Adjusting to the New AP Calculus Curriculum Framework

Adjusting to the New AP Calculus
Curriculum Framework
L’Hopital’s 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 βˆ’ 4
lim
πœ‹π‘₯
π‘₯β†’2
sin ( 2 )
Inspired by #3 AB 2016 International Multiple Choice
lim
3π‘₯ βˆ’ 4
π‘₯β†’βˆž √16π‘₯ 2
+1
√9π‘₯ 6 + 2
π‘₯β†’βˆ’βˆž π‘₯ 3 βˆ’ 1
lim
Inspired by #3 AB 2008 International Multiple Choice
Identify all asymptotes for
𝑦=
5+ 𝑒π‘₯
1βˆ’ 𝑒π‘₯
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 𝑓(2 + π‘₯ 2 )
π‘₯β†’0
ii) lim 𝑓(2 βˆ’ π‘₯ 2 )
π‘₯β†’0
iii) lim 𝑓(2 + π‘₯)
π‘₯β†’0
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.
𝟐
πŸ‘
1. ∫𝟏 (πŸ’π’™ + 𝟐) 𝒅𝒙
πŸ“
πŸ“
3. βˆ«πŸ• (πŸ‘π’™ + 𝟏) 𝒅𝒙
𝟐
5. βˆ«πŸ“ (πŸ’π’™ βˆ’ 𝟐) 𝒅𝒙
b. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (𝟐 + ) + 𝟐 ] ( )
𝒏
𝒏
_____
c. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [(𝟐 + ) + 𝟏 ] ( )
𝒏
𝒏
_____
d. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (𝟏 + ) + 𝟐 ] ( )
𝒏
𝒏
_____
e. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ‘ (πŸ• βˆ’ ) + 𝟏 ] ( )
𝒏
𝒏
_____
f. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (πŸ‘ βˆ’ ) + 𝟐 ] ( )
𝒏
𝒏
_____
g. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ‘ (𝟐 + ) + 𝟏 ] ( )
𝒏
𝒏
_____
h. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (πŸ“ + ) βˆ’ 𝟐 ] ( )
𝒏
𝒏
_____
i. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [(πŸ“ + ) + 𝟏 ] ( )
𝒏
𝒏
𝟐
πŸ“
6. ∫𝟐 (πŸ’π’™ + 𝟐) 𝒅𝒙
πŸ•
7. βˆ«πŸ“ (πŸ’π’™ βˆ’ 𝟐) 𝒅𝒙
𝟐
𝟏
8. βˆ«πŸ‘ (πŸ’π’™ + 𝟐) 𝒅𝒙
πŸ•
_____
𝟐
πŸ’
4. ∫𝟐 (πŸ’π’™ + 𝟐) 𝒅𝒙
πŸ‘
9. βˆ«πŸ“ (𝒙 + 𝟏) 𝒅𝒙
βˆ’πŸ‘
a. π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (πŸ“ βˆ’ ) βˆ’ 𝟐 ] ( )
𝒏
𝒏
πŸ‘
2. ∫𝟐 (𝒙 + 𝟏) 𝒅𝒙
πŸ‘π’‹
_____
π’β†’βˆž
πŸ‘π’‹ 𝟐
πŸ‘
π’β†’βˆž
πŸ‘π’‹ πŸ‘
πŸ‘
π’β†’βˆž
πŸπ’‹ 𝟐
𝟐
π’β†’βˆž
πŸπ’‹
βˆ’πŸ
π’β†’βˆž
πŸπ’‹ 𝟐
β€“πŸ
π’β†’βˆž
πŸ‘π’‹
πŸ‘
πŸπ’‹
𝟐
π’β†’βˆž
π’β†’βˆž
πŸπ’‹ πŸ‘
𝟐
π’β†’βˆž
πŸ“
10. ∫𝟐 (πŸ‘π’™ + 𝟏) 𝒅𝒙
πŸπ’‹ 𝟐
_____
𝟐
π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (𝟐 + ) + 𝟐 ] ( )
𝒏
𝒏
π’β†’βˆž
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.
𝟐
πŸ‘
1. ∫𝟏 (πŸ’π’™ + 𝟐) 𝒅𝒙
________________________________
2. __________________
π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (𝟐 + ) + 𝟐 ] ( )
𝒏
𝒏
πŸ“
πŸ‘π’‹ 𝟐
πŸ‘
π’β†’βˆž
3. βˆ«πŸ• (πŸ‘π’™ + 𝟏) 𝒅𝒙
________________________________
4. __________________
π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (𝟐 + ) + 𝟐 ] ( )
𝒏
𝒏
𝟐
πŸπ’‹ 𝟐
𝟐
π’β†’βˆž
5. βˆ«πŸ“ (πŸ’π’™ βˆ’ 𝟐) 𝒅𝒙
________________________________
6. __________________
π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ’ (πŸ‘ βˆ’ ) + 𝟐 ] ( )
𝒏
𝒏
πŸ•
πŸπ’‹ 𝟐
β€“πŸ
π’β†’βˆž
7. βˆ«πŸ“ (πŸ’π’™ βˆ’ 𝟐) 𝒅𝒙
________________________________
8. __________________
π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [πŸ‘ (𝟐 + ) + 𝟏 ] ( )
𝒏
𝒏
πŸ•
πŸ‘π’‹
πŸ‘
π’β†’βˆž
πŸ‘
9. βˆ«πŸ“ (𝒙 + 𝟏) 𝒅𝒙
________________________________
10. __________________
π₯𝐒𝐦 βˆ‘π’π’‹=𝟏 [(𝟐 + ) + 𝟏 ] ( )
𝒏
𝒏
πŸ‘π’‹ πŸ‘
π’β†’βˆž
πŸ‘
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. ∫ π‘₯ 2 (π‘₯ 3 + 1)2 𝑑π‘₯
2. A. ∫
3. A. ∫
π‘₯
π‘₯ 2 +5
1
𝑑π‘₯
π‘₯ 2 + 4π‘₯+ 5
π‘₯+2
B. ∫ π‘₯(π‘₯ 3 + 1)2 𝑑π‘₯
B. ∫ 2
𝑑π‘₯
π‘₯ + 4π‘₯+ 5
𝑑π‘₯
π‘₯+2
B. ∫ 2
𝑑π‘₯
π‘₯ + 4π‘₯+ 5
4. A. ∫
1
√4π‘₯βˆ’ π‘₯ 2 βˆ’3
𝑑π‘₯
5. A. ∫ π‘₯ (3π‘₯ βˆ’ 1)2 𝑑π‘₯
6. A. ∫
π‘₯2
π‘₯ 3 +1
𝑑π‘₯
B. ∫
√
2βˆ’π‘₯
4π‘₯βˆ’ π‘₯2 βˆ’3
𝑑π‘₯
8
B. ∫ π‘₯ (3π‘₯2 βˆ’ 1) 𝑑π‘₯
π‘₯3 +1
B. ∫ 2 𝑑π‘₯
π‘₯