Ex 7.9, 4 - Find definite integral 0 -> 2 integral x root(x+2)

Ex 7.9, 4 Evaluate the integrals using substitution ∫_0^2▒〖𝑥√(𝑥+2)〗⁡〖 (𝑝𝑢𝑡 𝑥+2=𝑡^2 )〗 ∫_0^2▒〖𝑥√(𝑥+2)〗⁡〖 𝑑𝑥〗 Put 𝑥+2=𝑡^2 Differentiating w.r.t. 𝑥 𝑑(𝑥 + 2)/𝑑𝑥=𝑑(𝑡^2 )/𝑑𝑡 ×𝑑𝑡/𝑑𝑥 1=2𝑡 × 𝑑𝑡/𝑑𝑥 𝑑𝑥=2𝑡 𝑑𝑡 Hence, when 𝑥 varies from 0 to 2, then t varies from √2 to 2 Therefore we can write ∫_0^2▒〖𝑥√(𝑥+2) 𝑑𝑥 =∫_(√2)^2▒〖(𝑡^2−2) √(𝑡^2 ) 2𝑡 𝑑𝑡〗〗 =∫_(√2)^2▒〖(𝑡^2−2)𝑡 ×2𝑡 𝑑𝑡〗 =2∫_(√2)^2▒〖(𝑡^2−2) 𝑡^2 𝑑𝑡〗 =2∫_(√2)^2▒〖(𝑡^4−2𝑡^2 ) 𝑑𝑡〗 =2[𝑡^(4+1)/(4+1)−2 𝑡^(2+1)/(2+1)]_(√2)^2 =2[𝑡^5/5−2 𝑡^3/3]_(√2)^2 =2× [2^5/5−2/3 2^3−((√2)^5/5−2/3 (√2)^3 )] =2× [𝑥^5/2−2/3 2^3−((√2)^5/2−2/3 (√2)^3 )] =2× [32/5−16/3−(4√2)/5+4/3 √2] =2× [(96 − 80 − 12√2 + 20√2)/15] =2× [(16 + 8√2)/15] =(2 × 8 √2 (√2 + 1))/15 =(𝟏𝟔√(𝟐 ) (√𝟐 + 𝟏))/𝟏𝟓

Ex 7.9, 4 - Chapter 7 Class 12 Integrals (Important Question)

Davneet Singh

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular