To evaluate the definite integral ∫0ln2eu1+eudu, use the substitution x=1+eu, which yields dx=eudu. Convert the limits of integration from u=0 and u=ln2 to x=2 and x=3, transforming the integral into ∫23xdx. Evaluating the antiderivative 32x3/2 at these new bounds gives the result 23−342.
Conditions: The integral is definite with bounds u=0 to u=ln2.; The integrand is eu1+eu.; Substitution x=1+eu is used.
To evaluate the definite integral ∫0ln2eu1+eudu, use the substitution x=1+eu, which yields dx=eudu. Convert the limits of integration from u=0 and u=ln2 to x=2 and x=3, transforming the integral into ∫23xdx. Evaluating the antiderivative 32x3/2 at these new bounds gives the result 23−342.
Conditions: The integral is definite with bounds u=0 to u=ln2.; The integrand is eu1+eu.; Substitution x=1+eu is used.
To derive φ(pk), we count the integers from 1 to pk that are coprime to pk. An integer is not coprime to pk if and only if it shares a prime factor with pk.
Conditions: p is a prime number.; k is a positive integer (implied by context).; The counting range is from 1 to pk inclusive.
To derive φ(pk), we count the integers from 1 to pk that are coprime to pk. An integer is not coprime to pk if and only if it shares a prime factor with pk.
Conditions: p is a prime number.; k is a positive integer (implied by context).; The counting range is from 1 to pk inclusive.