answer chapter 2 MAT235 module 1 (page 153 - page 168)

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Examples from Previous Semester Papers Example 1 : DEC 2019/ MAT235/ Q2a/ 6 marks 𝑥𝑒 2𝑥 −𝑥 𝑥→0 1−𝑐𝑜𝑠 2𝑥

Use L’Hopital’s Rule to evaluate lim Solution :

.

Use L’Hopital’s Rule :

➋

𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 lim ൬ ൰ 𝑥→0 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

𝑥𝑒 2𝑥 − 𝑥 𝑥𝑒 2𝑥 − 𝑥 lim ( ) = lim ( ) 𝑥→0 1 − 𝑐𝑜𝑠 2𝑥 𝑥→0 1 − 𝑐𝑜𝑠 2𝑥

➊

Direct substitution gives indeterminate 0 form 0

= lim ( 𝑥→0

= lim ( 𝑥→0

= lim ( 𝑥→0

= lim ( 𝑥→0

= lim ( 𝑥→0

= lim ( 𝑥→0

= =

𝑣𝑢′ + 𝑢𝑣′ − 1 ) 0 − (−2 𝑠𝑖𝑛 2𝑥)

𝑢=𝑥 𝑢′ = 1

➌

𝑒 2𝑥 + 2𝑥𝑒 2𝑥 − 1 ) 0 − (−2 𝑠𝑖𝑛 2𝑥)

➍

2𝑒 2𝑥 + 𝑣𝑢′ + 𝑢𝑣′ − 0 ) 4 𝑐𝑜𝑠 2𝑥

2𝑒 2𝑥 + 2𝑒 2𝑥 + 4𝑥𝑒 2𝑥 − 0 ) 4 𝑐𝑜𝑠 2𝑥

4𝑒 0 + 0 4 𝑐𝑜𝑠(0)

4(1) + 0 4(1)

Direct substitution still gives indeterminate 0 form 0

𝑒 2𝑥 + 2𝑥𝑒 2𝑥 − 1 ) 2 𝑠𝑖𝑛 2𝑥

4𝑒 2𝑥 + 4𝑥𝑒 2𝑥 ) 4 𝑐𝑜𝑠 2𝑥

𝑣 = 𝑒 2𝑥 𝑣′ = 2𝑒 2𝑥

➎

Again, use L’Hopital’s Rule : 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 lim ൬ ൰ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

𝑥→0

𝑢 = 2𝑥 𝑢′ = 2

𝑣 = 𝑒 2𝑥 𝑣′ = 2𝑒 2𝑥

no more indeterminate form. So, at this point, we can use direct substitution (no more L’Hopital’s Rule)

=1#

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153

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 2 : JUNE 2019/ MAT235/ Q2a/ 7 marks x

  x  Evaluate lim  sin   by using L’Hospital’s Rule. x →0   2 

➊

Solution :

➋ ➌

Take ln on both sides

𝑥 𝑥 𝑙𝑒𝑡 𝑦 = (𝑠𝑖𝑛 ( )) 2 𝑥 𝑥 𝑙𝑛(𝑦) = 𝑙𝑛 (𝑠𝑖𝑛 ( )) 2 𝑥 𝑙𝑛(𝑦) = 𝑥 𝑙𝑛 (𝑠𝑖𝑛 ( )) 2 𝑥 𝑙𝑛 (𝑠𝑖𝑛 ( )) 2 𝑙𝑛(𝑦) = 𝑥 −1

Direct substitution gives indeterminate form 00

➍

𝑥 𝑙𝑛 (𝑠𝑖𝑛 ( )) 2 ) lim (𝑙𝑛(𝑦)) = lim ( 𝑥→0 𝑥→0 𝑥 −1

➏ LHS : There is no variable ‘x’ to substitute, so after direct substitution, we got 𝑙𝑛(𝑦) ➋ (no more lim)

𝑙𝑛(𝑦) = lim

𝑥→0

𝑥 1 𝑥 ∙ 𝑐𝑜𝑠 (2) ∙ 2 𝑠𝑖𝑛 ( ) 2 −𝑥 −2

𝑥 𝑐𝑜𝑡 ( ) 1 𝑙𝑛(𝑦) = − lim ( −22 ) 2 𝑥→0 𝑥

𝑑𝑖𝑓𝑓 𝑛𝑢𝑚 lim ൬ ൰ 𝑥→0 𝑑𝑖𝑓𝑓 𝑑𝑒𝑛𝑜

𝑙𝑛(𝑦) = −lim (

2𝑥

𝑥 ) 𝑠𝑒𝑐 2 ( ) 2 𝑥 𝑙𝑛(𝑦) = −lim (2𝑥 𝑐𝑜𝑠 2 ( )) 𝑥→0 2 𝑥→0

𝑙𝑛(𝑦) = 0 𝑦 = 𝑒0 𝑦=1

© Amirah 2022

Note : Remember!..our goal now is to write into the form of 0 ∞ fraction or 0

∞

So, the next step we can solve the limits using L’Hopital’s Rule

Use L’Hopital’s Rule :

➑

𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 lim ൬ ൰ 𝑥→0 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟

Direct substitution gives indeterminate 0 form 0

𝐻𝑒𝑛𝑐𝑒,

∞

Direct substitution gives indeterminate ∞ form .

➐

1 𝑙𝑛(𝑦) = − lim ( ) 2 𝑥→0 𝑡𝑎𝑛 (𝑥 ) 2

Again, use L’Hopital’s Rule :

After we success write into fraction, the we take ‘lim’ on both sides

)

𝑥2

➒

∞

➎

1 𝑥 𝑙𝑛(𝑦) = − lim (𝑥 2 𝑐𝑜𝑡 ( )) 2 2 𝑥→0

2𝑥 1 𝑙𝑛(𝑦) = − lim ( ) 2 𝑥→0 1 𝑠𝑒𝑐 2 (𝑥 ) 2 2

0

If there are powers of f(x), first we need to take ln on both sides (so that we can change to multiplication using the properties of logarithm, then change into fraction)

1

(

So, try to adjust, until we can write into the 0 ∞ form of or

𝑥 𝑥 lim (𝑠𝑖𝑛 ( )) = lim (𝑦) 𝑥→0 𝑥→0 2 𝑥 𝑥 lim (𝑠𝑖𝑛 ( )) = lim (1) 𝑥→0 𝑥→0 2 𝑥 𝑥 lim (𝑠𝑖𝑛 ( )) = 1 # 𝑥→0 2

154

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 3a : DEC 2018/ MAT235/ Q2a/ 5 marks Find lim (x ln x ) by using L’Hospital’s Rule. x →0

Solution : 𝑙𝑛 𝑥 𝑙𝑖𝑚 (𝑥 𝑙𝑛 𝑥) = 𝑙𝑖𝑚 ൬ −1 ൰ 𝑥→0 𝑥→0 𝑥 1 ( ) = 𝑙𝑖𝑚 ( 𝑥−2 ) 𝑥→0 −𝑥

1 = −𝑙𝑖𝑚 (𝑥 2 ൬ ൰) 𝑥→0 𝑥

= −𝑙𝑖𝑚 (𝑥) 𝑥→0

= 0#

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155

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 3b : DEC 2018/ MAT235/ Q2b/ 6 marks 

Solve the improper integrals

 (x 1

x 2

+5

)

3

∞

dx .∫1

𝑥 𝑑𝑥 (𝑥 2 +5)3

Solution :

∫

(𝑥 2

𝑥 𝑑𝑥 = ∫ 𝑥(𝑥 2 + 5)−3 𝑑𝑥 + 5)3 = ∫ 𝑥 𝑢−3

𝑑𝑢 2𝑥

1 = ∫ 𝑢−3 𝑑𝑢 2

1 𝑢−2 = ( )+𝐶 2 −2 =−

=− ∞

𝑢 = 𝑥2 + 5

𝑑𝑢 = 2𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑥 = 2𝑥

1 +𝐶 4𝑢2

4(𝑥 2

1 +𝐶 + 5)2

𝑏 𝑥 𝑥 ∫ 𝑑𝑥 = 𝑙𝑖𝑚 ∫ 𝑑𝑥 2 3 𝑏→∞ 1 (𝑥 2 + 5)3 1 (𝑥 + 5)

= 𝑙𝑖𝑚 [− 𝑏→∞

=− =− =−

𝑏 1 ] 4(𝑥 2 + 5)2 1

𝑏 1 1 𝑙𝑖𝑚 [ 2 ] 4 𝑏→∞ (𝑥 + 5)2 1

1 1 1 𝑙𝑖𝑚 [ 2 − 2 ] 2 𝑏→∞ (𝑏 (1 + 5) + 5)2 4 1 1 1 𝑙𝑖𝑚 [ 2 − ] 4 𝑏→∞ (𝑏 + 5)2 36

1 1 = − [0 − ] 4 36

=

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1 # (𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠) 144

156

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 4 : JUNE 2018/ MAT235/ Q2a/ 7 marks Evaluate lim (sin(2 x )) by using L’Hospital’s Rule. x

x →0

Solution : 𝑙𝑒𝑡 𝑦 = (𝑠𝑖𝑛(2𝑥))𝑥

𝑙𝑛(𝑦) = 𝑙𝑛(𝑠𝑖𝑛(2𝑥))𝑥

𝑙𝑛(𝑦) = 𝑥 𝑙𝑛(𝑠𝑖𝑛(2𝑥)) 𝑙𝑛(𝑦) =

𝑙𝑛(𝑠𝑖𝑛(2𝑥)) 𝑥 −1

lim (𝑙𝑛(𝑦)) = lim (

𝑥→0

𝑥→0

𝑙𝑛(𝑠𝑖𝑛(2𝑥)) ) 𝑥 −1

1 ∙ 𝑐𝑜𝑠(2𝑥) ∙ 2 𝑠𝑖𝑛(2𝑥) 𝑙𝑛(𝑦) = lim ( ) 𝑥→0 −𝑥 −2 𝑙𝑛(𝑦) = −2 lim (

𝑐𝑜𝑡(2𝑥) ) 𝑥 −2

𝑙𝑛(𝑦) = −2 lim (

𝑥2 ) 𝑡𝑎𝑛(2𝑥)

𝑥→0

𝑙𝑛(𝑦) = −2 lim (𝑥 2 𝑐𝑜𝑡(2𝑥)) 𝑥→0 𝑥→0

2𝑥 𝑙𝑛(𝑦) = −2 lim ൬ ൰ 𝑥→0 2 𝑠𝑒𝑐 2 (2𝑥) 𝑥 ൰ 𝑙𝑛(𝑦) = −2 lim ൬ 2 𝑥→0 𝑠𝑒𝑐 (2𝑥) 𝑙𝑛(𝑦) = −2 lim (𝑥 𝑐𝑜𝑠 2 (2𝑥)) 𝑥→0

𝑙𝑛(𝑦) = 0 𝑦 = 𝑒0 𝑦=1

𝐻𝑒𝑛𝑐𝑒,

lim (𝑠𝑖𝑛(2𝑥))𝑥 = lim (𝑦)

𝑥→0

lim (𝑠𝑖𝑛(2𝑥))𝑥 𝑥→0

lim (𝑠𝑖𝑛(2𝑥))𝑥 𝑥→0

© Amirah 2022

𝑥→0

= lim (1) 𝑥→0

=1#

157

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 5 : JAN 2018/ MAT235/ Q2a/ 7 marks Find lim (sin 5 x )ln x . 1

x →0

+

Solution : 1

𝑙𝑒𝑡 𝑦 = (𝑠𝑖𝑛(5𝑥))𝑙𝑛𝑥

1

𝑙𝑛(𝑦) = 𝑙𝑛(𝑠𝑖𝑛(5𝑥))𝑙𝑛𝑥

1 𝑙𝑛(𝑠𝑖𝑛(5𝑥)) 𝑙𝑛𝑥 𝑙𝑛(𝑠𝑖𝑛(5𝑥)) 𝑙𝑛(𝑦) = 𝑙𝑛𝑥 𝑙𝑛(𝑦) =

lim+ (𝑙𝑛(𝑦)) = lim+ (

𝑥→0

𝑥→0

𝑙𝑛(𝑠𝑖𝑛(5𝑥)) ) 𝑙𝑛𝑥

1 ∙ 𝑐𝑜𝑠(5𝑥) ∙ 5 𝑠𝑖𝑛(5𝑥) 𝑙𝑛(𝑦) = lim+ ( ) 1 𝑥→0 ( ) 𝑥 𝑙𝑛(𝑦) = 5 lim+ ( 𝑥→0

𝑐𝑜𝑡(5𝑥) ) 1 ( ) 𝑥

𝑙𝑛(𝑦) = 5 lim+ (𝑥 𝑐𝑜𝑡(5𝑥)) 𝑥→0

𝑥 𝑙𝑛(𝑦) = 5 lim+ ൬ ൰ 𝑥→0 𝑡𝑎𝑛(5𝑥)

1 𝑙𝑛(𝑦) = 5 lim+ ൬ ൰ 𝑥→0 5 𝑠𝑒𝑐 2 (5𝑥)

𝑙𝑛(𝑦) = lim+ (𝑐𝑜𝑠 2 (5𝑥)) 𝑥→0

𝑙𝑛(𝑦) = 1 𝑦 = 𝑒1 𝑦=𝑒

𝐻𝑒𝑛𝑐𝑒,

1

𝑙𝑖𝑚+ (𝑠𝑖𝑛 5 𝑥)𝑙𝑛 𝑥 = lim+ (𝑦)

𝑥→0

𝑙𝑖𝑚

𝑥→0+

1 (𝑠𝑖𝑛 5 𝑥)𝑙𝑛 𝑥 1

𝑥→0

= lim+ (𝑒) 𝑥→0

𝑙𝑖𝑚+ (𝑠𝑖𝑛 5 𝑥)𝑙𝑛 𝑥 = 𝑒 #

𝑥→0

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158

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 6 : MAR 2017/ MAT235/ Q2a/ 7 marks 

Evaluate

 xe

−5x

dx .

0

Solution :

∫ 𝑥𝑒 −5𝑥 𝑑𝑥

Using Tabular Integration Sign

u (differentiate)

dv (integrate)

+

𝑥

𝑒 −5𝑥

−

1

+

0

1 − 𝑒 −5𝑥 5

1 1 ∫ 𝑥𝑒 −5𝑥 𝑑𝑥 = − 𝑥𝑒 −5𝑥 − 𝑒 −5𝑥 + ∫ 0 𝑑𝑥 5 25 1 1 ∫ 𝑥𝑒 −5𝑥 𝑑𝑥 = − 𝑥𝑒 −5𝑥 − 𝑒 −5𝑥 + 𝐶 5 25 ∞

∫ 𝑒 𝑎𝑥 𝑑𝑥 =

1 𝑎𝑥 𝑒 +𝐶 𝑎

1 −5𝑥 𝑒 25

𝑏

∫ 𝑥𝑒 −5𝑥 𝑑𝑥 = 𝑙𝑖𝑚 ∫ 𝑥𝑒 −5𝑥 𝑑𝑥 0

𝑏→∞ 0

𝑏 1 1 = 𝑙𝑖𝑚 [− 𝑥𝑒 −5𝑥 − 𝑒 −5𝑥 ] 𝑏→∞ 25 5 0

=

= = = =

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1 𝑙𝑖𝑚 [−5𝑥𝑒 −5𝑥 − 𝑒 −5𝑥 ]𝑏0 25 𝑏→∞

1 𝑙𝑖𝑚 [(−5𝑏𝑒 −5𝑏 − 𝑒 −5𝑏 ) − (0 − 𝑒 0 )] 25 𝑏→∞

1 𝑙𝑖𝑚 (−5𝑏𝑒 −5𝑏 − 𝑒 −5𝑏 + 1) 25 𝑏→∞

=

−5 1 1 (0 + 1) 𝑙𝑖𝑚 ൬ 5𝑏 ൰ + 25 25 𝑏→∞ 5𝑒

=

1 −5𝑏 1 1 𝑙𝑖𝑚 ൬ 5𝑏 ൰ + 𝑙𝑖𝑚 ൬− 5𝑏 + 1൰ 25 𝑏→∞ 𝑒 25 𝑏→∞ 𝑒

1 −1 1 𝑙𝑖𝑚 ൬ 5𝑏 ൰ + 25 𝑏→∞ 𝑒 25

=−

1 1 (0) + 25 25

1 # (𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠) 25

159

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 7a : OCT 2016/ MAT235/ Q2b/ 4 marks

ex − x − 1 . x → 0 cos x − 1

Find lim

Solution :

𝑒𝑥 − 𝑥 − 1 𝑒𝑥 − 1 − 0 = 𝑙𝑖𝑚 𝑥→0 𝑐𝑜𝑠 𝑥 − 1 𝑥→0 − 𝑠𝑖𝑛 𝑥 − 0 𝑙𝑖𝑚

𝑒𝑥 − 1 = 𝑙𝑖𝑚 𝑥→0 − 𝑠𝑖𝑛 𝑥

𝑒𝑥 − 0 𝑥→0 − 𝑐𝑜𝑠 𝑥

= 𝑙𝑖𝑚

𝑒𝑥 𝑥→0 𝑐𝑜𝑠 𝑥

= −𝑙𝑖𝑚

𝑒0 = −( ) 𝑐𝑜𝑠 0 1 = −൬ ൰ 1 = −1 #

© Amirah 2022

160

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 7b : OCT 2016/ MAT235/ Q2c/ 4 marks 

Find

 0

2x x2 + 1

dx .

Solution :

∫

2𝑥

√𝑥 2 + 1

1

𝑑𝑥 = ∫ 2𝑥(𝑥 2 + 1)−2 𝑑𝑥 1

= ∫ 2𝑥 𝑢−2 1

= ∫ 𝑢−2 𝑑𝑢

𝑑𝑢 2𝑥

𝑢 = 𝑥2 + 1

𝑑𝑢 = 2𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑥 = 2𝑥

1

𝑢2 +𝐶 = 1 ( ) 2

= 2√𝑢 + 𝐶

= 2√𝑥 2 + 1 + 𝐶 ∞

∫

0

2𝑥

√𝑥 2

+1

𝑏

𝑑𝑥 = 𝑙𝑖𝑚 ∫

𝑏→∞ 0

2𝑥

√𝑥 2 + 1

𝑑𝑥

𝑏

= 𝑙𝑖𝑚 [2√𝑥 2 + 1] 𝑏→∞

0

𝑏

= 2 𝑙𝑖𝑚 [√𝑥 2 + 1] 𝑏→∞

0

= 2 𝑙𝑖𝑚 [√𝑏 2 + 1 − √1] 𝑏→∞

= 2 𝑙𝑖𝑚 [√𝑏 2 + 1 − 1] 𝑏→∞

= 2[∞]

= ∞ # (𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠)

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161

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 8a : MAR 2016/ MAT235/ Q2c/ 6 marks 

Solve the improper integrals

ex

 1+ e

x

dx . [Hint : use u substitution, u = 1 + e x ]

0

Solution :

∫

𝑒𝑥 𝑒 𝑥 𝑑𝑢 𝑑𝑥 = ∫ ∙ 1 + 𝑒𝑥 𝑢 𝑒𝑥 1 = ∫ 𝑑𝑢 𝑢

= 𝑙𝑛|𝑢| + 𝐶

= 𝑙𝑛|1 + 𝑒 𝑥 | + 𝐶

𝑢 = 1 + 𝑒𝑥 𝑑𝑢 = 𝑒𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑥 = 𝑥 𝑒

∞

𝑏 𝑒𝑥 𝑒𝑥 ∫ 𝑙𝑖𝑚 ∫ 𝑥 𝑑𝑥 = 𝑏→∞ 𝑥 𝑑𝑥 0 1+𝑒 0 1+𝑒

= 𝑙𝑖𝑚 [𝑙𝑛|1 + 𝑒 𝑥 |]𝑏0 𝑏→∞

= 𝑙𝑖𝑚 [𝑙𝑛|1 + 𝑒 𝑏 | − 𝑙𝑛|1 + 𝑒 0 |] 𝑏→∞

= 𝑙𝑖𝑚 [𝑙𝑛|1 + 𝑒 𝑏 | − 𝑙𝑛|1 + 1|] 𝑏→∞

= ∞ − 𝑙𝑛(2)

= ∞ # (𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠)

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 8b : MAR 2016/ MAT235/ Q2d/ 4 marks Find lim

x →0

1 + 2x − 1 by using L’Hospital’s Rule. x

Solution :

1

(1 + 2𝑥)2 − 1 √1 + 2𝑥 − 1 𝑙𝑖𝑚 = 𝑙𝑖𝑚 𝑥→0 𝑥→0 𝑥 𝑥

1 1 (1 + 2𝑥)−2 ∙ 2 − 0 = 𝑙𝑖𝑚 2 𝑥→0 1 1

= 𝑙𝑖𝑚 (1 + 2𝑥)−2 𝑥→0

= 𝑙𝑖𝑚 =

1

𝑥→0 √1 +

1

2𝑥

√1 + 0

= −1 #

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 9a : SEP 2015/ MAT235/ Q1d/ 4 marks 𝑥 2 +10 𝑥→∞ 𝑒 𝑥 +2𝑥

Find lim

by using L’Hospital’s Rule.

Solution :

𝑥 2 + 10 2𝑥 + 0 𝑙𝑖𝑚 𝑥 = 𝑙𝑖𝑚 𝑥 𝑥→∞ 𝑒 + 2𝑥 𝑥→∞ 𝑒 + 2 = 𝑙𝑖𝑚

𝑥→∞ 𝑒 𝑥

= 𝑙𝑖𝑚 𝑥→∞

=

2 ∞

2 𝑒𝑥

2 +0

=0#

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 9b : SEP 2015/ MAT235/ Q2b/ 6 marks ∞

3

Solve the improper integral ∫0 𝑥 2 𝑒 −3𝑥 𝑑𝑥. Solution :

3

∫ 𝑥 2 𝑒 −3𝑥 𝑑𝑥 = ∫ 𝑥 2 𝑒 𝑢 ∙

𝑑𝑢 −9𝑥 2

𝑢 = −3𝑥 3

𝑑𝑢 = −9𝑥 2 𝑑𝑥 𝑑𝑢 𝑑𝑥 = −9𝑥 2

1 = − ∫ 𝑒 𝑢 𝑑𝑢 9 1 = − 𝑒𝑢 + 𝐶 9 =−

∞

1 −3𝑥 3 𝑒 +𝐶 9

𝑏

3

3

∫ 𝑥 2 𝑒 −3𝑥 𝑑𝑥 = 𝑙𝑖𝑚 ∫ 𝑥 2 𝑒 −3𝑥 𝑑𝑥 0

𝑏→∞ 0

= 𝑙𝑖𝑚 [− 𝑏→∞

=− =− =−

1 −3𝑥 3 𝑏 𝑒 ] 9 0

1 3 𝑏 𝑙𝑖𝑚 [ 𝑒 −3𝑥 ]0 9 𝑏→∞

1 3 𝑙𝑖𝑚 [𝑒 −3𝑏 − 𝑒 0 ] 9 𝑏→∞

1 1 𝑙𝑖𝑚 [ 3𝑏3 − 1] 𝑏→∞ 9 𝑒

1 = − [0 − 1] 9 =

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1 # (𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠) 9

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 10a : MAR 2015/ MAT235/ Q1c/ 4 marks 𝑥 3

Find lim+ 𝑙𝑛𝑥. 𝑥→0

Solution :

𝑥 1 𝑙𝑖𝑚+ 𝑙𝑛𝑥 = 𝑙𝑖𝑚+ (𝑥 𝑙𝑛𝑥) 𝑥→0 3 3 𝑥→0 =

1 𝑙𝑛𝑥 𝑙𝑖𝑚+ ൬ −1 ൰ 3 𝑥→0 𝑥

1 ( ) 1 = 𝑙𝑖𝑚+ 𝑥−2 3 𝑥→0 −𝑥 =−

=−

1 1 𝑙𝑖𝑚+ 𝑥 2 ൬ ൰ 𝑥 3 𝑥→0

1 𝑙𝑖𝑚 (𝑥) 3 𝑥→0+

1 = − (0) 3

= 0#

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MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 10b : MAR 2015/ MAT235/ Q2b/ 7 marks ∞

Solve the improper integral ∫0 𝑥𝑒 −𝑥 𝑑𝑥. Solution : ∫ 𝑥𝑒 −𝑥 𝑑𝑥

Using Tabular Integration Sign

u (differentiate)

dv (integrate)

+

𝑥

𝑒 −𝑥

−

1

−𝑒 −𝑥

+

0

𝑒 −𝑥

∫ 𝑒 𝑎𝑥 𝑑𝑥 =

1 𝑎𝑥 𝑒 +𝐶 𝑎

∫ 𝑥𝑒 5𝑥 𝑑𝑥 = −𝑥𝑒 −𝑥 − 𝑒 −𝑥 + ∫ 0 𝑑𝑥

∫ 𝑥𝑒 5𝑥 𝑑𝑥 = −𝑥𝑒 −𝑥 − 𝑒 −𝑥 + 𝐶 ∞

𝑏

∫ 𝑥𝑒 −𝑥 𝑑𝑥 = 𝑙𝑖𝑚 ∫ 𝑥𝑒 −𝑥 𝑑𝑥 0

𝑏→∞ 0

= 𝑙𝑖𝑚 [−𝑥𝑒 −𝑥 − 𝑒 −𝑥 ]𝑏0 𝑏→∞

= 𝑙𝑖𝑚 [(−𝑏𝑒 −𝑏 − 𝑒 −𝑏 ) − (0 − 𝑒 0 )] 𝑏→∞

= 𝑙𝑖𝑚 (−𝑏𝑒 −𝑏 − 𝑒 −𝑏 + 1) 𝑏→∞

= 𝑙𝑖𝑚 ൬ 𝑏→∞

= 𝑙𝑖𝑚 ൬ 𝑏→∞

= 𝑙𝑖𝑚 ൬ 𝑏→∞

= 0+1

−𝑏 1 ൰ + 𝑙𝑖𝑚 ൬− + 1൰ 𝑏→∞ 𝑒𝑏 𝑒𝑏 −𝑏 ൰ + (0 + 1) 𝑒𝑏

−1 ൰+1 𝑒𝑏

= 1 # (𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠)

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167

MAT235- CALCULUS II FOR ENGINEERS / Chapter 2 : Indeterminate Form and Improper Integral

Example 11 : JUL 2020/ MAT235/ Q1c/ 7 marks 1

Use L’Hopital’s Rule to evaluate 𝑙𝑖𝑚 (1 − 2𝑥)𝑥 . 𝑥→0

Solution :

1

𝑙𝑒𝑡 𝑦 = (1 − 2𝑥)𝑥

1

𝑙𝑛(𝑦) = 𝑙𝑛(1 − 2𝑥)𝑥

1 𝑙𝑛(1 − 2𝑥) 𝑥 𝑙𝑛(1 − 2𝑥) 𝑙𝑛(𝑦) = 𝑥 𝑙𝑛(𝑦) =

𝑙𝑖𝑚 (𝑙𝑛(𝑦)) = 𝑙𝑖𝑚 ( 𝑥→0

𝑥→0

𝑙𝑛(1 − 2𝑥) ) 𝑥

1 ∙ −2 1 − 2𝑥 𝑙𝑛(𝑦) = 𝑙𝑖𝑚 ( ) 𝑥→0 1 −2 𝑙𝑛(𝑦) = 𝑙𝑖𝑚 ൬ ൰ 𝑥→0 1 − 2𝑥 −2 𝑙𝑛(𝑦) = 1−0 𝑙𝑛(𝑦) = −2

𝑦 = 𝑒 −2

𝐻𝑒𝑛𝑐𝑒,

1

𝑙𝑖𝑚 (1 − 2𝑥)𝑥 = 𝑙𝑖𝑚 (𝑦)

𝑥→0

1 𝑙𝑖𝑚 (1 − 2𝑥)𝑥 𝑥→0 1

𝑥→0

= 𝑙𝑖𝑚 (𝑒 −2 ) 𝑥→0

𝑙𝑖𝑚 (1 − 2𝑥)𝑥 = 𝑒 −2

𝑥→0

1

𝑙𝑖𝑚 (1 − 2𝑥)𝑥 = 0.1353 #

𝑥→0

END OF CHAPTER 2

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