What's my mistake applying the Euler-Maclaurin formula?

Question

I tried to calculate $$I = \int_{3}^{7} x^3 dx$$ using the Euler-Maclaurin formula. I figured the function $f(x) = x^3$ was continiously differentiable $2$ times, so my formula became $$S - I = \sum_{k=1}^{2} (\frac{B_k}{k!}(f^{(k-1)}(7) - f^{(k-1)}(3))) + R_p$$ $$S - I = \frac{B_1}{1!} \cdot (f^{(0)}(7) - f^{(0)}(3)) + \frac{B_2}{2!} \cdot (f^{(1)}(3) + f^{(1)}(7)) + R_2$$ And calculating $S$ was rather easy ($S = \sum_{k=4}^{7}k^3 = 748$) so I continued $$S - I =\frac{1}{2} \cdot (7^3 - 3^3) + \frac{1}{12} \cdot (3 \cdot 7^2 - 3 \cdot 3^2) + R_2 =$$ $$\frac{1}{2} \cdot 316 + \frac{1}{12} \cdot 120 + R_2 = $$ $$158 + 10 + R_2 = 168 + R_2$$ And $R_2$ could be calculated as $$R_2 = (-1)^{2+1} \cdot \int_{3}^{7} |f^{(2)}(x)|dx = $$ $$-1 \cdot \int_{3}^{7} |6x| dx = $$ $$-1 \cdot (3\cdot7^2 - 3\cdot3^2)) = -120$$ so $S - I$ became $$S - I = 168 - 120 = 48$$ $$I = S - 48 = 748 - 48 = 700$$ However, after some further calculations and even checking it with WolframAlpha, I came to the conclusion that $$\int_{3}^{7} f(x)dx = 580$$ which isn't the same as what I got. Does anyone know where I went wrong?