Question

Знайти середнє значення функції.

Answer 1

Ответ:

$\displaystyle A(x) = \frac{1}{2} \bigg(ln(\frac{6}{5} )\bigg)$

Объяснение:

$\displaystyle A(x) = \frac{\displaystyle \int\limits^{3/2}_1 {\frac{1}{x^2+x} } \, dx }{3/2-1}$

Разберемся с интегралом

$\displaystyle \int\limits^{3.2}_1 {\frac{1}{x^2+x} } \, dx \\\\\\\frac{1}{x^2+x} = \frac{A}{x}^{\setminus x+1} +\frac{B}{x+1}^{\setminus x} \\\\\\\frac{1}{x^2+x} =\frac{Ax+A+Bx}{x(x+1)} \\\\\left \{ {{A=1 \hfill\atop {A+B=0}} \right. \quad \Rightarrow A=1;\quad B=-1\\\\\\\int\limits^{3/2}_1 {\frac{1}{x^2+x} } \, dx =\int\limits^{3/2}_1\frac{1 }{x} \; dx-\int\limits^{3/2}_1\frac{1 }{x+1} \; dx=ln(x)\bigg|_1^{3.2}-ln (x+1)\bigg|_1^{3.2}=$

$\displaystyle =ln\bigg(\frac{3}{5} \bigg)+ln(2) = ln \bigg(\frac{6}{5} \bigg)$

$\displaystyle A(x) = \frac{1}{2} \bigg(ln(\frac{6}{5} )\bigg)$