Question

If $\int\limits_0^1\left(x^{21}+x^{14}+x^7\right)\left(2 x^{14}+3 x^7+6\right)^{1 / 7} d x=\frac{1}{l}(11)^{m / n}$ where $l, m, n \in N , m$ and $n$ are coprime then $l+m+n$ is equal to ______

Answer 1

Correct Answer: 63

The correct answer is 63.
$\int \left(x^{20}+x^{13}+x^6\right){\left(2x^{21}+3x^{14}+6x^7\right)}^{1/7}dx$
$2x^{21}+3x^{14}+6x^7=t$
$42\left(x^{20}+x^{13}+x^6\right)dx=dt$
$\frac{1}{42}\underset{0}{\overset{11}{\int }}{t}^{\frac{1}{7}}dt={\left(\frac{t^{\frac{8}{7}}}{\frac{8}{7}}\times \frac{1}{42}\right)}_0^{11}$
$=\frac{1}{48}{\left(t^{\frac{8}{7}}\right)}_0^{11}=\frac{1}{48}(11{)}^{8/7}$
$l=48,m=8,n=7$
$l+m+n=63$

Answer 2

Applying integral transformation: \[ \int_{0}^{1} x^l (2x^{14} + 3x^7 + 6)^{1/7} dx \] Setting \( t = 42(x^{20} + x^{13} + x^6) dx \), \[ \frac{1}{42} \int_0^{11} t^7 dt \] \[ = \frac{1}{48} (11^{8/7}) \] \[ l = 48, \quad m = 8, \quad n = 7 \] \[ l + m + n = 63 \]