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STUDY NOTES Week 3 Differentiation Outcomes: Integration by means of completing the square

Integration by completing the square Steps used to complete the square 1. 2. 3. 4. 5. 6.

Rearrange, so that you start with 𝑥 2 Factor out coefficient of 𝑥 2 Next, half the coefficient of x Square the (half coeff. of x) Complete the square: (x + ½ coeff)2 Multiple the coefficient back

Formulas used

∫

1

4 5 6

∫ √𝑎2 − 𝑏2 𝑥 2 𝑑𝑥

Integrate the following 1. ∫ √24 − 4𝑥 − 4𝑥 2 𝑑𝑥 𝑑𝑥 2. ∫ 2 √5+4𝑥−𝑥 𝑑𝑥

∫

𝑑𝑥 𝑎2 − 𝑏2 𝑥 2

∫ √𝑥 2 ± 𝑏2 𝑑𝑥

Examples

3. ∫ 𝑥 2−4𝑥+8

𝑑𝑥

√𝑎2 − 𝑏 2 𝑥 2 𝑑𝑥 ∫ 2 𝑎 + 𝑏2 𝑥 2

2 3

𝑦

∫

𝑑𝑥

√𝑏 2 𝑥 2 ± 𝑎2

∫ 𝑦 𝑑𝑥

1 −1 𝑏𝑥 sin +𝑐 𝑏 𝑎 𝑏𝑥 1 tan−1 +𝑐 𝑎𝑏 𝑎 𝑏𝑥 𝑥 𝑎2 sin−1 + √𝑎2 − 𝑏2 𝑥 2 + 𝑐 2𝑏 𝑎 2 𝑎 + 𝑏𝑥 1 ln ( )+𝑐 2𝑎𝑏 𝑎 − 𝑏𝑥 2 𝑏 𝑥 2 √𝑥 ± 𝑏2 ± ln [𝑥 + √𝑥 2 ± 𝑏2 ] + 𝑐 2 2 1 ln [𝑏𝑥 + √𝑏 2 𝑥 2 ± 𝑎2 ] + 𝑐 𝑏

Solution 1. ∫ √24 − 4𝑥 − 4𝑥 2 𝑑𝑥

Completing the square: 24 − 4𝑥 − 4𝑥 2

= −4𝑥 2 − 4𝑥 + 24

= −4(𝑥 2 + 𝑥 − 6)

𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 2

𝑎𝑑𝑑 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 (ℎ𝑎𝑙𝑓 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 )2

1 1 = −4 (𝑥 2 + 𝑥 + ( )2 − ( )2 − 6) 2 2 1 2 1 2 = −4 [(𝑥 + ) − ( ) − 6 ] 2 2 2 25 1 ] = −4 [(𝑥 + ) − 2 4

1 2 = −4 (𝑥 + ) + 25 2

𝑟𝑒𝑚𝑜𝑣𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠

1 2 = 25 − 4 (𝑥 + ) 2

∴ ∫ √24 − 4𝑥 − 4𝑥2 𝑑𝑥

= ∫ √25 − 4 (𝑥 + =

1 2 ) 𝑑𝑥 2

𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 3

𝑎2 𝑏𝑥 𝑥 2 sin−1 + √𝑎 − 𝑏2 𝑥 2 + 𝑐 2𝑏 𝑎 2

1 𝑎2 = 25 𝑎 = 5 𝑏2 = 4 𝑏 = 2 𝑥 = (𝑥 + ) 2

1 1 2 2𝑥 (𝑥 + 2) 25 −1 √25 − 4 (𝑥 + ) + 𝑐 sin + = 2 2(2) 5 2 2. ∫

𝑑𝑥

√5+4𝑥−𝑥2

Complete the square 5 + 4𝑥 − 𝑥 2

1𝑠𝑡 𝑠𝑡𝑒𝑝 𝑖𝑠 𝑡𝑜 𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒 = −𝑥 2 + 4𝑥 + 5

𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑢𝑡 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥 (𝑖𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒 = −1)

= −[𝑥 2 − 4𝑥 − 5]

2 1 𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝑥) 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒 𝑐𝑜𝑒𝑓𝑓 = −4 𝑎𝑑𝑑 𝑎𝑛𝑑 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡 ( 2

= −[𝑥 2 − 4𝑥 + (−2)2 − (−2)2 − 5] = −[𝑥 2 − 4𝑥 + 4 − 4 − 5] = −[(𝑥 − 2)2 − 9]

𝑓𝑖𝑟𝑠𝑡 𝑡ℎ𝑟𝑒𝑒 𝑡𝑒𝑟𝑚𝑠 𝑔𝑖𝑣𝑒 𝑢𝑠 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒:

𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑐𝑜𝑒𝑓𝑓 𝑖𝑛𝑠𝑖𝑑𝑒 𝑏𝑟𝑎𝑐𝑘𝑒𝑡

= 9 − (𝑥 − 2)2 𝑑𝑥

∴∫

=∫

√5 + 4𝑥 − 𝑥2 𝑑𝑥

√9 − (𝑥 − 2)2

𝑡ℎ𝑖𝑠 𝑖𝑠 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 1

𝑎2 = 9 𝑎 = 3 𝑏 2 = 1(𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑛𝑢𝑚𝑏𝑒𝑟) 𝑏 = 1 =

=

𝑏𝑥 1 sin−1 + 𝑐 𝑏 𝑎

1 1(𝑥 − 2) +𝑐 sin−1 3 1

= sin−1 𝑑𝑥

(𝑥 − 2) +𝑐 3

3. ∫ 𝑥 2−4𝑥+8

Complete the square 𝑥 2 − 4𝑥 + 8

= 𝑥 2 − 4𝑥 + 4 − 4 + 8 = (𝑥 − 2)2 + 4

∴∫

=∫ =

𝑥2

𝑑𝑥 − 4𝑥 + 8

𝑑𝑥 (𝑥 − 2)2 + 4

1 𝑥−2 tan−1 ( )+𝑐 2 2

𝑥 = (𝑥 − 2)...