Question

So I messed around with WolframAlpha and I think it miscalculated \(\displaystyle\int\int\int\int\int\int~x~~dx~dx~dx~dx~dx~dx\) ?

http://www.wolframalpha.com/input/?i=%E2%88%AB%E2%88%AB%E2%88%AB%E2%88%AB%E2%88%AB%E2%88%ABx+dxdxdxdxdxdx

Answer 1

term with power of 6 is missing?

Answer 2

well computers are stupid :DD

Answer 3

No mistake:\[\int\left(\int\left(\int\left(\int\left(\int\left(\int x~dx\right)~dx\right)~dx\right)~dx\right)~dx\right)~dx\\
=\int\left(\int\left(\int\left(\int\left(\int\left(\frac{1}{2}x^2+C_1\right)~dx\right)~dx\right)~dx\right)~dx\right)~dx\\
=\int\left(\int\left(\int\left(\int\left(\frac{1}{3\cdot2}x^3+C_1x+C_2\right)~dx\right)~dx\right)~dx\right)~dx\\
=\int\left(\int\left(\int\left(\frac{1}{4\cdot3\cdot2}x^4+\frac{C_1}{2}x^2+C_2x+C_3\right)~dx\right)~dx\right)~dx\\
=\int\left(\int\left(\frac{1}{5\cdot4\cdot3\cdot2}x^5+\frac{C_1}{3\cdot2}x^3+\frac{C_2}{2}x^2+C_3x+C_4\right)~dx\right)~dx\\
=\int\left(\frac{1}{6\cdot5\cdot4\cdot3\cdot2}x^6+\frac{C_1}{4\cdot3\cdot2}x^4+\frac{C_2}{3\cdot2}x^3+\frac{C_3}{2}x^2+C_4x+C_5\right)~dx\\
=\frac{1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}x^7+\frac{C_1}{5\cdot4\cdot3\cdot2}x^5+\frac{C_2}{4\cdot3\cdot2}x^4+\frac{C_3}{3\cdot2}x^3+\frac{C_4}{2}x^2+C_5x+C_6\]
All hail the infallible WA.

There's no 6-th power term because there's a difference of 2 in the degrees between the highest and second highest-power terms.

Answer 4

ohh. that explains the "missing power"

thanks for clear things up!

Answer 5

You're welcome!