Integrales

26. Hallar el área de la región limitada por las curvas: y = | x – 1 |; y = x2 – 2x; x = 0; x = 2
Tenemos las funciones:
y = | x – 1 |
y = x2 – 2x
x = 0
x = 2

Entonces elÁrea de la región será:
As = 02x-1dx+ 02x2-2xdx
As = 01x-1dx+ 12x-1dx+ 02x2-2xdx
As = x- x22|01+x22-x|12-(x33- x2)|02
As = [1- 12+(0--12-83-4]
As = 12+ 12+ 43= 73 μ2

27. Hallar elárea de la región limitada por: y = | x – 2 | - | x – 6 |; x - y = 4
Tenemos las funciones:
y = | x – 2 | - | x – 6 |
x - y = 4

Entonces el Área de la región será:
As =[04x-4dx-04x-2-x-6dx]+[48x-2-x-6dx- 48x-4dx]
As = [04x-4dx- 02x-2-x-6dx- 24x-2-x-6dx]+[46x-2-x-6dx+ 68x-2-x-6dx- 48x-4dx]
As = [04x-4dx- 02-4dx- 242x-8]+[462x-8+ 684dx- 48x-4dx]
As = x22-4x|04+4x|02-x2-8x|24+x2-8x|46+4x|68-x22- 4x|48
As = [8-16+8--16--12+[-12--16+8-8]
As = 8μ2
28. Hallar el área de la región limitada por: y = | x – 2 |; y + x2 = 0; x = 1; x = 3
Tenemos las funciones:
y = | x– 2 | x = 1
y + x2 = 0 x = 3

As = 13|x-2|dx-13(-x2)dx
As = [12x-2dx+23|x-2|dx+13x2dx
As = [122-xdx+23(x-2)dx+13x2dx
As = 2x-x22|12+x22-2x|23+x33|13
As = 2-32+-32+2+263
As = 293 μ229. Hallar el área de la región limitada por las curvas : y = |x – 5|-| x + 3 |; x + y = 2
Tenemos las funciones:
y = | x – 5 |- | x + 3 |
x + y = 2As = [-60x-5-|x+3|dx--60(2-x)dx]+[010(2-x)dx- 010|x-5|-|x+3|dx]
As = [-6-3x-5-|x+3|dx+-30x-5-|x+3|dx- -602-xdx]+[0102-xdx-05x-5-|x+3|dx- 510xx-5-|x+3|dx]
As = [-6-38dx+-302-2xdx--602-x]+[0102-xdx-05(2-2x)dx- 510(-8)dx]
As = 8x|-6-3+2x-x2|-30-2x- x22|-60+2x-x22|010-2x-x2|05+8x|510
As = [24-15+30]+[-30+15+40]
As = 64 μ2
29. Hallar el área de la región limitada por: y = x – x2; y = -xTenemos las funciones:
y = x – x2
y = -x

As = 02(x-x2)dx-02(-x)dx
As = x22-x33|02+x22|02
As = -23+2
As = 43 μ2

IV. Hallar el volumen de los sólidos siguientes:
1. El volumen...