🔥 Day 137 — 2025 MIT Integration Bee (Qualifying) Q12 Evaluate: ∫ sqrt( x * cbrt( x * 4th-root( x * 5th-root( x * … ) ) ) ) dx This “infinite nested radical” looks wild, but it has a self-similar structure. If you call the whole expression y, the nest shows that y = sqrt(x * y). Solve that tiny equation first—then the integral is instant. 👉 Hints • Let y be the entire radical. • Self-similarity ⇒ y = sqrt(x * y) ⇒ y^2 = x y. • Nonzero branch gives y = x (for x ≥ 0). Would you spot the self-sim
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🔥 Day 137 — 2025 MIT Integration Bee (Qualifying) Q12 Evaluate: ∫ sqrt( x * cbrt( x * 4th-root( x * 5th-root( x * … ) ) ) ) dx This “infinite nested radical” looks wild, but it has a self-similar structure. If you call the whole expression y, the nest shows that y = sqrt(x * y). Solve that tiny equation first—then the integral is instant. 👉 Hints • Let y be the entire radical. • Self-similarity ⇒ y = sqrt(x * y) ⇒ y^2 = x y. • Nonzero branch gives y = x (for x ≥ 0). Would you spot the self-sim
206.5K views4 months ago
TikTokbrainrot.mathematics
🔥 Day 152 — 2026 MIT Integration Bee (Qualifying) Q18 365 Problems for 365 Days Evaluate: Integral of (sin^2(x)/x^2 − sin(2x)/x) dx This one is a “recognize the derivative” speed problem. The integrand looks messy, but it’s literally the derivative of a super simple expression if you think “quotient rule trig identity.” If you try to integrate directly, it’s pain. If you spot the structure, it’s instant. Hints: \t•\tLook for something like (sin^2 x)/x \t•\tDifferentiate it with the quotient rul
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🔥 Day 152 — 2026 MIT Integration Bee (Qualifying) Q18 365 Problems for 365 Days Evaluate: Integral of (sin^2(x)/x^2 − sin(2x)/x) dx This one is a “recognize the derivative” speed problem. The integrand looks messy, but it’s literally the derivative of a super simple expression if you think “quotient rule trig identity.” If you try to integrate directly, it’s pain. If you spot the structure, it’s instant. Hints: \t•\tLook for something like (sin^2 x)/x \t•\tDifferentiate it with the quotient rul
216.5K views4 months ago
TikTokbrainrot.mathematics
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🔥 Day 133 — 2025 MIT Integration Bee (Qualifying) Q17 Evaluate: ∫ sin(x) * sinh(x) dx Wild mix of circular and hyperbolic functions. The fastest path is a classic double “integration by parts” loop: set up I = ∫ sin(x) sinh(x) dx, integrate by parts twice, and watch I appear on both sides so you can solve for it. 👉 Hints • First parts: u = sin x, dv = sinh x dx. • Second parts (on the new integral): u = cos x, dv = cosh x dx. • You’ll end with 2I = (sin x) cosh x − (cos x) sinh x. Would you ca
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🔥 Day 133 — 2025 MIT Integration Bee (Qualifying) Q17 Evaluate: ∫ sin(x) * sinh(x) dx Wild mix of circular and hyperbolic functions. The fastest path is a classic double “integration by parts” loop: set up I = ∫ sin(x) sinh(x) dx, integrate by parts twice, and watch I appear on both sides so you can solve for it. 👉 Hints • First parts: u = sin x, dv = sinh x dx. • Second parts (on the new integral): u = cos x, dv = cosh x dx. • You’ll end with 2I = (sin x) cosh x − (cos x) sinh x. Would you ca
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TikTokbrainrot.mathematics
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🔥 Day 128 — 2025 MIT Integration Bee (Qualifying) Q11 Evaluate: Integral from 0 to 20 of floor(x)^2 dx This is a classic step-function trap. Don’t try to “anti-differentiate” floor(x). Instead, split the interval into unit chunks and add rectangles. Once you see the pattern, it turns into a clean sum of squares. 👉 Hints • On [k, k 1), floor(x) = k. • Integral becomes sum_{k=0}^{19} k^2. • Use the formula for 1^2 2^2 … n^2. Would you spot the step-function strategy fast enough? ⏱️ 💭 Comment yo
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🔥 Day 132 — 2025 MIT Integration Bee (Qualifying) Q16 Evaluate: ∫ [ sin(4x) * cos(x) ] / [ cos(2x) * sin(x) ] dx It looks nasty, but this one collapses fast. Expand sin(4x), cancel what you can, and the integrand turns into a simple power of cos. From there it’s a one-liner with the double-angle identity. 👉 Hints • sin(4x) = 2 sin(2x) cos(2x) = 4 sin x cos x cos(2x) • Most things cancel → 4 cos^2 x • Use cos^2 x = (1 cos 2x)/2 Would you spot the cancellations under time pressure? ⏱️ 💭 Comment
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🔥 Day 146 — 2026 MIT Integration Bee (Qualifying) Q12 Evaluate: Integral from 0 to 1 of sqrt(x^2 x sqrt(x^2 x sqrt(x^2 x …))) dx This is an “infinite nested radical” problem that looks impossible… until you use self-similarity. Let the whole radical be y. Then the inside contains y again, which gives you a clean equation to solve before you integrate. Hints: \t•\tLet y = sqrt(x^2 x sqrt(x^2 x …)) \t•\tThen y = sqrt(x^2 x y) \t•\tSolve for y on [0, 1], then integrate 📣 Project Mentor: free guid
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TikTokbrainrot.mathematics
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