Question

How do you evaluate the definite integral `int t sqrt( t^2+1dt)` bounded by `[0,sqrt7]`?

Answer 1

It is

#int_0^sqrt7 t*sqrt(t^2+1)dt=int_0^sqrt7 1/2*(t^2+1)'*sqrt(t^2+1)dt= int_0^sqrt7 1/2*[(t^2+1)^(3/2)/(3/2)]'dt= 1/3*[(t^2+1)^(3/2)]_0^sqrt7= 1/3 (16 sqrt(2)-1) ~~ 7.2091#

Answer 2

`int_0^sqrt7 tsqrt(t^2+1)" " dt=7.209138999`

Explanation:

From the given
`int tsqrt(t^2+1)" " dt=` bounded by `[0, sqrt7]`

`int_0^sqrt7 tsqrt(t^2+1)" " dt=1/2int_0^sqrt7 2t(t^2+1)^(1/2)" "dt`

`int_0^sqrt7 tsqrt(t^2+1)" " dt=1/3*(t^2+1)^(3/2)` from 0 to `sqrt7`

`int_0^sqrt7 tsqrt(t^2+1)" " dt=1/3[(sqrt7^2+1)^(3/2)-(0^2+1)^(3/2)]`

`int_0^sqrt7 tsqrt(t^2+1)" " dt=1/3[(8)^(3/2)-(+1)^(3/2)]`

`int_0^sqrt7 tsqrt(t^2+1)" " dt=7.209138999`

God bless...I hope the explanation is useful.