Apibrėžtinio integralo reikšmė

Įrodykite,  kad  [tex]\int_1^2 x⋅2^{\lg x}dx=\dfrac{2^{\lg2+2}−1}{\lg2+2};[/tex]

Sprendimas: [tex]x\cdot 2^{\lg x}=x\cdot2^{\frac{\log_2x}{\log_2 10}}=x\cdot (2^{\log_2x})^{\frac{1}{\log_2 10}}=x\cdot x^{\lg 2}=x^{\lg 2+1}.[/tex]
$$\int_1^2 x⋅2^{\lg x}dx=\int_1^2 x^{\lg 2+1}dx=\dfrac{x^{\lg 2+2}}{\lg 2+2}\Biggr|_{1}^{2}=\dfrac{2^{\lg 2+2}}{\lg 2+2}-\dfrac{1^{\lg 2+2}}{\lg 2+2}=\dfrac{2^{\lg 2+2}-1}{\lg 2+2}.$$