1.

-Hallar el área limitada por la curva y=x²-3x+2

a= 1
b= -3
c= 2
v (3/2, -1/4)
x1 (2,0)
x2 (1,0)
y (0,2)
−b −(−3) 3
x= = =
2a 2( 1) 2
4 ac−b
2
4 (1)(2)−(−3)2 −1
y= = =
4a 4 (1) 4

−b+ √ b2−4 ac −(−3 ) + √ (−3)2−4(1)(2) 3+ √ 9−8 3+ √ 1 3+1 4

x 1= = = = = = =2
2a 2(1) 2 2 2 2

−b− √ b 2−4 ac −(−3 )− √ (−3)2−4 (1)(2) 3−√ 9−8 3−√ 1 3−1 2

x 2= = = = = = =1
2a 2(1) 2 2 2 2

y= (0) ²-3(0) +2
y=2

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

2
A=∫ [ 0− ( X −3 X + 2 )−0 ] ⅆx
2

2
A=∫ [ −X +3 X −2 ] ⅆx
2

1
 2 2 2
A=−∫ X ⅆx +3 ∫ X dx−2∫ ⅆx
2

1 1 1

2 3
−X
−∫ X ⅆx =
2

1 3
2 2
3x
3∫ xⅆx =
1 2
2
−2∫ ⅆx =-2x
1

[ ]
2
−x3 3 x 2
A= + −2 x
3 2 1

[ ][ ]
3 2 3 2
−( 2 ) +3 ( 2 ) −( 1 ) +3 ( 1 ) +5 1
A= −2(2) - −2(1) =¿ ¿ = U²
3 2 3 2 6 6

2.-Determinar el área limitada por la curva y=-x²+4x-3

a= -1
b= 4
c= -3
v (2,1)
x1 (1,0)
x2 (3,0)
y= (0,-3)
−b −4 −4
x= = = =2
2 a 2 (−1 ) −2

4 ac−b
2
4 (−1)(−3)−(4)2 12−16 −4
y= = = = =1
4a 4(−1) −4 −4

−b+ √ b2−4 ac −4 + √ (4)2−4 (−1)(−3) −4 + √ 16−12 −4 + √ 4 −4 +2 −2

x 1= = = = = =
2a 2(−1) −2 −2 −2 −2
=1

−b− √ b 2−4 ac −4−√(4)2−4(−1)(−3) −4−√ 16−12 −4−√ 4 −4−2

x 2= = = = = =
2a 2(−1) −2 −2 −2
−6
=3
−2
y=- (0) ²+4(0) -3
y=-3

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

3
A=∫ [ (−X + 4 X −3 )−0 ] ⅆx
2

3
A=∫ [ −X +4 X−3 ] ⅆx
2

3 3 3
A=−∫ X ⅆx + 4∫ X dx−3 ∫ ⅆx
2

1 1 1

3 3
−X
−∫ X ⅆx =
2

1 3
3 2
4x
4 ∫ xⅆ x =
1 2
3
−3 ∫ ⅆx =-3x
1

[ ]
3
−x3
A= +2 x ²−3 x
3 1

[ ][ ]
3 3
−( 3 ) −( 1 ) 4
A= +2(3)²−3(3) - +2(1)²−3(1) = U²
3 3 3

3.-Hallar el área formada por la curva y=x² y las rectas verticales

cuando x1=0; x2=-2
a= 1
b= 0
c= 0
v (0,0)
px1 (0,0)
px2 (0,0)
py= (0,0)
−b 0 0
x= = = =0
2 a 2( 1) 2

4 ac−b
2
4 (1)(0)−(0)2 0
y= = = =0
4a 4 (1) 4

−b+ √ b2−4 ac 0+ √ (0)2−4 (1)(0) 0+ √ 0−0

x 1= = = =0
2a 2(1) 2

−b− √ b 2−4 ac 0− √ (0)2−4 (1)(0) 0− √ 0−0

x 2= = = =0
2a 2(1) 2

y=- (0) ²
y=0

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

0
A=∫ [ X −0 ] ⅆx
2

−2

0
A=∫ [ X ] ⅆx
2

−2

0
A=∫ X ⅆx
2

−2
0 3
X
∫ X ⅆx =
2

3
−2

[ ]
0
x3
A=
3 −2

A= [ ][ ]
( 0 )3
3
-
(−2 )3
3

8 8
A=0+ = U²
3 3

4.-Hallar el área encerrada por la curva y=x²+2x+1 y las rectas

verticales cuando X1=-1; X2=-4
a= 1
b= 2
c= 1
v (-1,0)
px1 (-1,0)
px2 (-1,0)
py= (0,1)
vértice
−b −2 −2
x= = = = -1
2 a 2( 1) 2
2 2
4 ac−b 4 (1)(1)−(2) 4−4 0
y= = = = =0
4a 4 (1) 4 4

−b+ √ b2−4 ac −2+ √ (2)2−4 (1)(1) −2+ √ 4−4 −2+ √ 0

x 1= = = = = -1
2a 2(1) 2 2

−b− √ b 2−4 ac −2−√(2)2−4(1)(1) −2−√ 4−4 −2−√ 0

x 2= = = = = -1
2a 2(1) 2 2

y=- (0) ²+2(0) +1

y=1
 b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

−1
A=∫ [ ( X +2 X +1 ) −0 ] ⅆx
2

−4

−1 −1 −1
A=∫ X ⅆx +2 ∫ X dx+ 1 ∫ ⅆx
2

−4 −4 −4

−1 3
∫ X 2 ⅆx = X3
−4

−1
2 ∫ xⅆx =x²
−4

−1
1 ∫ ⅆx =x
−4

[ ]
−1
x3 2
A= + x + x
3 −4

A= [ (−1 )3
3
2
+ (−1 ) +(−1) - ][
(−4 )3
3
2
+ (−4 ) +(−4) ]
−1 28
A= + ¿ 9 U²
3 3

5.-Hallar el área limitada por la curva y=x²+x+1, y las rectas verticales

cuando x1=0; x2= 2
a= 1
b= 1
c= 1
v (-1/2,3/4)
x1 (∞,0)
x2 (∞,0)
y= (0,1)
−b −1 −1
x= = =
2 a 2( 1) 2

4 ac−b
2
4 (1)(1)−(1)2 4−1 3
y= = = =
4a 4 (1) 4 4

−b+ √ b2−4 ac −1+ √ (1)2−4 (1)(1) −1+ √ 1−4 −1+ √−3

x 1= = = = =∞
2a 2(1) 2 2

−b− √ b 2−4 ac −1−√(1)2−4(1)(1) −1−√ 1−4 −1−√−3

x 2= = = = =∞
2a 2(1) 2 2

y=- (0) ²+(0) +1

y=1

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

2
A=∫ [ ( X + X + 1 )−0 ] ⅆx
2

2 2 2
A=∫ X ⅆx +∫ xⅆx +1∫ ⅆx
2

0 0 0
2 3
X
∫ X ⅆx =2
3
0

2 2
x
∫ xⅆx = 2
0

2
1∫ ⅆx =x
0

[ ]
2
x3 x2
A= + + x
3 2 0

A=
3 [
( 2 )3 ( 2 )2
+
2
+(2) -
3
+
2 ][
( 0 )3 ( 0 )2
+(0) ]
20
A= U²
3

6.-calcula el área encerrada por las funciones y=x²+x+1, y=x, y las

rectas verticales x1=0; x2=1
y=x²+x+1
a= 1
b= 1
c= 1
v (-1/2,3/4)
px1 (∞,0
px2 (∞,0)
py= (0,1)
−b −1 −1
x= = =
2 a 2( 1) 2

4 ac−b
2
4 (1)(1)−(1)2 4−1 3
y= = = =
4a 4 (1) 4 4

−b+ √ b2−4 ac −1+ √ (1) −4 (1)(1) −1+ √ 1−4 −1+ √−3

2
x 1= = = = =∞
2a 2(1) 2 2

−b− √ b 2−4 ac −1−√(1) −4(1)(1) −1−√ 1−4 −1−√−3

2
x 2= = = = =∞
2a 2(1) 2 2

y=- (0) ²+(0) +1

y=1
y=x
x=0
y= (0)
y=0
px= (0,0)
py= (0,0)

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

1
A=∫ [ ( X + X + 1 )−( x) ] ⅆx
2

1
A=∫ [ ( X +1 ) ] ⅆx
2

1 1
A=∫ X ⅆx +1∫ ⅆx
2

0 0

1 3
X
∫ X ⅆx =2

3
0

1
1∫ ⅆx =x
0

[ ]
1
x3
A= + x
3 0

A= [ ][( 1 )3
3
+(1) -
( 0 )3
3
+(0) ]
4
A= U²
3
7.-calcular el área limitada entre la recta y=x-3 y el eje de las x´, con las
rectas verticales x1=1; x2=2
y=x-3
x-3=0
y= (0)-3
x=3
y=-3
px= (3,0)
py= (0,-3)

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

2
A=∫ [ 0−( X−3 ) −0 ] ⅆx
1

2
A=∫ [ (−X +3 ) ] ⅆx
1

2 2
A=−∫ Xⅆx +3∫ ⅆx
1 1

2 2
X
-∫ Xⅆx −
1 2
2
3∫ ⅆx =3x
1

[ ]
2
−x2
A= +3 x
2 1

[ ][ ]
2 2
−( 2 ) −( 1 ) 5 3
A= +3(2) - +3(1) =4− = U²
2 2 2 2
8.-calcular el área limitada entre la recta y=x+6 y el eje de las x´, con
las rectas verticales x1=4; x2=8
Función lineal
y=x+6
x+6=0
y= (0) +6
x=-6
y=6
px= (-6,0)
py= (0,6)

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

8
A=∫ [ ( X +6 )−0 ] ⅆx
4

8
A=∫ [ ( X +6 ) ] ⅆx
4

8 8
A=∫ Xⅆx + 6∫ ⅆx
4 4

8 2
∫ Xⅆx= X2
4
8
6 ∫ ⅆx =6x
4

[ ]
8
x2
A= +6 x
2 4
A=
2[
( 8 )2
+6 (8) -
( 4 )2
2 ][
+6( 4) ]
A=80−32 =48 U²

9.-Calcular el área de la superficie entre las curvas y=x²-4; y=8-2x²

y=x²-4
a= 1
b= 0
c= -4
v (0, - 4)
px1 (2,0)
px2 (-2,0)
py= (0,-4)
−b 0 0
x= = = =0
2 a 2( 1) 2
2
4 ac−b
2
4 (1)(−4)−(0) −16−0 −16
y= = = = =-4
4a 4 (1) 4 4

−b+ √ b2−4 ac 0+ √ (0) −4 (1)(−4) 0+ √ 0+16 0+ √ 16 0+4

2
x 1= = = = = =2
2a 2 (1) 2 2 2

== √
−b− √ b 2−4 ac 0− (0)2−4 (1)(−4) 0− √0+16 0− √16 0−4
x 2= = = = = -2
2a 2(1) 2 2 2
y= (0) ²-4
y=-4

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

2
A 1=∫ [ 0−( X −4 ) −0 ] ⅆx
2

−2

2
A 1 ∫ [−X + 4 ] ⅆx
2

−2

2 2
A 1=−∫ X ⅆx + 4 ∫ d x
2

−2 −2

2 3
−x
−∫ x ² ⅆx =
−2 3
 2
4 ∫ dx =4x
−2

[ ]
2
−x 3
A 1= +4 x
3 −2

[ ][ ]
3 3
−( 2 ) −(−2 )
A 1=¿ +4 (2) - +4−(2)
3 3
16 16 32
A 1= + = U²
3 3 3

Y=8-2X²
a=-2 b= 0 c= 8
v= (0, 8)
px1= (-2,0)
px2 =(2,0)
py= (0,8)
−b 0 0
x= = = =0
2 a 2 (−2 ) −4
4 ac−b
2
4 (−2)(8)−(0)2 −64−0 −64
y= = = = =8
4a 4 (−2) −8 −8

−b+ √ b2−4 ac 0+ √ (0) −4 (−2)(8) 0+ √ 0+64 0+ √ 64 0+8 8

2
x 1= = = = = = =- 2
2a 2(−2) −4 2 −4 −4
−b− √ b 2−4 ac 0− √(0) −4 (−2)(8) 0− √0+ 64 0− √ 64 0−8 −8
2
x 2= = = = = = =2
2a 2(−2) −4 2 −4 −4
y= 8-2(0) ²
y=8
b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

2
A 2=∫ [ ( 8−2 X )−0 ] ⅆx
2

−2

2
A 2=∫ [ 8−2 X ] ⅆx
2

−2

2 2
A 2=8 ∫ ⅆx −2 ∫ X ² dx
−2 −2
 2
8 ∫ dx=8x
−2
2 3
−2 x
−2 ∫ x ² ⅆx =
−2 3

[ ]
2
2 x3
A 2= 8 x−
3 −2

[ ][ ]
3 3
2(2) 2(−2)
A 2=¿ 8(2)− - 8(−2)−
3 3
32 32 64
A 2= + = U²
3 3 3
64 32
AT= + =32U²
3 3

10.-calcular el área limitada entre la función y=2x-2 y el eje de las x´,

con las rectas verticales x1=1; x2=3
y=2x-2
2x-2=0
2
x=
2
x=1
y=-2
px= (1,0)
py= (0,-2)

b
A=∫ [ f ( x )−g ( X ) ] ⅆx
a

3
A=∫ [ ( 2 X−2 )−0 ] ⅆx
1

3
A=∫ [ ( 2 X−2 ) ] ⅆx
1

3 3
A=2∫ Xⅆx −2∫ ⅆx
1 1

3 2
2X
2 ∫ X ⅆx =
1 2
3
−2∫ ⅆx =-2x
1

3
A=[ X 2−2 x ]1

A=¿[(3)2−2(3) ¿- [(1)2−2(1)¿

A=3+1=4U²