Question

Alguien sabe como hallar esa área por rectángulo diferencial horizontal

Answer 1

Tienes que realizar 4 integrales:
$Area= \int\limits^{0.5}_0 {(1-x- \sqrt{x/2} )} \, dx + \int\limits^1_{0.5} { (\sqrt{x/2} -(1-x))} \, dx+\\ \int\limits^2_1 { (\sqrt{x/2}-(x-1)) } \, dx + \int\limits^{2.5}_2 {((x-1)- \sqrt{x/2} )} \, dx \\=(x-x^2/2-(1/3)(2x^3)^{1/2})|^{0.5}_0+((1/3)(2x^3)^{1/2}-x+x^2/2)|^1_{0.5}\\+((1/3)(2x^3)^{1/2}-x^2/2+x)|^2_1+(x^2/2-x-(1/3)(2x^3)^{1/2})|^{2.5}_2\\\\=1/2-1/8-1/6+(1/3)(2)^{1/2}-1+1/2-1/6+1/2-1/8\\+4/3-2+2-(1/3)(2)^{1/2}+1/2-1+25/8-5/2-(5/6)(5)^{1/2}-2\\+2+4/3= \frac{65}{24}- \frac{5 \sqrt{5} }{6}=0.84$

Saludos!