x=4 to 6

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1. Let 2x -7 = u
so dx = du/2
Integrand becomes
2/u^2*(du/2) = du/u^2 from u = 1 to 5.
= -1/u @ u = 5 -
(-1/u @ u = 1)
= 1 - 1/5 = 4/5

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2. 👎 0
2. 2/((2x-7)^2)
= 2(2x-7)^-2

integral of that is
-(2x-7)^-1 or -1/(2x-7)

and its value form x=4 to 6
= -1/5 + 1
= 4/5

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