Question

Sa se calculeze integrala din poza

Answer 1

Răspuns:

$\frac{1}{3}$

Explicație pas cu pas:

$\displaystyle \int_{-1}^{1}\frac{x^2}{e^x+1}dx = I\\ \\ \textrm{Integrare prin parti:}\\ \\ u = x^2\implies du = 2xdx\\ \\ dv = \frac{1}{e^x+1}dx \implies v = \int \frac{1}{e^x+1}dx \\ \\ \int \frac{1}{e^x+1}dx \\ \\ t = e^x + 1\implies dt = e^xdx \implies dx = \frac{dt}{e^x} = \frac{dt}{t-1} \\ \\ \implies \int \frac{1}{e^x+1}dx = \int \frac{1}{t}\cdot \frac{dt}{t-1} = \int \frac{dt}{t(t-1)}\\ \\ \frac{1}{t(t-1)} = \frac{1}{t-1} - \frac{1}{t} \implies \int \frac{dt}{t-1} - \int \frac{dt}{t} = \ln(t-1) - \ln(t) = \ln(e^x+1-1) - \ln(e^x+1) = x - \ln(e^x+1) = v$

$\displaystyle \implies I = uv\Biggr|_{-1}^{1} - \int_{-1}^{1} v du = x^2\Big(x - \ln(e^x+1)\Big) \Biggr|_{-1}^1- \int_{-1}^{1} \Big(x - \ln(e^x+1)\Big)\cdot 2xdx = x^3\Biggr|_{-1}^{1} - x^2\ln(e^x+1)\Biggr|_{-1}^{1} - 2\int_{-1}^{1}x^2dx + \int_{-1}^{1} 2x\ln(e^x+1)dx = x^3\Biggr|_{-1}^{1} - \frac{2}{3}x^3\Biggr|_{-1}^{1} - \Bigg(\underbrace{x^2\ln(e^x+1) \Biggr|_{-1}^1 - \int_{-1}^1 \ln(e^x+1)\cdot 2xdx}_{\textstyle\begin{gathered} \int_{-1}^1 \frac{x^2 \cdot e^x}{e^x+1}dx\end{gathered}}\Bigg)$

$\displaystyle I = \int_{-1}^1 \frac{x^2}{e^x+1}dx\\ \\ \textrm{Domeniul de integrare este simetric: [-1; 1]}\implies \textrm{Putem inlocui x cu -x si valoarea este aceasi}\\ \\ I = \int_{-1}^1 \frac{(-x)^2}{e^{-x} + 1}dx = \int_{-1}^1\frac{x^2}{\frac{1}{e^x} + 1}dx = \boxed{\int_{-1}^1 \frac{x^2 \cdot e^x}{e^x + 1} = I}$

$\displaystyle I = \Bigg(x^3\Biggr|_{-1}^{1}\Bigg) \cdot \Bigg(1 - \frac{2}{3}\Bigg) - \int_{-1}^1 \frac{x^2 \cdot e^x}{e^x + 1} \implies I = \Bigg(x^3\Biggr|_{-1}^{1}\Bigg) \cdot \Bigg(1 - \frac{2}{3}\Bigg) - I\\ \\ 2I = \Big(1 - (-1)\Big)\cdot \frac{1}{3}\\ \\ I = \frac{2}{6} = \boxed{\frac{1}{3}}$