<P> Substituting back into the integral, you find that for A over x to x </P> <Dl> <Dd> A = F 0 / x 0 m m + 1 ⋅ (x 1 m + 1 − x 0 m + 1) (\ displaystyle A = (\ frac (F_ (0) / x_ (0) ^ (m)) (m + 1)) \ cdot (x_ (1) ^ (m + 1) - x_ (0) ^ (m + 1))) </Dd> </Dl> <Dd> A = F 0 / x 0 m m + 1 ⋅ (x 1 m + 1 − x 0 m + 1) (\ displaystyle A = (\ frac (F_ (0) / x_ (0) ^ (m)) (m + 1)) \ cdot (x_ (1) ^ (m + 1) - x_ (0) ^ (m + 1))) </Dd> <Dl> <Dd> log ⁡ A = log ⁡ (F 0 / x 0 m m + 1 ⋅ (x 1 m + 1 − x 0 m + 1)) = log ⁡ F 0 m + 1 − log ⁡ 1 x 0 m + log ⁡ (x 1 m + 1 − x 0 m + 1) (\ displaystyle \ log A = \ log \ left ((\ frac (F_ (0) / x_ (0) ^ (m)) (m + 1)) \ cdot (x_ (1) ^ (m + 1) - x_ (0) ^ (m + 1)) \ right) = \ log (\ frac (F_ (0)) (m + 1)) - \ log (\ frac (1) (x_ (0) ^ (m))) + \ log (x_ (1) ^ (m + 1) - x_ (0) ^ (m + 1))) </Dd> </Dl>

How to find the slope of a logarithmic graph