Question

Evaluate the integral by using substitution.

`int_0^2 xsqrt(x+2)`  (Put x + 2 = `t^2`)

Answer 1

Let `I = int_0^2 x sqrt (x + 2) dx`

Put x + 2 = t

⇒ dx = dt

When x = 0, t = 2 and when x = 2, t = 4

∴ `I = int_2^4 (t - 2) sqrtt  dt `

`= int_2^4 (t^(3/2) - 2t^(1/2)) dt`

`= [2/5 t^(5/2) - 2 xx 2/3 t^(3/2)]_2^4`

`= [2/5 (4)^(5/2) - 4/3 t^(3/2)]_2^4`

`= [2/5 (4)^(5/2) - 4/3 (4)^(3/2)] - [2/5 (2)^(5/2) = 4/3 (2)^(3/2)]`

`= 2/5 (2)^5 - 4/3 (2)^3 - 2/5 xx 4sqrt2 + 4/3 xx 2sqrt2`

`= 2/5 xx 32 - 4/3 xx 8 - 8/5 sqrt2 + 8/3 sqrt2`

`= 64/5 - 32/3 - (8/5 sqrt2 - 8/3 sqrt2)`

`= (192 - 160)/15 - ((24sqrt2 - 40sqrt2))/15`

`= 32/15 + (16sqrt2)/15`

`= 16/15 (2+sqrt2)`

or `(16sqrt2)/15 (sqrt2+1)`