Attempt this tough 4u question (1 Viewer)

[Dr Mathematics]

This question was written by me.

 

[WeiWeiMan]

This question was written by me.

what the fuck am i reading

 

[Average Boreduser]

what the fuck am i reading

watch the mf solutions be one line

 

[liamkk112]

This question was written by me.

for the first part, we know that aR = bQ; so
aR -ab^2 = bQ-ab^2
a(R-b^2) = b(Q-ab)
thus a/b = (Q-ab)/(R-b^2)

 

[WeiWeiMan]

This question was written by me.

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)

Spoiler: my answer

I=ln4

(c)

Spoiler: solution steps

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 dx

Spoiler: my answer

I=1/2 ln2

 

[Sisyphus1015]

nice wei

 

[Dr Mathematics]

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)

Spoiler: my answer

I=ln4

(c)

Spoiler: solution steps

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 dx

Spoiler: my answer

I=1/2 ln2

This is correct.

 

[Dr Mathematics]

Here is another interesting one:

 

[WeiWeiMan]

Here is another interesting one:View attachment 43774

Spoiler: possibly dodgy solution

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]

 

[Dr Mathematics]

Spoiler: possibly dodgy solution

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]

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.
Nonetheless your solution is still valid and shows great depth of understanding.

 

[Dr Mathematics]

Another problem that is interesting.