JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 14)
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 $$l, m, n \in \mathbb{N}, m$$ and $$n$$ are coprime then $$l+m+n$$ is equal to ____________.
Answer
63
Explanation
$I=\int_{0}^{1}\left(x^{21}+x^{14}+x^{7}\right)\left(2 x^{14}+3 x^{7}+6\right)^{1 / 7} d x$
$I=\int_{0}^{1}\left(x^{20}+x^{13}+x^{6}\right)\left(2 x^{21}+3 x^{14}+6 x^{7}\right)^{1 / 7} d x$
Let $2 x^{21}+3 x^{14}+6 x^{7}=t$
$\Rightarrow 42\left(x^{20}+x^{13}+x^{6}\right) d x=d t$
$I=\frac{1}{42} \int_{0}^{11} t^{1 / 7} d t=\frac{1}{42} \frac{7}{8}\left[t^{8 / 7}\right]_{0}^{11}$
$=\frac{1}{48} (11)^{8/7}$
$\therefore \quad I=48, m=8, n=7$
$\therefore \quad l+m+n=63$
$I=\int_{0}^{1}\left(x^{20}+x^{13}+x^{6}\right)\left(2 x^{21}+3 x^{14}+6 x^{7}\right)^{1 / 7} d x$
Let $2 x^{21}+3 x^{14}+6 x^{7}=t$
$\Rightarrow 42\left(x^{20}+x^{13}+x^{6}\right) d x=d t$
$I=\frac{1}{42} \int_{0}^{11} t^{1 / 7} d t=\frac{1}{42} \frac{7}{8}\left[t^{8 / 7}\right]_{0}^{11}$
$=\frac{1}{48} (11)^{8/7}$
$\therefore \quad I=48, m=8, n=7$
$\therefore \quad l+m+n=63$
Comments (0)
