Given `int_0^1root(3)(1+7x)dx`
Integrate using the u-substitution method.
Let `u=1+7x`
`(du)/dx=7`
`dx=(du)/7`
`=int_0^1u^(1/3)*(du)/7`
`=1/7int_0^1u^(1/3)du`
`=1/7*u^(4/3)/(4/3)` Evaluated from x=0 to x=1.
`=1/7*3/4*(1+7x)^(4/3)` Evaluated from x=0 to x=1.
`=3/28
[(1+7*1)^(4/3)-(1+7*0)^(4/3)]`
`=3/28[8^(4/3)-1^(4/3)]`
`=3/28[16-1]`
`=3/28[15]`
`=45/28`
`=1.607`
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