Adjusting to the New AP Calculus Curriculum Framework LβHopitalβs Rule is coming! Should we panic? Keep calm and BC on! #4 BC 2016 Operational Free Response #5 BC 2013 Operational Free Response Inspired by #1 BC 2016 International Multiple Choice π₯2 β 4 lim ππ₯ π₯β2 sin ( 2 ) Inspired by #3 AB 2016 International Multiple Choice lim 3π₯ β 4 π₯ββ β16π₯ 2 +1 β9π₯ 6 + 2 π₯βββ π₯ 3 β 1 lim Inspired by #3 AB 2008 International Multiple Choice Identify all asymptotes for π¦= 5+ ππ₯ 1β ππ₯ Inspired by #88 BC 2016 International Multiple Choice (only 9% of BC students answered correctly!) Find each of the following limits using the graph at left. i) lim π(2 + π₯ 2 ) π₯β0 ii) lim π(2 β π₯ 2 ) π₯β0 iii) lim π(2 + π₯) π₯β0 LO 3.2A Interpret the definite integral as the limit of a Riemann sum. Express the limit of a Riemann sum in integral notation. Instructions: Match the integral expression in the left column with the appropriate limit of a Riemann sum in the right column. π π 1. β«π (ππ + π) π π π π 3. β«π (ππ + π) π π π 5. β«π (ππ β π) π π b. π₯π’π¦ βππ=π [π (π + ) + π ] ( ) π π _____ c. π₯π’π¦ βππ=π [(π + ) + π ] ( ) π π _____ d. π₯π’π¦ βππ=π [π (π + ) + π ] ( ) π π _____ e. π₯π’π¦ βππ=π [π (π β ) + π ] ( ) π π _____ f. π₯π’π¦ βππ=π [π (π β ) + π ] ( ) π π _____ g. π₯π’π¦ βππ=π [π (π + ) + π ] ( ) π π _____ h. π₯π’π¦ βππ=π [π (π + ) β π ] ( ) π π _____ i. π₯π’π¦ βππ=π [(π + ) + π ] ( ) π π π π 6. β«π (ππ + π) π π π 7. β«π (ππ β π) π π π π 8. β«π (ππ + π) π π π _____ π π 4. β«π (ππ + π) π π π 9. β«π (π + π) π π βπ a. π₯π’π¦ βππ=π [π (π β ) β π ] ( ) π π π 2. β«π (π + π) π π ππ _____ πββ ππ π π πββ ππ π π πββ ππ π π πββ ππ βπ πββ ππ π βπ πββ ππ π ππ π πββ πββ ππ π π πββ π 10. β«π (ππ + π) π π ππ π _____ π π₯π’π¦ βππ=π [π (π + ) + π ] ( ) π π πββ j. LO 3.2A Interpret the definite integral as the limit of a Riemann sum. Express the limit of a Riemann sum in integral notation. Instructions: Fill in the missing integral expression in the left column or the appropriate limit of a Riemann sum in the right column. π π 1. β«π (ππ + π) π π ________________________________ 2. __________________ π₯π’π¦ βππ=π [π (π + ) + π ] ( ) π π π ππ π π πββ 3. β«π (ππ + π) π π ________________________________ 4. __________________ π₯π’π¦ βππ=π [π (π + ) + π ] ( ) π π π ππ π π πββ 5. β«π (ππ β π) π π ________________________________ 6. __________________ π₯π’π¦ βππ=π [π (π β ) + π ] ( ) π π π ππ π βπ πββ 7. β«π (ππ β π) π π ________________________________ 8. __________________ π₯π’π¦ βππ=π [π (π + ) + π ] ( ) π π π ππ π πββ π 9. β«π (π + π) π π ________________________________ 10. __________________ π₯π’π¦ βππ=π [(π + ) + π ] ( ) π π ππ π πββ π Integration Techniques In each of the pairs of similar problems below, one of the problems can be solved using a usubstitution and the other can be solved using an algebraic manipulation to re-write the integrand. Determine the necessary technique and find each integral. 1. A. β« π₯ 2 (π₯ 3 + 1)2 ππ₯ 2. A. β« 3. A. β« π₯ π₯ 2 +5 1 ππ₯ π₯ 2 + 4π₯+ 5 π₯+2 B. β« π₯(π₯ 3 + 1)2 ππ₯ B. β« 2 ππ₯ π₯ + 4π₯+ 5 ππ₯ π₯+2 B. β« 2 ππ₯ π₯ + 4π₯+ 5 4. A. β« 1 β4π₯β π₯ 2 β3 ππ₯ 5. A. β« π₯ (3π₯ β 1)2 ππ₯ 6. A. β« π₯2 π₯ 3 +1 ππ₯ B. β« β 2βπ₯ 4π₯β π₯2 β3 ππ₯ 8 B. β« π₯ (3π₯2 β 1) ππ₯ π₯3 +1 B. β« 2 ππ₯ π₯
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