Dr Mathematics
New Member
- Joined
- Aug 4, 2024
- Messages
- 6
- Gender
- Undisclosed
- HSC
- N/A
This question was written by me.
Attachments
-
102.8 KB Views: 59
what the fuck am i readingThis question was written by me.
watch the mf solutions be one linewhat the fuck am i reading
for the first part, we know that aR = bQ; soThis question was written by me.
nglThis question was written by me.
This is correct.ngl
(a) what liamkk112 did
(b) maybe say cross multiply instead of partial fractions (idk how you use partial fractions but cross multiplying worked for me)
(c)I=ln4
the denominator is an annoying ass factorisation basically (a-b)(a^n-1+a^n-2b+a^n-3b^2 +...+b^n-1) = a^n-b^n
integral ends up being ∫_0^pi/4 tanx dxI=1/2 ln2
Here is another interesting one:View attachment 43774
A simpler solution is to consider areas: Notice the first integral from b to a represents the area under the curve from b to a (by definition). Similarly the second integral represents the area between the curve and the y axis from d to c (by definition). Since f(b)=d and f(a)=c, these 2 areas are a larger continuous area. This area is a large rectangle with the corner nearest to the origin missing. The missing component has base length b and height d as these were the areas not included in our integral ( namely 0 to b and 0-d). Thus the area is the large rectangle minus bd. The large rectangle simply has side lengths a and c (as our integrals only went up to a and c) thus its area is ac. Hence the total area is ac-bd.monotonically increasing hence one to one
thus f^-1(y) = f^-1(f(x)) = x,
blah blah blah somehow show that f^-1'(x) = 1/f'(x) type shit
using substitution u=f^-1(y) on the second integral
∫_c^d f^-1(y)dy = ∫_b^a uf'(u) du = ∫_b^a xf'(x) dx = [xf(x)]_b^a - ∫_b^a f(x) dx
adding first and second integrals:
∫_b^a f(x) dx+∫_c^d f^-1(y)dy=[xf(x)]_b^a - ∫_b^a f(x) dx+∫_b^a f(x) dx=[xf(x)]_b^a = af(a)-bf(b) = ac-bd [quod erat demonstrandum type shit]