If _0^1(x^21+x^14+x^7)(2x^14+3x^7+6)^1/7,dx = 1l(11)^m/n, where l,m,nN, m,n coprime, then …

Question

If $\displaystyle\int_0^1(x^{21}+x^{14}+x^7)(2x^{14}+3x^7+6)^{1/7}\,dx = \dfrac1l(11)^{m/n}$, where $l,m,n\in\mathbb{N}$, $m,n$ coprime, then $l+m+n$ is equal to \_\_\_\_\_\_\_\_.

Answer: 63

Accepted Answer

Note $x^{21}+x^{14}+x^7 = \dfrac1{42}\cdot x^6(2x^{21}\cdots)$; more directly, with $t=2x^{21}+3x^{14}+6x^7$ arranged so $dt=42(x^{20}+x^{13}+x^6)dx$ (matching a rescaled integrand), the integral reduces to

$$\frac1{42}\int_0^{11}t^{1/7}\,dt = \frac1{42}\cdot\frac{7}{8}\left[t^{8/7}\right]_0^{11} = \frac1{48}(11)^{8/7}$$

So $l=48,\,m=8,\,n=7$, giving $l+m+n=63$.

Solutions · 1