Question

O valor aproximado da integral e^3 e^2 x In (x) dx é Escolha uma: a. 463,34 b. 235,68 c. 156,77 d. 529,47 e. 392,52
alguém ajuda por favor!

Answer 1

Temos o seguinte:

$\int\limits^{e^3}_{e^2} {x \cdot ln(x)} \, dx$

Resolvendo a integral indefinida, com integral por partes:

$u = ln(x) \therefore du = \frac{1}{x} dx \\ \\ dv = x \therefore v = \int x \, dx \therefore v = \frac{x^2}{2} \\ \\ \int {x \cdot ln(x)} \, dx = ln(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \\ \\ \int {x \cdot ln(x)} \, dx = ln(x) \cdot \frac{x^2}{2} - \frac{1}{2} \int x\, dx \\ \\ \int {x \cdot ln(x)} \, dx = ln(x) \cdot \frac{x^2}{2} - \frac{1}{2} \cdot \frac{x^2}{2} \\ \\ \int {x \cdot ln(x)} \, dx = \frac{x^2}{2} (ln(x) - \frac{1}{2} )$

Agora aplicando os limites:

$\lim_{x \to e^3} \frac{x^2}{2} (ln(x) - \frac{1}{2} ) = \frac{(e^3)^2}{2} (ln(e^3) - \frac{1}{2} ) = \frac{e^6}{2} (3- \frac{1}{2} ) = \frac{5e^6}{4} \\ \\ \\ \lim_{x \to e^2} \frac{x^2}{2} (ln(x) - \frac{1}{2} ) = \frac{(e^2)^2}{2} (ln(e^2) - \frac{1}{2} ) = \frac{e^4}{2} (2 - \frac{1}{2} ) = \frac{3e^4}{4}$

Logo temos:

$\int\limits^{e^3}_{e^2} {x \cdot ln(x)} \, dx = \frac{5e^6}{4} - \frac{3e^4}{4} = \frac{5e^6 - 3e^4}{4} \approx 463.34$