Puzzling
Favorite Answer
Let's start with the ranges:
u = 8x - 2
The lower limit was x = 3. Plug that in to find the lower limit for u:
u = 8(3) - 2 = 22
The upper limit was x = 6. Plug that in to find the upper limit for u:
u = 8(6) - 2 = 46
Apparently you had these but the screenshot only shows the first digit of each.
Next calculate du/dx:
d(8x - 2)/dx = 8
If du/dx = 8, then dx = du/8
Replace 8x - 2 with u.
Replace dx with du/8
Original:
9 / (8x - 2)² dx
Substituted:
9 / u² * du/8
= (9/8) 1/u² du
Answer:
integral from 22 to 46 of (9/8) 1/u² du
I'll leave the evaluation to you so you hopefully learn how to do this for yourself next time.
rotchm
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Wayne DeguMan
If u = 8x - 2, then du = 8.dx
Hence, dx = du/8
so, ∫ [9/(8x - 2)²] dx => ∫ (9/u²)(du/8)
i.e. (9/8) ∫ (1/u²).du
As this is in terms of u, we need to convert the original limits of x in terms of u
so, with x = 6 we have, u = 8(6) - 2 = 46
with x = 3 we have, u = 8(3) - 2 = 22
:)>
Ray
I think the definite integral becomes ∫ 9/(8u²) du and the new limits should be 22 and 46.