]]>
LearnNext
##### Get a free home demo of LearnNext

Available for CBSE, ICSE and State Board syllabus.
Call our LearnNext Expert on 1800 419 1234 (tollfree)
OR submit details below for a call back

clear

# Evaluation of Definite integration by method of the substitution

2,230 Views
Have a doubt? Clear it now.
live_help Have a doubt, Ask our Expert
format_list_bulleted Take this Lesson Test

#### Evaluation of Definite integration by method of the substitution - Lesson Summary

Steps to evaluate âˆ«ab f(x) dx by the method of substitution:

âˆ«01 (2x + 3)âˆš(3 - 2x) dx

Step - 1) Consider the integral without taking the given limits.

Consider âˆ« (2x + 3)âˆš(3 - 2x) dx

Step -2) Substitute t = f(x) or x = g(t) to reduce the given integral to a known form.

Let (3 - 2x) = t2

â‡’ 2x = 3 - t2

â‡’ 2x + 3 = 6 - t2

2 dx = -2t dt

â‡’dx = -t dt

âˆ« (2x + 3)âˆš(3 - 2x) dx = âˆ« (6 - t2)âˆšt2 x (-tdt)

âˆ« - (6 - t2)t2 dt

= âˆ« (t4 - 6t2) dt

Step -3) Integrate the new integrand with respect to the new variable without placing the constant of integration.

= (t5/5) - 6(t3/3)

= t5/5 - 6t3/3

= t5/5 - 2t3

âˆ« (2x + 3)âˆš(3 - 2x) dx = [ t5/5 - 2t3]

Step - 4) Write the answer in terms of the original variable by re-substituting the new variable.

âˆ« (2x + 3)âˆš(3 - 2x) dx = [ (âˆš(3 - 2x))5/5 - 2(âˆš(3 - 2x))3]

Step - 5) Find the values of the answers obtained in Step - 4 at the given upper and lower limits.

âˆ«01 (2x + 3)âˆš(3 - 2x) dx = [ (âˆš(3 - 2x))5/5 - 2(âˆš(3 - 2x))3]01

Step - 6) Subtract the value at the lower limit from the value of the upper limit to obtain the required definite integral.

= [ (âˆš(3 - 2(1)))5/5 - 2(âˆš(3 - 2(1)))3] - [ (âˆš(3 - 2(0)))5/5 - 2(âˆš(3 - 2(0)))3]

= 1/5 - 2 - 9âˆš3/5 + 6âˆš3

= (1 - 10 - 9âˆš3 + 30âˆš3)/5

= (-9 + 21âˆš3)/5

âˆ«01 (2x + 3)âˆš(3 - 2x) dx = (-9 + 21âˆš3)/5

Evaluate âˆ«ab âˆš(x-a/b-x) dx

Step - 1) Consider the integral without taking the given limits.

Step -2) Substitute t = f(x) or x = g(t) to reduce the given integral to a known form.

Put x = a cos2 Î¸ + b sin2 Î¸

dx = a(-2 cos Î¸ sin Î¸) dÎ¸ + b(2 sin Î¸ cos Î¸)

= -a sin 2Î¸ dÎ¸ + b sin 2Î¸ dÎ¸

= (b - a) sin 2Î¸ dÎ¸

x - a = (a cos2 Î¸ + b sin2 Î¸ - a)

= b sin2 Î¸ - a(1 - cos2 Î¸)

= b sin2 Î¸ - a sin2 Î¸

= (b - a) sin2 Î¸

b - x = (b - acos2 Î¸ - bsin2 Î¸)

= b(1 - sin2 Î¸) - acos2 Î¸

= b cos2 Î¸ - a cos2 Î¸

= (b - a) cos2 Î¸

When x = a, a cos2 Î¸ + b sin2 Î¸ = a

â‡’ b sin2 Î¸ = a (1 - cos2 Î¸)

â‡’ b sin2 Î¸ = a sin2 Î¸

â‡’ (b - a) sin2 Î¸ = 0

As (b - a) â‰  0, sin Î¸ = 0

â‡’ Î¸ = 0o

âˆ´ When x = a; Î¸ = 0o

âˆ´ When x = b, a cos2 Î¸ + bsin2 Î¸ = b

â‡’ a cos2 Î¸ = b(1 - sin2 Î¸)

â‡’ a cos2 Î¸ = b cos2 Î¸

â‡’ cos2 Î¸ (a - b) = 0

As a - b â‰  0 , cos Î¸ = 0

â‡’ Î¸ = Ï€/2

âˆ´ When x = b,; Î¸ = Ï€/2

Step -3) Integrate the new integrand with respect to the new variable without placing the constant of integration.

âˆ«ab âˆš(x-a/b-x) dx

Step-4) Keep the integral in the new variable itself and change the limits of the integral accordingly.

= âˆ«0Ï€/2 âˆš((b-a)sin2Î¸/(b-a)cos2Î¸) x (b - a) sin 2Î¸ dÎ¸

= âˆ«0Ï€/2 (b-a) sin Î¸/cos Î¸ x 2 sin Î¸ cos Î¸ dÎ¸

= 2(b - a) âˆ«0Ï€/2 sin2Î¸ dÎ¸

= 2(b - a) âˆ«0Ï€/2 (1 - cos2Î¸)/2 dÎ¸

= (b - a) [ âˆ«0Ï€/2 dÎ¸ - âˆ«0Ï€/2 cos2Î¸ dÎ¸

Step - 5) Find the values of the answers obtained in the previous step at the given upper and lower limits.

= (b - a) [ [Î¸]0Ï€/2 - [sin2Î¸/2]0Ï€/2 ]

Step - 6) Subtract the value at the lower limit from the value of the upper limit to obtain the required definite integral.

= (b-a) [ (Ï€/2 - 0) - (sin(2xÏ€/2)/2 - sin(2x0)/2)]

= 2(b - a)[Ï€/4 - 1/2.(0)]

= 2(b - a) x Ï€/4

= (b - a) Ï€/2

Hence,
âˆ«ab âˆš(x-a/b-x) dx = (b - a) Ï€/2

Hence, using these two methods, we can evaluate a given definite integral by the method of substitution.