you need u-substitution.
let u=x^3
therefore, du= 3x^2 dx.
replace the x^3 in e^x^3 with u, so its e^u.
x^2 is replaced with du, but since there is a factor of 3 missing, you balance it out by bringing a 1/3 to the outside of the integral.
1/3 * integral from 0 to 1 of (e^u)du
integrate e^u, which is still e^u. Plug back in u with x^3.
f(x) = 1/3(e^x^3) from 0 to 1. Let it be f(1)-f(0).
1/3(e^1 - e^0)
1/3(e-1), or (e-1)/3
let u=x^3
therefore, du= 3x^2 dx.
replace the x^3 in e^x^3 with u, so its e^u.
x^2 is replaced with du, but since there is a factor of 3 missing, you balance it out by bringing a 1/3 to the outside of the integral.
1/3 * integral from 0 to 1 of (e^u)du
integrate e^u, which is still e^u. Plug back in u with x^3.
f(x) = 1/3(e^x^3) from 0 to 1. Let it be f(1)-f(0).
1/3(e^1 - e^0)
1/3(e-1), or (e-1)/3