`int_0^1 (3t - 1)^50 dt` Evaluate the definite integral.

Given: `int_0^1(3t-1)^50dt`

Integrate using the Substitution Rule.

Let `u=3t-1`

`(du)/dt=3`

`dt=(du)/3`

`=int_0^1u^50*(du)/3`

`=1/3int_0^1u^50du`

`=1/3 *u^51/51`  evaluated from t=0 to t=1

`=1/3*(3t-1)^51/51`  evaluated from t=0 to t=1

`=1/3[(3*1-1)^51/51-(3*0-1)^51/51]`

`=1/3[2^51/51-((-1)^51)/51]`

`=1/3[[2^51+1^51]/51]`

`=(2^51+1)/153`

`=1.472xx10^13` 

Given: `int_0^1(3t-1)^50dt`

Integrate using the Substitution Rule.

Let `u=3t-1`

`(du)/dt=3`

`dt=(du)/3`

`=int_0^1u^50*(du)/3`

`=1/3int_0^1u^50du`

`=1/3 *u^51/51`  evaluated from t=0 to t=1

`=1/3*(3t-1)^51/51`  evaluated from t=0 to t=1

`=1/3[(3*1-1)^51/51-(3*0-1)^51/51]`

`=1/3[2^51/51-((-1)^51)/51]`

`=1/3[[2^51+1^51]/51]`

`=(2^51+1)/153`

`=1.472xx10^13`