Question

Find the definite integral of
`2^(lnx)` from 0 to 1.

Answer 1

Let y =
`2^(lnx)`
lnx = log₂y
lnx =
`(lny)/(ln2)`
lnx·ln2 = lny

y =
`e^(lnxln2)`
y =
`x^(ln2)`

∴ the integral of x^(㏑2) is
`(x^(ln2+1))/(ln2 + 1)`
So, with the bounds, it becomes:


`(1)/(ln2 + 1)`