Question

Essa é uma questão de calculo, pode ajudar?

Answer 1

$\displaystyle I=\int_{0}^adx\int_{x/a}^x\frac{x}{x^2+y^2}dy\\ \\ I=\int_{0}^ax\,dx\int_{x/a}^x\frac{1}{x^2+y^2}dy\\ \\ I=\int_{0}^ax\left\left[\dfrac{1}{x}\arctan\left(\frac{y}{x}\right)\right]\right|_{y=x/a}^{y=x}\,dx\\ \\ I=\int_{0}^ax\left[\dfrac{1}{x}\arctan\left(\frac{x}{x}\right)-\dfrac{1}{x}\arctan\left(\frac{x/a}{x}\right)\right]dx\\ \\ I=\int_{0}^a \frac{\pi}{4}-\arctan \frac{1}{a}\, dx\\ \\ \\ \boxed{I=\frac{\pi a}{4}-a\cdot\arctan \frac{1}{a}}$

2)

$\displaystyle I=\int_{0}^{a}y\,dy \int _{y-a}^{2y}x\,dx\\ \\ \\ I=\int_{0}^{a}\left.\left(\frac{x^2}{2}\right)\right|_{y-a}^{2y}y\,dy\\ \\ I=\int_{0}^{a}y\left(\frac{4y^2}{2}-\frac{(y-a)^2}{2}\right)\,dy\\ \\ I=\int_{0}^{a}\frac{3y^3}{2}+ay^2-\frac{a^2y}{2}\,dy\\ \\ I=\left.\left(\frac{3y^4}{8}+\frac{ay^3}{3}-\frac{a^2y^2}{4}\right)\right|_{0}^{a}$

$\displaystyle I=\frac{3a^4}{8}+\frac{a^4}{3}-\frac{a^4}{4}\\ \\ \boxed{I=\frac{11a^4}{24}}$