Question

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mă poate ajuta cineva cu asta
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Answer 1

Notam:

$I=\int\limits^a_{-a}f(x)\ dx=\int\limits^a_{-a}f(-x) \dx\\\\I=\int\limits^a_{-a} {\frac{x^4}{e^x+1} } \, dx =\int\limits^a_{-a} {\frac{(-x)^4}{e^{-x}+1} } \, dx$

$I=\int\limits^a_{-a} {\frac{x^4}{e^{-x}+1} } \, dx=\int\limits^a_{-a}\frac{x^4}{\frac{1}{e^x} +1} \ dx\\\\I=\int\limits^a_{-a}\frac{x^4\cdot e^x}{e^x+1}\ dx$

Adunam cele doua integrale:

$2I=\int\limits^a_{-a}\frac{x^4+x^4\cdot e^x}{e^x+1}\ dx\\\\ 2I=\int\limits^a_{-a}\frac{x^4(1+e^x)}{e^x+1} \ dx\\\\2I=\int\limits^a_{-a}x^4\ dx\\\\2I=\frac{x^5}{5}\ |_{-a}^a$

$2\cdot (-\frac{32}{5})= \frac{a^5-(-a^5)}{5} \\\\2\cdot (-\frac{32}{5})=\frac{2a^5}{5}\ \ \ |:\frac{2}{5}\\\\ -32=a^5\\\\ a=-2$

Raspuns: e) -2

Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/956588

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