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Day 153 — 2026 MIT Integration Bee Qualifying (Q19) 365 Problems for 365 Days Today’s problem is a nested logarithm integration question that shows up a lot in AP Calculus and contest math: you’re looking for the “perfect substitution” hidden in the denominator xlog(x). If you choose the right u-sub, the integral turns into a clean integration-by-parts problem with ulog(u). Problem: Integral of [ log(log x) * log(log(log x)) ] / [ x * log x ] dx High-yield hint: \t•\tIf you see dx/(x log x), thi
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Day 153 — 2026 MIT Integration Bee Qualifying (Q19) 365 Problems for 365 Days Today’s problem is a nested logarithm integration question that shows up a lot in AP Calculus and contest math: you’re looking for the “perfect substitution” hidden in the denominator xlog(x). If you choose the right u-sub, the integral turns into a clean integration-by-parts problem with ulog(u). Problem: Integral of [ log(log x) * log(log(log x)) ] / [ x * log x ] dx High-yield hint: \t•\tIf you see dx/(x log x), thi
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The milk and coffee riddle: The answer is more surprising than you think!
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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 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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MIT integration bee gave students 3 minutes for this. (sol. 85) #calculus #integral #engineering #math #mit
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