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MIT Integration bee PROBLEMS #math #mathematics #manim #integrals
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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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Solve This Challenging MIT Entrance Exam Question
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🔥 Day 139 — 2026 MIT Integration Bee (Qualifying) Q2 Evaluate: ∫ e^(2026·e^x) · e^x dx This is a perfect u-sub “inside-the-exponential” setup. If you set u = 2026·e^x, then du = 2026·e^x dx and the integral collapses immediately. 👉 Hints • u = 2026·e^x • du = 2026·e^x dx → (e^x dx) = du/2026 • Integral becomes (1/2026) ∫ e^u du Would you catch the substitution in time? ⏱️ 💭 Drop your answer below 👇 ✅ Full step-by-step solution is on my page. ➡️ Follow for the rest of the 2026 Qualifying set!
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🔥 Day 139 — 2026 MIT Integration Bee (Qualifying) Q2 Evaluate: ∫ e^(2026·e^x) · e^x dx This is a perfect u-sub “inside-the-exponential” setup. If you set u = 2026·e^x, then du = 2026·e^x dx and the integral collapses immediately. 👉 Hints • u = 2026·e^x • du = 2026·e^x dx → (e^x dx) = du/2026 • Integral becomes (1/2026) ∫ e^u du Would you catch the substitution in time? ⏱️ 💭 Drop your answer below 👇 ✅ Full step-by-step solution is on my page. ➡️ Follow for the rest of the 2026 Qualifying set!
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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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🔥 Day 138 — 2026 MIT Integration Bee (Qualifying) Q1 Evaluate: ∫ from −pi to pi of sin^(2025)(x) * cos^(2026)(x) dx This looks impossible at first glance, but there’s a 2-second parity trick. Power counts matter: odd vs even makes or breaks symmetry on [−a, a]. 👉 Hints • sin(x) odd → odd^2025 stays odd • cos(x) even → even^2026 stays even • odd * even = odd; integral of odd over [−a, a] is 0 Would you spot the parity in time? ⏱️ 💭 Comment your reasoning below 👇 ✅ Full step-by-step solution i
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TikTokbrainrot.mathematics
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