Alguien sabe como hallar esa área por rectángulo diferencial horizontal
Adjuntos:
![](https://es-static.z-dn.net/files/de9/2eeeb6282ff6f9226748aed7e517da63.jpg)
Drefius:
buen problema, por ello es difícil e.e
Respuestas
Respuesta dada por:
0
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 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](https://tex.z-dn.net/?f=Area%3D+%5Cint%5Climits%5E%7B0.5%7D_0+%7B%281-x-+%5Csqrt%7Bx%2F2%7D+%29%7D+%5C%2C+dx+%2B+%5Cint%5Climits%5E1_%7B0.5%7D+%7B+%28%5Csqrt%7Bx%2F2%7D+-%281-x%29%29%7D+%5C%2C+dx%2B%5C%5C+%5Cint%5Climits%5E2_1+%7B+%28%5Csqrt%7Bx%2F2%7D-%28x-1%29%29+%7D+%5C%2C+dx+%2B+%5Cint%5Climits%5E%7B2.5%7D_2+%7B%28%28x-1%29-+%5Csqrt%7Bx%2F2%7D+%29%7D+%5C%2C+dx++%5C%5C%3D%28x-x%5E2%2F2-%281%2F3%29%282x%5E3%29%5E%7B1%2F2%7D%29%7C%5E%7B0.5%7D_0%2B%28%281%2F3%29%282x%5E3%29%5E%7B1%2F2%7D-x%2Bx%5E2%2F2%29%7C%5E1_%7B0.5%7D%5C%5C%2B%28%281%2F3%29%282x%5E3%29%5E%7B1%2F2%7D-x%5E2%2F2%2Bx%29%7C%5E2_1%2B%28x%5E2%2F2-x-%281%2F3%29%282x%5E3%29%5E%7B1%2F2%7D%29%7C%5E%7B2.5%7D_2%5C%5C%5C%5C%3D1%2F2-1%2F8-1%2F6%2B%281%2F3%29%282%29%5E%7B1%2F2%7D-1%2B1%2F2-1%2F6%2B1%2F2-1%2F8%5C%5C%2B4%2F3-2%2B2-%281%2F3%29%282%29%5E%7B1%2F2%7D%2B1%2F2-1%2B25%2F8-5%2F2-%285%2F6%29%285%29%5E%7B1%2F2%7D-2%5C%5C%2B2%2B4%2F3%3D+%5Cfrac%7B65%7D%7B24%7D-+%5Cfrac%7B5+%5Csqrt%7B5%7D+%7D%7B6%7D%3D0.84++)
Saludos!
Saludos!
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