Let √(4+5x)=t and x=(t? -4)/5

dx=2t/5 dt

x∈

Original formula =∫(2 to 3) (t? -4)/5*t*2t/5 dt

=2/25* ∫(2 to 3) (t 4-4t? )dt

= 2/25 * (t 5/5-4t 3/3) | (2 to 3)

=2/25×[ (243/5-36)-(32/5-32/3)]

=422/ 125-72/36+6/75

=(422× 108-72× 125×3+6× 180)/( 125× 108)

=6 177/4500

2)

The original formula =∫(2 to 3) (3-x)dx+∫(3 to 4) (x-3)dx.

=( 3x-x？ /2)|(2 to 3)+(x? /2-3 times) |(3 to 4)

=[9-9/2-(6-2)]+[ 8- 12-(9/2-9) ]

= 1