University of Co-operative and Management, Sagaing Department of Mathematics

Second Year B.BSc (Accounting and Finance, Applied Statistics, Marketing Management & Social Enterprise Management Specializes)

C Math 2001 (Business Mathematics) Second Semester

Chapter -5 Techniques of Integration • Integration by Substitution • Integration by Parts • Improper Integrals

5-1 INTEGRATION BY SUBSTITUTION 1. ‫𝑐 ׬‬. 𝑓 𝑥 𝑑𝑥 = 𝑐. ‫𝑥𝑑 𝑥 𝑓 ׬‬

2. ‫ 𝑥 𝑓 ׬‬+ 𝑔(𝑥) 𝑑𝑥 = ‫ 𝑥𝑑 𝑥 𝑓 ׬‬+ ‫𝑥𝑑 𝑥 𝑔 ׬‬ 3. ‫ 𝑥 𝑓 ׬‬− 𝑔(𝑥) 𝑑𝑥 = ‫ 𝑥𝑑 𝑥 𝑓 ׬‬− ‫𝑥𝑑 𝑥 𝑔 ׬‬ 𝑛

4. ‫= 𝑥𝑑 𝑥 ׬‬ 5.

1 ‫𝑥𝑑 𝑥 ׬‬

𝑎𝑥

1 𝑥 𝑛+1 𝑛+1

+ 𝐶, 𝑛 ≠ −1

= 𝑙𝑛𝑥 + 𝐶

6. ‫= 𝑥𝑑 𝑒 ׬‬

1 𝑎𝑥 𝑒 𝑎

+𝐶

Example (1) Find ‫ ׬‬5𝑥 + 1 𝑑𝑥 1Τ Solution; ‫(׬‬5𝑥 + 1) 2 𝑑𝑥 Let, 𝑢 = 5𝑥 + 1and 𝑑𝑢 = 5𝑑𝑥 1 𝑑𝑢 = 𝑑𝑥 5 න(5𝑥

1ൗ + 1) 2 𝑑𝑥

=

1ൗ 1 න 𝑢 2 𝑑𝑢

1 1ൗ = න 𝑢 2 𝑑𝑢 5 3ൗ 𝑢 2



1 = +𝐶 5 3ൗ 2 2 3ൗ = 𝑢 2+𝐶 15

5

Example (2) Find ‫(׬‬6𝑥 − 8)3 𝑑𝑥 Solution; Let 𝑢 = 6𝑥 − 8 and 𝑑𝑢 = 6𝑑𝑥 1 𝑑𝑢 = 𝑑𝑥 6 1 3 3 න(6𝑥 − 8) 𝑑𝑥 = න 𝑢 . 𝑑𝑢 6 1 = න 𝑢3 𝑑𝑢 6



1 𝑢4 = +𝐶 6 4 1 = (6𝑥 − 8)4 +𝐶 24 1

Example (3) Find ‫ ׬‬10𝑥 5𝑥 2 + 1 𝑑𝑥 Solution; Let 𝑢 = 5𝑥 2 + 1 and 𝑑𝑢 = 10𝑥 𝑑𝑥 න 10𝑥

5𝑥 2

+ 1 𝑑𝑥 =

1ൗ න 𝑢 2 𝑑𝑢

=



3ൗ 𝑢 2

+𝐶

3ൗ 2 2 3ൗ = 𝑢 2+𝐶 3 2 3ൗ 2 = (5𝑥 + 1) 2 +𝐶 3 2

3

Example (4) Find ‫ 𝑥(׬‬3 + 2)4 𝑥 2 𝑑𝑥 Let 𝑢 = 𝑥 3 + 2and 𝑑𝑢 = 3𝑥 2 𝑑𝑥 1 𝑑𝑢 = 𝑥 2 𝑑𝑥 3 1 3 4 2 4 න(𝑥 + 2) 𝑥 𝑑𝑥 = න 𝑢 . 𝑑𝑢 3 1 = න 𝑢4 𝑑𝑢 3



1 𝑢5 = +𝐶 3 5 (𝑥 3 + 2)5 = +𝐶 15 3

5

Example (5) 𝑥2 Find ‫𝑥𝑑 𝑒𝑥 ׬‬ Solution; Let 𝑢 = 𝑥 2 and 𝑑𝑢 = 2𝑥𝑑𝑥 1 𝑑𝑢 = 𝑥𝑑𝑥 2 1 𝑥2 𝑢 න 𝑥𝑒 𝑑𝑥 = න 𝑒 𝑑𝑢 2 1 = න 𝑒 𝑢 𝑑𝑢 2



1 𝑢 = 𝑒 +𝐶 2 1 𝑥2 = 𝑒 +𝐶

Example(6) Find

2𝑥+3 ‫ 𝑥( ׬‬2 +3𝑥+6)4 𝑑𝑥

Solution; Let 𝑢 = 𝑥 2 + 3𝑥 + 6 and 𝑑𝑢 = 2𝑥 + 3 𝑑𝑥 2𝑥 + 3 1 න 2 𝑑𝑥 = න 4 𝑑𝑢 4 (𝑥 + 3𝑥 + 6) 𝑢 = න 𝑢−4 𝑑𝑢

𝑢−3 = +𝐶 −3



1 2 = − 𝑥 + 3𝑥 + 6 3 1

−3

+C

Example (7) What is

4𝑥 ‫ 𝑥 ׬‬2 +6 𝑑𝑥

?

Solution; Let 𝑢 = 𝑥 2 + 6and 𝑑𝑢 = 2𝑥 𝑑𝑥 2𝑑𝑢 = 4𝑥𝑑𝑥 4𝑥 2 න 2 𝑑𝑥 = න 𝑑𝑢 𝑥 +6 𝑢 1 = 2 න 𝑑𝑢 𝑢



= 2 ln 𝑢 + 𝐶 = 2 ln 𝑥 2 + 6 + 𝐶 4𝑥 න 2 𝑑𝑥 = 2 ln 𝑥 2 + 6 + 𝐶

Example (8) 𝑒 2𝑥 ‫ 𝑒 ׬‬2𝑥 −5 𝑑𝑥

Find Solution; Let 𝑢 = 𝑒 2𝑥 − 5 and 𝑑𝑢 = 2𝑒 2𝑥 𝑑𝑥 1 𝑑𝑢 = 𝑒 2𝑥 𝑑𝑥 2 𝑒 2𝑥 1 1 න 2𝑥 𝑑𝑥 = න . 𝑑𝑢 𝑒 −5 𝑢 2 1 1 = න 𝑑𝑢 2 𝑢



1 = ln 𝑢 + 𝐶 2 1 = ln 𝑒 2𝑥 −5 + 𝐶

Example (9) Find

(8+ln 𝑥)2 ‫׬‬ 𝑥

𝑑𝑥

Solution; Let 𝑢 = 8 + ln 𝑥 and 𝑑𝑢 = (8 + ln 𝑥)2 න 𝑑𝑥 = න 𝑢2 𝑑𝑢 𝑥



1 𝑑𝑥 𝑥

𝑢3 = +𝐶 3 (8 + ln 𝑥)3 = +𝐶 3 (8 + ln 𝑥)2 (8 + ln 𝑥)3 න 𝑑𝑥 = +𝐶

SUBSTITUTION AND THE DEFINITE INTEGRAL Example (10) Compute

4 ‫׬‬0 2𝑥

𝑥 2 + 9 𝑑𝑥

Solution; Let 𝑢 = 𝑥 2 + 9 and 𝑑𝑢 = 2𝑥𝑑𝑥 When, 𝑥 = 0 → 𝑢 = 02 + 9 = 9 𝑥 = 4 → 𝑢 = 42 + 9 = 25 4

න 2𝑥 0

25

𝑥2

+ 9 𝑑𝑥 = න 9

1ൗ 𝑢 2

𝑑𝑢

25 3ൗ 𝑢 2 = 3ൗ 2 9 2 = 3

 4

3ൗ 25 2

−

3ൗ 9 2

2 3 = 5 − 33 3 2 = 125 − 27 3 196 = 3

න 2𝑥 𝑥 2 + 9 𝑑𝑥 =

196

Example (11) Find ‫ 𝑥 𝑥 ׬‬+ 5 𝑑𝑥 Solution; Let 𝑢 = 𝑥 + 5 and 𝑑𝑢 = 𝑑𝑥 𝑢−5=𝑥 න 𝑥 𝑥 + 5 𝑑𝑥

= න(𝑢 − =න



=

3ൗ 𝑢 2

5ൗ 𝑢 2

1ൗ 5)(𝑢) 2 𝑑𝑢

−

−5

1ൗ 5𝑢 2 3ൗ 𝑢 2

𝑑𝑢

+𝐶

3ൗ 2 2 10 5ൗ = (𝑥 + 5) 2 − 𝑥+5 5ൗ 2

3ൗ 2

+𝐶

Example (12) 𝑥3 ‫ 𝑥 ׬‬2 +9 𝑑𝑥

Find Solution; Let 𝑢 = 𝑥 2 + 9and 𝑑𝑢 = 2𝑥𝑑𝑥 1 2 𝑢 − 9 = 𝑥 𝑑𝑢 = 𝑥𝑑𝑥 2 𝑥3 1 1 න 2 𝑑𝑥 = න (𝑢 − 9) 𝑑𝑢 𝑥 +9 𝑢 2 1 9 = න 1− 𝑑𝑢 2 𝑢



=

1

1 = (𝑢 − 9 ln 𝑢 ) + 𝐶 2 𝑥 2 + 9 − 9 ln 𝑥 2 + 9 + 𝐶

5-2 INTEGRATION BY PARTS The integration by parts formula

න 𝑢 𝑑𝑣 = 𝑢𝑣 − න 𝑣 𝑑𝑢

Example (13) Find ‫ 𝑥 ׬‬2 ln 𝑥 𝑑𝑥 Solution; Let 𝑢 = ln 𝑥, 𝑑𝑢 =

1 𝑑𝑥, 𝑥

𝑑𝑣 = 𝑥 2 𝑑𝑥 𝑣=

𝑥3 3

න 𝑢 𝑑𝑣 = 𝑢𝑣 − න 𝑣 𝑑𝑢 3 3 1 𝑥 𝑥 න 𝑥 2 ln 𝑥 𝑑𝑥 = ln 𝑥. −න . 𝑑𝑥 3 3 𝑥

𝑥 3 ln 𝑥 1 = − න 𝑥 2 𝑑𝑥 3 3



𝑥 3 ln 𝑥 1 𝑥3 = − . +𝐶 3 3 3 𝑥 3 ln 𝑥 𝑥3

Example (14) Find ‫ 𝑒𝑥 ׬‬3𝑥 𝑑𝑥 Solution; Let 𝑢 = 𝑥, 𝑑𝑢 = 𝑑𝑥,

𝑑𝑣 = 𝑒 3𝑥 𝑑𝑥 𝑣=

𝑒 3𝑥 3

න 𝑢 𝑑𝑣 = 𝑢𝑣 − න 𝑣 𝑑𝑢 3𝑥 3𝑥 𝑒 𝑒 න 𝑥𝑒 3𝑥 𝑑𝑥 = 𝑥 . − න 𝑑𝑥 3 3

=

𝑥𝑒 3𝑥 3 𝑥𝑒 3𝑥 3

−

= −

1 𝑒 3𝑥 . +C 3 3 𝑒 3𝑥 +C 9 𝑥𝑒 3𝑥 𝑒 3𝑥

Example (15) Find ‫ 𝑥 ׬‬2 𝑒 4𝑥 𝑑𝑥 Solution; Let 𝑢 = 𝑥 2 , 𝑑𝑢 = 2𝑥𝑑𝑥,

𝑑𝑣 = 𝑒 4𝑥 𝑑𝑥 𝑣=

𝑒 4𝑥 4

න 𝑢 𝑑𝑣 = 𝑢𝑣 − න 𝑣 𝑑𝑢 4𝑥 4𝑥 𝑒 𝑒 න 𝑥 2 𝑒 4𝑥 𝑑𝑥 = 𝑥 2 . − න 2𝑥 𝑑𝑥 4 4

𝑥 2 𝑒 4𝑥 1 = − න 𝑒 4𝑥 𝑥 𝑑𝑥 4 2

𝑢 = 𝑥,

Let

𝑑𝑢 = 𝑑𝑥,

𝑑𝑣 = 𝑒 4𝑥 𝑑𝑥 𝑒 4𝑥 4

𝑣= 4𝑥 4𝑥 𝑒 𝑒 න 𝑥𝑒 4𝑥 𝑑𝑥 = 𝑥 . − න 𝑑𝑥 4 4 𝑥𝑒 4𝑥 1 𝑒 4𝑥 = − +𝐶 4 4 4 𝑥𝑒 4𝑥 𝑒 4𝑥 = − +𝐶 4 16 2 4𝑥 4𝑥 4𝑥 𝑥 𝑒 1 𝑥𝑒 𝑒 න 𝑥 2 𝑒 4𝑥 𝑑𝑥 = − − 4 2 4 16



𝑥 2 𝑒 4𝑥 𝑥𝑒 4𝑥 𝑒 4𝑥 = − + +𝐶 4 8 32 2 4𝑥

4𝑥

4𝑥

+C

Example (16) Find ‫ ׬‬ln 𝑥 𝑑𝑥 Solution; Let 𝑢 = ln 𝑥, 𝑑𝑢 =

1 𝑑𝑥, 𝑥

𝑑𝑣 = 𝑑𝑥 𝑣=𝑥

න 𝑢 𝑑𝑣 = 𝑢𝑣 − න 𝑣 𝑑𝑢 1 න ln 𝑥 𝑑𝑥 = ln 𝑥 . 𝑥 − න 𝑥 . 𝑑𝑥 𝑥 = 𝑥 ln 𝑥 − න 𝑑𝑥



= 𝑥 ln 𝑥 − 𝑥 + 𝐶

Example (17) Find ‫( 𝑥 ׬‬ln 𝑥 2 )𝑑𝑥 Solution; Let 𝑢 = ln 𝑥 2 , 𝑑𝑢 = 2 ln 𝑥

1 𝑑𝑥,𝑣 𝑥

𝑑𝑣 = 𝑥 𝑑𝑥 =

𝑥2 2

න 𝑢 𝑑𝑣 = 𝑢𝑣 − න 𝑣 𝑑𝑢 2 2 𝑥 𝑥 1 2 2 න 𝑥 (ln 𝑥 )𝑑𝑥 = ln 𝑥 . −න 2 ln 𝑥 𝑑𝑥 2 2 𝑥

𝑥 2 ln 𝑥 2 = − න 𝑥 ln 𝑥 𝑑𝑥 2

Let 𝑑𝑢 =

𝑢 = ln 𝑥 , 1 𝑑𝑥, 𝑥

𝑑𝑣 = 𝑥 𝑑𝑥 𝑥2 2 2

𝑣= 𝑥 𝑥2 1 න 𝑥 ln 𝑥 𝑑𝑥 = ln 𝑥 . − න . 𝑑𝑥 2 2 𝑥

𝑥 2 ln 𝑥 1 = − න 𝑥 𝑑𝑥 2 2



𝑥 2 ln 𝑥 1 𝑥2 = − +𝐶 2 2 2 𝑥 2 ln 𝑥 𝑥2 = − +𝐶 2 4 2 2 𝑥 ln 𝑥 𝑥 න 𝑥 (ln 𝑥 2 )𝑑𝑥 = − +𝐶

References: 1.Howard L. Rolf. Raymond J. Cannon and Gareth Willams, Mathematics for Management, Social & Life Sciences, Wm.C. Brown Publishers, 1991. 2.Raymond A. Barnet & Michael R. Ziegler, Finite Mathematics, Dellen Publishing Company, 1990.