Question

$\text{Let}\lambda ={\int }_1^9(\frac{1}{2}+\sqrt{\frac{1}{4}+{\log }_{3}x})dx+{\int }_1^2\frac{3^{x^2-\left[x\right]}}{3^{\left\{x\right\}}}dx$ then $2\lambda \text{equals}$ [where [.] denotes the greatest integer function and {.} denotes the fractional part function]

Answer 1

$\because \left\{x\right\}=x-\left[x\right]\Rightarrow \lambda ={\int }_1^9(\frac{1}{2}+\sqrt{\frac{1}{4}+lo{g}_{3}x})dx+{\int }_1^2{3}^{x^2-x}dx{f}^{-1}\left(x\right)=3^{x^2-x}andf\left(x\right)=\frac{1}{2}+\sqrt{\frac{1}{4}+lo{g}_{3}x}\lambda ={\int }_1^{9}f\left(x\right)dx+{\int }_1^2{f}^{-1}\left(x\right)dx{f}^{-1}\left(x\right)=t\Rightarrow x=f\left(x\right)dx=f^{\prime }\left(t\right)dt-\left(1\right)\lambda ={\int }_1^{9}f\left(x\right)dx+{\int }_1^{9}t{f}^{\prime }\left(t\right)dt\lambda ={\int }_1^{9}f\left(x\right)dx+\left(tf\right(t){)}_1^9-{\int }_1^{9}f(t\left)dt\right(\text{using (1))}\text{we get,}\lambda =9.f\left(9\right)-1.f\left(1\right)\lambda =9\times 2-1\times 1=172\lambda =34$