Encuentra el área bajo la curva

1.−f ( x )=2x2+3 [−1,3]

3

∫−1

3

(2 x2+3 )dx=2x3

3+3 x=¿

-1

( 2(3)3

3+3 (3))−( 2 (−1 )3

3+3 (−1 ))=30.66

2.−f ( x )= x2−1 [0,1 ]

1

∫0

1

( x2−1)dx= x2

4−x=¿

0

( (1 )2

4−1)−( (0 )2

4−(0 ))=−0.75

Encuentra el área entre las curvas

f ( x )=5 x2+3(−12 ,0.74)g ( x )=x+5

∫−0.5

0.74

( x+5 x )dx−∫−0.5

0.74

(5 x2+3 )dx=¿

0.74 0.74

x2

2+5 x−5 x

3

3+3 x=¿

-0.5 -0.5

[( (0.74 )2

2+5(0.74))−( (−0.5 )2

2+5(−0.5))]−¿

[( 5 (0.74 )3

3+3 (0.74 ))−( 5 (−0.5 )3

3+3 (−0.5 ))]=¿

(3.9738−(−2.375 ) )−(2.8953− (−1.7083 ) )=¿

¿6.3488−4.6036=1.74

Longitud del arco

y=x 2– 3 x+1 [0 ,1 ]

dydx

=2x−3

( dydx )2

= (2X−3 )2=4 x2+9

∫0

1

√1+( dydx )2

dx=∫0

1

√1+4 x2+9dx=¿¿

1

∫0

11+4 x2+9

32

32

=21+4 x2+9

32

3=¿

0

1

21+4 X+9

32

3 = [2 1+4 (1)+9323 ]−[2 1+4 (0 )+9

32

3 ]=

0

34.92−21.08=13.84