7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

$ = \int_{\frac{1}{n}}^{a - \frac{1}{n}} {\frac{{\sqrt {\frac{1}{n} + a - \frac{1}{n} - x} \,\,\,\,\,\,\,\,\,\,\,dx}}{{\sqrt {a - \left( {\frac{1}{n} + a - \frac{1}{n} - x} \right) + } \sqrt {\frac{1}{n} + a - \frac{1}{n} - x} }}} $

$\left[ \because \int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)\,dx}} \right]$

$I = \int_{\frac{1}{n}}^{a - \frac{1}{n}} {\frac{{\sqrt {a - x} }}{{\sqrt x + \sqrt {a - x} }}dx} $.....$(ii)$

Adding $(i)$ and $(ii),$ we get

$2I = \int_{\,1/n}^{\,a - (1/n)} {1\,dx = \left[ {\,x} \right]_{\,1/n}^{\,a - \frac{1}{n}}} $

$ \Rightarrow 2I = a - \frac{1}{n} - \frac{1}{n} = \frac{{na - 2}}{n}$

$ \Rightarrow I = \frac{{na - 2}}{{2n}}$.