### Definite Integration Question 22

#### Question 22 - 01 February - Shift 1

If

$\int_0^{1}(x^{21}+x^{14}+x^{7})(2 x^{14}+3 x^{7}+6)^{1 / 7} d x=\frac{1}{l}(11)^{m / n}$

where $1, m, n \in \mathbb{N}, m$ and $n$ are coprime then $1+m+n$ is equal to

## Show Answer

#### Answer: 63

#### Solution:

#### Formula: Integration by substitution, Standard formulas for Indefinite Integration

$ \begin{aligned} & I = \int(x^{20}+x^{13}+x^{6})(2 x^{21}+3 x^{14}+6 x^{7})^{1 / 7} d x \\ & \text{subsitute} \quad 2 x^{21}+3 x^{14}+6 x^{7}=t \Rightarrow 42(x^{20}+x^{13}+x^{6}) d x=d t \\ & I=\frac{1}{42} \int_0^{11} t^{\frac{1}{7}} d t\\ & I=[\frac{t^{\frac{8}{7}}}{\frac{8}{7}} \times \frac{1}{42}]_0^{11} \\ & I =\frac{1}{48}(t^{\frac{8}{7}})_0^{11}\\ & I =\frac{1}{48}(11)^{8 / 7} = \frac{1}{l}(11)^{m / n} \\ \end{aligned} $

On comparing, we get

$ l=48, m=8, n=7 $

$ l+m+n=63$