( 32 x )= 53 arcsen ( 32 x ) sen x +7 cos x 1 dx. x x. e 2 x +1 dx. 5x 7 dx. x sen x dx. x 4 x x 1 dx. x 2 dx. dxx. x x x dx. 1 x.

I.E.S. Juan Carlos I

Ciempozuelos (Madrid)

Matemáticas II

*

Análisis III: Integrales

*

1. Integrales inmediatas (o casi inmediatas):

a)

_∫



4x

2

_{– 5x}

_

₇

_

_dx

_{=4· 1} 3x 3_{– 5· 1} 2x 2_{+7 x +C= 4 x}3 3 – 5 x 2 2 +7 x + C

b)

∫

3 x

3

_{– 5 x}

2

+

3 x +4 dx

=3· 1 4x 4_{– 5· 1} 3x 3_{+3· 1} 2x 2_{+4 x +C= 3 x}4 4 – 5 x 3 3 + 3 x2 2 +4 x + C

c)

∫

1

2 x



7

dx

= 1 2ln∣2 x + 7∣+ C =ln

√

∣2 x +7∣+C

d)

∫

5

dx



x

∫

x −1₅ dx = 5 4x 4 5_{+C= 5} 5

√

x4 4 +C

e)

∫



x – sen x



dx

∫

x dx+

_∫

−sen x dx=1 2x 2 +cos x +C

f)

∫

sen x+7 cos x – 1 dx

∫

sen x dx+7

∫

cos x dx−

∫

dx=−cos x +7 sen x − x +C

g)

∫

tg

2

x dx

∫

1+tg2x −1dx=

∫

1+tg2x dx−

∫

dx=tg x − x +C

h)

_∫

_√

x – 2 dx

∫

x 1 2_dx−2

_∫

_dx=2 3x 3 2_{−2x= 2}

√

x3 3 −2 x +C

i)

∫

2

√

x

dx

∫

4 2

√

xdx= 4

∫

1 2

√

xdx=4

√

x +C

j)

∫

e

2 x +1

dx

∫

1₂2e2 x + 1_dx=1 2

∫

2e 2 x +1_dx=1 2e 2 x +1 +C

k)

∫

cos(a x +b)dx

∫

1_aa· cos(a x +b)dx=1

a

∫

a· cos(a x +b)dx= 1 asen(a x+ b)+C

l)

∫



x

2



4 x



x

2

−

1



dx

=

∫

x4_{+4 x}3 −x2_{−4 x dx=}1 5x 5 +x4 −1 3x 3 −2 x2 +C

m)

_∫

3



x

2

dx

= 1 3

√

2

∫

x 1 3_dx= 1 3

√

2 3 4x 4 3_+C=3 4 3

√

x4 2 +C

n)

∫

_x−1

dx

=ln∣x− 1∣+ C

o)

∫

x





x

x

2

dx

=

∫

1_x+

√

_xx2dx=

∫

1 xdx +

∫

x −3 2_{dx=ln∣x∣− 2 x}− 1 2_+C

p)

∫

3

1



x

2

dx

=3

∫

1 1+x2dx=3arctg x +C

q)

∫

5



2 – 3 x

2

dx

= 5

√

2

∫

1

√

1−3 2x 2 dx= 5

√

2

∫

1

√

1−

(

√

3 2x

)

2dx= 5

√

2

√

2 3arcsen

(

√

3 2x

)

= 5

_√

3arcsen

(

√

3 2x

)

r)

∫

1

(

x−2)

2

dx

=

∫

(x−2) −2_{dx=−( x−2)}−1_+C=− 1 x −2+C

s)

∫

e

2x −3

_dx

₌1 2e 2 x −3 +C

t)

∫

5 dx

1+7 x

2 =5

∫

1 1+(

√

7x )2dx= 5

√

7arctan(

√

7x )+C

u)

_∫

x

2

– 5 x+4

x +1

dx

x2 = −5 x+ 4=(x−6 )(x +1)+ 10

∫

x−6+ 10 x+1dx= 1 2x 2_{−6x +10 ln∣x+1∣+C}

v)

∫

x

3

– 3x

2



x – 1

x

−

2

dx

x3 = −3 x2 +x+1=(x−2)(x2 +x+3)+5

∫

x 2 +x+3+ 5 x −2dx= x 3 3+ x 2 2+3x +5ln∣x− 2∣+C

w)

∫

x – 1

√

2 x –

√

x +1

dx

=

∫

(x –1)(

√

2x +

√

x +1) (

_√

2x –

√

x +1)(

√

2 x+

√

x+1)dx =

∫

(x –1)(

√

2x +

√

x +1) x−1 dx=

∫

(

√

2x +

√

x +1)dx= 2

√

2 3 x 3 2₊2 3(x+ 1) 3 2_+C

x)

∫

e

x

(

x−1) – e

x

(

x−1)

2

dx

Derivada de un cociente= ex x −1+C

y)

∫

2 e

2 x

+e

2 x

e

x

dx

=

_∫

3 ex_{dx =3e}x_+C

z)

∫

3

11

x

dx

=3

∫

e −ln 11· x_{dx =−} 3 ln11e −ln11 · x_{+C =} 3 11x_ln11+C

2. Integrales de funciones compuestas / cambio de variable:

a)

_∫

(2 x+6)

5

dx

= t=2 x + 6 d t =2dx

∫

t5dt 2= 12

∫

t 5_{dt= 1} 12t 6 +C=(2x +6)6 12 +C

b)

∫

5

7 x – 9

dx

t=7 x −9= d t=7 dx

∫

5 t dt 7= 57

∫

1 t dt= 57ln∣t∣+C= 57ln∣7x −9∣+C

c)

∫

3 x e

5 x2−7

dx

= t =5 x2₋₇ d t =10 x dx

∫

₁₀3 et_{dt =} 3 10e t_+C=3 10e 5 x2 −7_+C

d)

∫

x

(

5x+3)(5x−3)

dx

∫

x 25 x2₋₉dx t =25 x=2₋₉ d t =50 x dx

∫

₅₀1 ·1 tdt = 1 50ln∣t∣+C= 1 50ln

∣

25 x 2₋₉

_∣

_+C

e)

∫

cos x sen

3

_{x dx}

₌ t=sen x d t =cos x dx

∫

t3_{dt = 1} 4t 4_{+C= sen}4x 4 +C

f)

∫

x

3

√

x

4

– 1

dx

t2==x4₊₁ 2 t d t= 4 x3 dx

∫

_2tt dt=1 2

∫

dt= 1 2t +C=

√

x4₋₁ 2 +C

g)

∫

sen x cos x dx

= t=sen x d t =cos x dx

∫

t dt = 1 2t 2_{+C= sen}2x 2 +C

h)

∫

4 x – 3

2 x

2

_{– 3 x−14}

dx

t =2 x2=_{−3 x −14} d t =(4 x−3)dx

∫

1_tdt =ln∣t∣+C=ln

∣

2 x2_{−3 x −14}

_∣

_+C

i)

_∫

1

x ln x

dx

= t =lnx d t =dx x

∫

1_tdt=ln∣t∣+C=ln

(

∣ln∣x∣∣

)

+C

j)

∫

x

(

x

2

+

3)

5

dx

t =x=2 +3 d t =2 x dx

∫

_t15 dt 2= 1 2

∫

t −5_{dt =−}1 8t −4 +C=− 1 8( x2₊₃₎4+C

k)

_∫

1

x

ln

3

x dx

t =lnx= d t =dx x

∫

t3_{dt =}1 4t 4_+C=1 4ln 4_∣x∣+C

l)

∫

2 x dx



9−x

2

dx

t=9− x= 2 d t =−2xdx

∫

−1

_√

_tdt =−2

√

t +C=−2

√

9− x2_+C

m)

∫

sen x dx

_cos

5

x

d t=−senx dxt =cos x=

∫

−1 t5dt =−

∫

t −5_{dt =}1 4t −4 +C= 1 4cos4_x+C

n)

∫



x−1



x

2

– 2 x dx

= t2 =x2_{−2 x} 2 t d t =(2 x −2)dx t dt =(x−1)dx

∫

t2_{dt= 1} 3t 3_+C=(x2−2 x ) 3 2 3 +C

o)

_∫

arcsen x



1 – x

2

dx

= t =arcsenx d t = dx √1−x2

∫

t dt= 1 2t 2 +C= arcsen2x 2 +C

p)

_∫



1



ln x



2

x

dx

= t=1 +ln x d t = dx x

∫

t2_{dt= 1} 3t 3_+C=(1 +ln x )3 3 +C

q)

∫

x

2



x

3

−

1



3 5

_dx

₌ t =x3 −1 d t=3 x2_dx

∫

t 3 5dt 3= 524t 8 5_+C=5( x3−1) 8 5 24 +C

r)

_∫

x

4

_e

x5_3

dx

= t =x5₊₃ d t=5 x4_dx

∫

etdt 5= 1 5e t_+C=ex 5 +3 5 +C

s)

_∫

x sen



x

2

_

_

_

_dx

₌ t =x2 + π d t =2 x dx

∫

sen t dt 2=− 12cos t +C=− cos ( x2_{+ π )} 2 +C

t)

_∫

x ln



x

2



3



x

2



3

dx

= t =ln(x2 +3 ) d t = 2 x x2₊₃dx

∫

t dt 2= 14t 2_+C=ln2(x2+3) 4 +C

u)

∫

e

x

1



e

2x

dx

= t =ex d t =ex_dx

∫

1 1+t2dt=arctan t +C =arctan(e x_)+C

v)

∫

1



tg

2

_x

tg x

dx

d t =(1 +tgt =tg x=2_{x )dx}

∫

1_tdt =ln∣t∣+C=ln∣tg x∣+ C

w)

∫

tg x dx

∫

sen x_cosxdx _{t =cosx}=

d t =−senx dx

∫

−1 tdt =−ln∣t∣+C=−ln∣cos x∣+C=ln∣sec x∣+C

x)

∫

3

x

1



9

x

dx

∫

3x 1 +(3x₎2dx _{t =3}=x d t=ln 3· 3x_dx

∫

_ln31 · 1 1+t2dt= 1_ln3arctant +C= arctan(3x ) ln 3 +C

y)

_∫

x

3

_sen

_

_x

4

_{– 3 }

_

_dx

₌ t =x4_{−3 π} d t= 4 x3 dx

∫

sen t dt 4=− 14cos t +C=− cos( x4_{−3π )} 4 +C

z)

_∫

e

x

5



x

dx

= t =√x d t = dx 2√x

∫

₅2et_{dt = 2} 5e t_{+C= 2e}√x 5 +C

3. Integrales por partes:

a)

∫

2 x e

−x

_dx

= u=2 x →u ' =2 v '=e−x →v =−e−x −2x e−x₊

∫

2e−x_{dx=−2x e}−x_{– 2e}−x_{+C=−2( x +1) e}−x_+C

b)

∫

x e

− x3

_dx

u =x =→u '=1 v '=e− x3 _{→v=−3 e}− x3 −3x e− x 3₊

_∫

_3e− x 3_{dx=−3 x e}− x 3_–9e− x 3_{+C=−3( x +3)e}− x 3_+C

c)

∫

x

2

e

2 x

dx

= u=x2 →u ' =2x v ' =e2 x →v = 1 2e 2 x 1 2x 2_e2x −

∫

x e2x_dx = u= x →u '=1 v '=e2 x →v= 1 2e 2 x 1 2x 2_e2x −1 2x e 2x +1 2

∫

e 2 x_dx= =1 2x 2_e2x₋1 2x e 2x₊1 4e 2x_+C=2 x2−2 x +1 4 e 2 x_+C

d)

∫

[

x e

x

]

3

dx

=

∫

x3_e3 x_dx = u =x3 →u '=3 x2 v ' =e3 x →v =1 3e 3 x 1 3x 3_e3 x −

∫

x2_e3x_dx = u =x2 →u ' =2 x v '=e3 x →v =1 3e 3 x =1 3x 3_e3 x₋1 3x 2_e3 x₊2 3

∫

x e 3 x_{dx =} u= x →u ' =1 v '=e3 x →v=1 3e 3 x 1 3x 3_e3 x₋1 3x 2_e3 x₊2 9x e 3 x₋2 9

∫

e 3 x_dx = 1 3x 3_e3 x_{− 1} 3x 2_e3 x_{+ 2} 9x e 3 x_{− 2} 27e 3 x_{+C =9 x}3−9 x2+6 x− 2 27 e 3 x_+C

e)

∫

x ln x dx

u=lnx=→u ' =1_x v ' =x →v =1 2x 2 1 2x 2_{ln x− 1} 2

∫

x dx = 12x 2_{ln x − 1} 4x 2_{= x}2 4 (2ln∣x∣− 1)+C

f)

∫

x

2

_{ln xdx}

_{u =lnx}= →u ' = 1_x v '= x2 →v =1 3x 3 1 3x 3_{ln x − 1} 3

∫

x 2_{dx = 1} 3x 3_{ln x− 1} 9x 3_{= x}3 9 (3 ln∣x∣−1)+C

g)

∫

ln x

_x

dx

u =lnx=→u '=1 x v ' =1 x →v=ln x ln2_x−

_∫

ln x x dx

⏟

I

→I=ln2_{x −I →2 I =ln}2_{x+C →I =}ln2∣x∣

2 +C

h)

∫

x

√

x +1 dx

u= x =→u ' =1 v '=√x+ 1 →v =2 3(x +1) 3 2 2 3x ( x+1) 3 2₋2 3

∫

(x +1) 3 2_{dx =}2 3x ( x +1) 3 2₋ 4 15(x +1 ) 5 2_+C

i)

∫

2 x

2 3

√

x +1 dx

= u =2 x2 →u '=4 x v '=3 √x+ 1 →v = 3₄(x +1 ) 4 3 3 2x 2 (x +1) 4 3 −

∫

3 x (x +1) 4 3_dx = u=3 x →u '=3 v '=(x +1 ) 4 3 →v= 3₇(x +1 ) 7 3 =3 2x 2_(x+1)34₋9 7x ( x+1) 7 3₊9 7

∫

(x +1) 7 3_dx=3 2x 2_{(x +1)}43₋9 7x ( x +1) 7 3₊27 70(x+ 1) 10 3 _+C

j)

∫

e

x

_{sen x dx}

= u =ex _{→u '=e}x v ' =sen x →v =−cos x −cos x ex₊

∫

ex_{cos x d x}

⏟

u=ex →u '=ex v ' =cosx →v=sen x =−cos x ex_{+sen x e}x₋

∫

ex_{sen x dx}

⏟

I

I=−cos x ex+sen x ex−I→ 2I=(sen x−cos x)ex+C→I =1

2(sen x −cos x ) e x +C

k)

_∫

_e

x

₂

x+1

_dx

= u =2x + 1 →u '=ln 2 ·2x+ 1 v '=ex _→v=ex ln 2·ex₂x +1_{−ln 2}

_∫

_ex₂x +1_dx

⏟

I →I =ln2ex₂x +1_{−ln2· I →I=} ln2 1+ln2e x₂x +1

l)

∫

3 x cos x dx

u=3 x =→u' =3 v ' =cosx →v =sen x

m)

_∫

sen

2

_{x dx}

=

∫

_⏟

sen x · senx dx I = u= sen x →u ' =cosx v '=sen x →v =−cosx

−sen x cos x +

_∫

cos x · cos x dx=−sen x cos x +

_∫

cos2_{x dx=}

=−sen x cos x +

_∫

(1−sen2_{x)dx=−sen x cosx +}

_∫

_dx−

_∫

_sen2_{x dx}

⏟

I

→ 2I=−senx cos x +x → I=−1

2sen x cos x + x 2+C

n)

∫

arctan x dx

= u =arctanx →u '= 1 √1+x2 v '=1 →v= x x · arctanx −

_∫

x

√

1+ x2dx t =√=1 +x2 dt = x √1+x2dx

x · arctan x−

_∫

dt=x · arctan x−t =x · arctan x−

√

1+ x2_+C

o)

∫

arccos x dx

= u =arccos x →u '=− 1 √1−x2 v ' =1 →v =x x · arccos x +

_∫

x

√

1−x2dx t =√=1− x2 dt =− x √1 −x2dx

x ·arccos x −

_∫

dt =x · arccosx −t =x ·arccos x−

√

1− x2_+C

p)

_∫

_x

_

_{ln x}

_

2

dx

= u=ln2 x →u '=2 lnx_x v ' =x →v =1 2x 2 1 2x 2_ln2_{x −}

_∫

_{x ln x dx =} u =lnx →u ' =1_x v '= x →v=1 2x 2 1 2x 2_ln2_x−1 2x 2_{ln x +}1 2

∫

x d x = =1 2 x 2_ln2 ∣x∣−1 2 x 2_ln∣x∣+1 4x 2 +C

q)

_∫

_

_x

_

₁

_

2

e

x

_dx

= u =( x+1)2 _{→u ' =2(x +1)} v '=ex →v=ex (x +1)2_ex₋

∫

2(x +1)ex_{dx =} u =2(x +1) →u ' =2 v '=ex →v =ex (x +1)2_ex_{−2(x +1)e}x₊

∫

2ex_dx=

=(x +1)2_ex_{−2( x+1)e}x_+2ex_=(x2_+1)ex_+C

r)

_∫



x

2

_{– 3 x}



2



ln x dx

= u=ln x →u ' =1 x v ' =x2_{−3 x+ 2 →v =1} 3x 3 −3 2x 2_{+2 x}

(

1 3x 3_{− 3} 2x 2_{+2 x}

)

ln x−

_∫

(

1 3x 2_{− 3} 2x+2

)

dx =

(

1 3x 3_{− 3} 2x 2_{+2 x}

)

ln∣x∣− 1 9x 3_{+ 3} 4x 2_{−2 x +C}

s)

_∫

ln



1



x







1



x



3

dx

=

∫

(1+ x)− 3 2_{ln(1+ x)dx =} u=ln(1+x ) →u '= 1_1+x v ' =(1+ x)− 3 2 _{→v =−2 (1+x )}− 1 2 −2(1+x )− 1 2_{ln(1+x )+2}

_∫

_{(1+ x)}− 3 2_dx= =−2(1+x )− 1 2_{ln(1+x )−4(1+x )}− 1 2₌₋ln(1+ x )2+4

√

1+ x +C

4. Integrales racionales:

a)

∫

1

x

2



x –6

dx

=

_∫

1 (x +3)(x−2)dx=

∫

(

A x+3+ B x−2

)

dx=

∫

A(x−2)+B(x +3) (x +3)(x −2) dx x=2 → 5 B=1= x=−3 → −5 A=1 =

∫

−1 5· 1 x+3+ 1 5· 1 x−2dx=− 1 5ln( x+3)+ 1 5ln(x −2)= 1 5·ln

(

∣

x −2 x +3

∣

)

+C

b)

∫

3 x

3

x

2

−

4

dx

3 x3_{:( x}2_{−4)=3x (Resto : 12 x) →}

∫

(

3 x+ 12x (x +2)(x−2)

)

dx=

∫

(

3 x + Ax +2+ Bx −2

)

dx = =

∫

(

3 x +A( x+2)+B(x −2) (x +2)(x−2)

)

dx x= 2 → 4A= 24= x=−2 → −4 B=−24

∫

(

3x + 6 x+2+ 6 x −2

)

dx= 3 2x 2_{+6ln(x +2)+6ln (x−2)=}3 x2 2 +ln

(

x 2₋₄

₎

6_+C

c)

∫

1

x

3

– 4x

2

– 25 x



100

dx

=

∫

1 (x− 4)(x +5 )(x−5)dx =

∫

(

A x−4+ B x+5+ C x−5

)

dx =

∫

A(x +5)(x −5 )+B (x−4)(x−5)+C (x−4 )(x +5 ) (x −4)(x +5)(x −5 ) dx x=4 → −9 A=1= x =−5 → 90B =1 x=5 → 10 C=1 =

∫

(

−1 9· 1x−4+ 190· 1x +5+ 110· 1x−5

)

dx =− 19ln∣x −4∣+ 190ln∣x +5∣+ 110ln∣x − 5∣+ C

d)

∫

_x

2

x

2



x

−

2

dx

=

∫

x 2_{+x−2−x +2} x2_{+x −2} dx=

∫

(

1+ −x +2 (x +2)(x−1)

)

dx=

∫

(

1+ Ax +2+ Bx−1

)

dx= =

∫

(

1+A(x −1)+B( x+2) (x+2)(x−1)

)

dx x =−2 → −3 A=4= x=1 → 3 B=1

∫

(

1− 4 3· 1x+2+ 13· 1x−1

)

dx= =x −4 3ln(x +2)+ 1 3ln(x−1)= x+ 1 3ln

∣

x−1 (x +2)4

∣

+C

e)

∫

x

3

_{– 2x}

2

_

_{x – 1}

x

2

– 3 x



2

dx

(x3_{−2 x}2_{+x−1): (x}2_{−3x +2)=x +1 (Resto : 2 x−3) →}

∫

(

x +1+ 2 x−3 (x−2)(x−1)

)

dx= =

_∫

(

x+1+ A x −2+ B x −1

)

dx=

∫

(

x+1+ A( x−1)+B (x−2) (x−2)(x−1)

)

dx x=1 → −B =−1= x=2 → A=1 =

∫

(

x+1+ 1 x −2+ 1x −1

)

dx= 12x 2 +x +ln(x −2)+ln(x−1)= x2 2+x +ln

∣

x 2 −3 x+ 2

∣

+C

f)

∫

1

x

3

+

x

2

– x−1

dx

=

∫

1 (x +1)2_(x−1)dx=

∫

(

A x +1+ B (x+1)2+ C x−1

)

dx =

∫

A(x +1)(x−1)+B(x−1)+C(x +1)2 (x+1)2_(x−1) dx x =1 → 4 C=1= x =−1 → −2 B=1 x=0 → − A−B +C=1 =

∫

(

−1 4· 1 x +1− 1 2· 1 (x +1)2+ 1 4· 1 x −1

)

dx=− 1 4ln(x+1)+ 1 2· 1 x+1+ 1 4ln( x−1)= 1 2( x +1)+ 1 4ln

∣

x−1 x+1

∣

+C

g)

∫

2x – 4



x

−

1



2



x



3



dx

=

∫

(

A x−1+ B (x −1)2+ C x +3

)

dx=

∫

A(x−1)(x+3)+B(x +3)+C (x−1)2 (x−1)2_(x+3) dx x=1 → 4 B=−2= x=−3 → 16 C=−10 x =0 → −3 A+ 3 B+ C=−4 =

∫

(

5 8· 1 x −1− 1 2· 1 (x−1)2− 5 8· 1 x+3

)

dx= 5 8ln( x−1)+ 1 2· 1 x−1− 5 8ln(x +3)= 1 2( x− 1)+ 5 8ln

∣

x− 1 x +3

∣

+C

h)

∫

1



x

−

1



x



3



2

dx

=

∫

(

A x−1+ B x +3+ C (x +3)2

)

dx=

∫

A(x +3)2_{+B(x−1)(x+3)+C(x−1)} (x −1)(x+3)2 dx x=1 → 16 A=1= x=−3 → −4 C=1 x=0 → 9 A−3 B−C=1 =

∫

(

1 16· 1 x−1− 1 16· 1 x+3− 1 4· 1 (x +3)2

)

dx = 1 16ln( x−1)− 1 16· ln(x+3)+ 1 4· 1 x +3= 1 4( x +3)+ 1 16ln

∣

x− 1 x +3

∣

+C

i)

∫

3 x−2

_x

2

₋₄

dx

=

_∫

3x−2 (x +2)(x−2)dx=

∫

(

A x+2+ B x−2

)

dx=

∫

A(x−2)+B (x+2) (x +2)(x −2) dx x =2 → 4 B=4= x =−2 → − 4 A=−8 =

∫

(

2 x +2+ 1 x −2

)

dx=2ln(x+2)+ln(x −2 )=ln

∣

(x+ 2) 2 (x− 2)

∣

+C

j)

∫

2x

2

_

_{5 x –1}

x

3



x

2

– 2 x

dx

=

∫

2 x 2_{+5 x−1} x (x−1)( x+2)dx =

∫

(

A x+ B x −1+ C x +2

)

dx=

∫

A(x−1)(x +2)+B x (x+2)+C x (x−1) x (x−1)( x+2) dx x =0 → −2 A=−1= x=1 → 3 B=6 x =−2 → 6C=−3 =

∫

(

1 2· 1 x+ 2 x −1− 1 2· 1 x+2

)

dx= 1 2ln x+2ln(x−1)− 1 2ln(x +2)=2ln∣x −1∣+ 1 2ln

∣

x x+ 2

∣

+C

k)

∫

_x

2

−

3

– 3 x

dx

=

∫

−3 x (x−3)dx=

∫

(

A x+ B x−3

)

dx=

∫

A(x−3)+B x x (x−3) dx x =0 → −3 A=−3= x =3 → 3 B=−3 =

∫

(

1 x− 1 x−3

)

dx =ln x −ln( x−3)=ln

∣

x x− 3

∣

+C

l)

∫

−

3 x



2

x

3

−

x

dx

=

_∫

−3 x+2 x (x +1)(x−1)dx =

∫

(

A x+ B x +1+ C x−1

)

dx =

∫

A(x +1)(x−1)+B x (x−1)+C x (x +1) x (x +1)(x−1) dx x=0 → −A=2= x=1 → 2 C=−1 x=−1 → 2 B=5 =

∫

(

− 2 x+ 52· 1x +1− 12· 1x−1

)

dx=−2 ln∣x∣+ 52ln∣x +1∣− 12ln∣x −1∣+C

m)

∫

−

4 x

2

_

_{16 x}

_

₁₆



x

2

_{– 4}

_

2

dx

=

∫

−4x 2_{+16 x +16} (x+2)2_(x−2)2 dx=

∫

(

A x+2+ B (x +2)2+ C x −2+ D (x −2)2

)

dx= =

_∫

A(x +2)(x−2) 2_{+B (x−2)}2_{+C(x +2)}2_{(x−2)+D(x +2)}2 (x+2)2 (x−2)2 dx x= 2 → 16 D=32= x=−2 → 16B =−32 x=0 → 8 A+4 B−8 C+4D=16 x =1 → 3 A+ B−9C+ 9 D=28 =

∫

(

1 x+2− 2 (x+2)2− 1 x−2+ 2 (x−2)2

)

dx=ln( x+2)+ 2 x+2−ln(x−2)− 2 x−2=− 8 x2₋₄+ln

∣

x+ 2 x− 2

∣

+C

n)

∫

5 x

2

x

3

– 3 x

2



3 x – 1

dx

=

∫

5x 2 (x−1)3dx=

∫

(

A x−1+ B (x−1)2+ C (x−1)3

)

dx= =

∫

A(x−1) 2_{+B(x −1)+C} (x−1)3 dx x=1 → C=5= x =0 → A−B+C =0 x=−1 → 4 A−2B +C=5 =

∫

(

5 x−1+ 10 (x −1)2+ 5 (x −1)3

)

dx=5ln(x−1)− 10 x −1− 5 2· 1 (x−1)2=− 20 x−15 2( x−1)2+5ln∣x−1∣+C

o)

∫

x

4

_

_{2 x – 6}

x

3



x

2

– 2 x

dx

(x4_+2x−6):(x3_+x2_{−2 x)= x−1 (Resto : 3 x}2_{−6) →}

∫

(

x−1+ 3 x 2₋₆ x (x +2)(x −1)

)

dx= =

∫

(

x−1+A x+ B x +2+ C x−1

)

dx=

∫

(

x−1+ A( x+2)(x−1)+Bx(x −1)+Cx (x+2) x (x +2)(x −1)

)

dx x=0 → −2 A=−6= x =−2 → 6 B=6 x =1 → 3 C=−3 =

∫

(

x−1+3 x+ 1 x +2− 1 x −1

)

dx= 1 2x 2_{−x+3ln x +ln(x +2)−ln(x−1)=}x2 2−x +ln

∣

x4_{+2 x}3 x −1

∣

+C

p)

∫

2 x – 3

x

3

– 2 x

2

– 9 x



18

dx

=

∫

2 x−3 (x−2)(x +3)(x−3)dx =

∫

(

A x −2+ B x +3+ C x −3

)

dx =

∫

A(x+3)(x−3)+B(x−2)(x−3)+C (x−2)(x +3) (x−2)(x +3)(x −3 ) dx x =2 → −5A=1= x=−3 → 30 B=−9 x =3 → 6 C=3 =

∫

(

−1 5· 1x−2− 310· 1x +3+ 12· 1x−3

)

dx =− 15ln∣x −2∣− 310ln∣x +3∣+ 12ln∣x −3∣+C

q)

∫

1

x

3

+

x

dx

=

∫

1 x (x2₊₁₎dx =

∫

(

A x+ B x+C x2₊₁

)

dx =

∫

A(x2_+1)+Bx2_{+C x} x (x2₊₁₎ dx =

∫

(A+B)x2_{+C x+ A} x (x2₊₁₎ dx =A+ B=0 C =0 A=1 =

∫

(

1 x− x x2₊₁

)

dx =ln∣x∣− 1 2ln( x 2_+1)+C

r)

∫

x

2

_x

+

3

₊

x+1

_x

dx

=

_∫

x 2_+x+1 x (x2₊₁₎dx =

∫

(

A x+ B x+C x2₊₁

)

dx =

∫

A(x2_+1)+Bx2_{+C x} x (x2₊₁₎ dx =

∫

(A+B)x2_{+C x+ A} x (x2₊₁₎ dx =A+ B=1 C =1 A=1 =

∫

(

1 x+ 1 x2₊₁

)

dx =ln∣x∣+arctan x +C

s)

∫

x+3

x

2

+6 x+10

dx

=

∫

x +3 x2+6 x +9+1dx=

∫

x+3 (x +3)2+1dx =t = x +3 dt=dx

∫

t t2+1dt =12ln(t 2₊₁₎₌₁ 2ln

(

(x +3) 2₊₁

₎

_+C

t)

∫

_x

1− x

3

+

x

dx

=

∫

1−x x (x2₊₁₎dx =

∫

(

A x+ B x+C x2₊₁

)

dx =

∫

A(x2_+1)+Bx2_{+C x} x (x2₊₁₎ dx =

∫

(A+B)x2_{+C x+ A} x (x2₊₁₎ dx =A+ B=0 C=−1 A=1 =

∫

(

1 x− x +1 x2₊₁

)

dx =

∫

(

1 x− x x2₊₁− 1 x2₊₁

)

dx =ln∣x∣− 1 2ln( x 2_{+1)− arctan x + C}

u)

∫

8 x

2

+

x

(

x−2)( 4 x

2

₊₁₎

dx

=

∫

(

A x −2+ B x +C 4 x2₊₁

)

dx=

∫

A(4 x2_{+1)+Bx (x−2)+C (x−2)} (x−2)(4 x2₊₁₎ dx=

∫

(4 A+B)x2_{+(C−2B)x +A−2C} x (x2₊₁₎ dx 4 A+ B=8= C−2B =1 A−2 C=0 =

∫

(

2 x −2+ 1 4 x2₊₁

)

dx=

∫

2 x −2dx+

∫

1 (2 x)2₊₁dx

⏟

t =2x → dt=2 dx =

∫

2 x−2dx + 1 2

∫

1 t2₊₁dt=ln( x− 2) 2₊1 2arctan(2 x )+C

v)

∫

2

x

4

−1

dx

=

∫

2 (x +1)(x−1)( x2₊₁₎dx=

∫

(

A x +1+ B x−1+ C x +D x2₊₁

)

dx= =

∫

A(x−1)(x 2_{+1)+B(x +1)(x}2_{+1)+C x (x +1)(x−1)+D(x+1)(x−1)} (x+1)(x −1)(x2₊₁₎ dx x=1 → 4 B=2= x =−1 → −4 A=2 x =0 → − A+B−D=2 x =2 → 5 A+ 15B + 6C+3D=2 =

∫

(

−1 2· 1 x +1+ 1 2· 1 x−1− 1 x2₊₁

)

dx= 1 2ln

∣

x−1 x+ 1

∣

−arctan x+ C

w)

∫

3 x+ 4

x

3

– 3 x

2

+4 x−12

dx

=

∫

3x +4 (x−3)(x2₊₄₎dx

∫

(

A x−3+ B x+C x2₊₄

)

dx =

∫

A(x2_{+4)+Bx( x−3)+C (x−3)} (x −2 )(4x2₊₁₎ dx =

∫

(A+B) x2_{+(C −3B) x+4A −3C} (x−2)( x2₊₄₎ dx = = A+B =0 C −3 B=3 4 A−3C =4

∫

(

_x−31 − x x2₊₄

)

dx=

∫

1 x −3dx −

∫

x x2₊₄dx

⏟

t =x2_{+4 → dt =2 x dx} =

_∫

1 x −3dx − 1 2

∫

1 tdt=ln∣x− 3∣− 1 2ln( x 2_{+4 )+C}

x)

∫

3 x

2

+

x+10

x

3

_{– 2 x}

2

_{+4 x−8}

dx

=

∫

3 x 2_{+x +10} (x−2)(x2₊₄₎dx

∫

(

A x −2+ B x +C x2₊₄

)

dx =

∫

A(x2_{+4)+Bx (x−2)+C (x−2)} (x −2)(x2₊₄₎ dx =

∫

(A+B) x2_{+(C −2B)x +4A−2C} (x−2)(x2₊₄₎ dx= = A+B=3 C −2 B=1 4 A−2 C =10

∫

(

_x−23 + 1 x2₊₄

)

dx =

∫

3 x−2dx +

∫

1 4

(

(

x 2

)

2 +1

)

dx

⏟

t= x 2 → d t = 12dx =

_∫

3 x−2dx + 1 2

∫

1 t2₊₁dt=3 ln∣x− 2∣+ 1 2arctan

(

x 2

)

+C

y)

∫

x

_x

23

+6 x+3

₊

_x

2

_{– 2}

dx

=

_∫

x 2_{+6 x +3} (x−1)(x2_{+2 x+2)}dx

∫

(

A x−1+ B x +C x2_{+2x +2}

)

dx=

∫

A(x2_{+2x +2)+Bx (x −1)+C(x −1)} (x −1)(x2_{+2 x+2)} dx= =

∫

(A+B) x 2_{+(2 A−B+C) x+2A−C} (x −2)(x2₊₄₎ dx A+B =1= 2 A−B+ C=6 2 A−C=3

∫

(

_x−12 + −x+1 x2_{+2x +2}

)

dx =

∫

2 x−1dx+

∫

−x+1 x2_{+2x +1+1}dx= =

∫

2 x−1dx+

∫

−(x +1)+2 (x+1)2₊₁dx

⏟

t =x +1 → dt =dx =

∫

2 x−1dx−

∫

t t2₊₁dt +

∫

2 t2₊₁dt=2ln∣x−1∣− 1 2ln

(

(x+1) 2₊₁

₎

_{+2arctan( x+1 )+ C}

z)

∫

3 x – 3

x

2

– x +2

dx

∫

3 x –3 x2_{– 2 ·}1 2· x + 1 4− 1 4+2 dx =

_∫

3 x –3 2− 3 2

(

x –1 2

)

2 +7 4 dx=

_∫

3

(

x –1 2

)

− 3 2 7 4

[

(

2

√

7

(

x – 1 2

)

)

2 +1

]

dx = t = 2 √7

(

x− 12

)

dt= 2 √7dx =

∫

3 t − 3

√

7 7 t2₊₁ dt =

∫

3 t t2₊₁dt −

∫

3

√

7 7 t2₊₁dt = 3 2ln

(

4 7

(

x− 1 2

)

2 +1

)

−3

√

7 7 arctan

(

2

√

7

(

x − 1 2

)

)

+C

5. Integrales trigonométricas:

a)

∫

cos

7

_{x dx}

=

_∫

(

cos2_x

)

3cos x dx =

_∫

(

1−sen2_x

)

3cos x dx

⏟

t =sen x → dt = cosx dx =

_∫

(1−t2₎3_{dt =}

∫

(1−3 t2_+3t4_−t6_{)dt =} =t −t3₊3 5t 5₋1 7t 7_{+C=sen x −sen}3_{x + 3} 5sen 5_{x − 1} 7sen 7_{x +C}

b)

∫

cotg x dx

=

∫

cos x sen xdx

⏟

t = sen x → dt =cos x dx =

∫

1 tdt =lnt +C=ln( sen x)+ C

c)

∫

3 sen x cos x dx

= t =sen x dt=cos x dx

∫

3t dt=3 2t 2_+C=3 2sen 2_{x + C}

d)

∫

4 sen

3

_{x cos}

2

_x

=

∫

4sen2_cos2_{x sen x dx =}

_∫

₄

₍

_1−cos2_x

₎

_cos2_{x sen x dx}

⏟

t =cosx → dt =−sen x dx =−

∫

4(1−t2_)t2_{dt =} =

∫

(4t4_−4t2_)dt=4 5t 5₋4 3t 3_+C=4 5cos 5_x−4 3cos 3_{x + C}

e)

∫

sen x

cos

2

x

dx

= t =cos x dt=−sen x dx

∫

−1 t2dt =

∫

−t −2_dt=t−1 +C= 1 cos x+C

f)

∫

sen x – tg x

_{cos x}

dx

=

∫

sen x−sen x cos x cos x dx =

∫

1− 1 cos x cos x sen x dx

⏟

t =cos x → dt =−sen xdx =

∫

1 t−1 t dt =

∫

(

1 t2− 1 t

)

dt= =−1 t−ln t +C=− 1 cos x−ln∣cos x∣+C

g)

_∫

sen

5

_xcos

3

_{x dx}

=

∫

sen5x cos2x cos x dx=

_∫

_⏟

sen5x (1−sen2x) cos x dx

t =sen x → dt =cosx dx =

∫

t5(1−t2)dt =

_∫

(t5−t7)dt = =1 6t 6₋1 8t 8_+C=1 6sen 6_{x − 1} 8sen 8_{x +C}

h)

_∫

cos

3

x

1– sen x

dx

=

∫

cos 2_x 1−sen xcos x dx=

∫

1−sen2_x 1−sen x cos x dx

⏟

t =sen x dt= cosx dx =

∫

1−t 2 1−t dt =

∫

(1+t )(1−t ) 1−t dt = =

∫

(1+t )dt =t +1 2t 2_{+C=sen x +}1 2sen 2_{x +C}

i)

∫

₁

dx



sen x

=

∫

1−sen x (1+sen x)(1−sen x )dx =

∫

1−sen x 1−sen2_xdx=

∫

1−sen x cos2_x dx=

∫

1 cos2_xd x +

∫

−sen x cos2_x dx

⏟

t =cos x dt =−sen x dx = =tan x +

_∫

1 t2dt =tan x− 1 t+C=tan x− 1

cos x+C=tan x−sec x +C =

sen x−1

j)

∫

dx

sen

2

xcos

2

x

=

∫

(

1 sen2_x+ 1 cos2_x

)

dx=

∫

1 sen2_xdx

⏟

t= π 2−x → dt =−dx +

∫

1 cos2_xdx=

∫

−1 sen2

(

π2−t

)

dt

⏟

sen₍π₂−t₎=cost +

∫

1 cos2_xdx=

∫

−1 cos2_tdt +

∫

1 cos2_xdx=

=−tant +tan x+C=−tan

₍

π

2−x

)

+tan x +C=tan x −

1 tan x+C=

sen x−cos x

sen x cos x +C

También con el cambio t = tan x → d t= dx

cos2_x sen 2_{x =1−cos}2_{x =1−} 1 1+tan2_x= tan2_x 1+ tan2_x= t2 1+t2

∫

_sen2d x_{x cos}2_xt =tan x=

∫

1+t2 t2 dt=

∫

(

1 t2+1

)

dt= −1 t +t +C=tan x− 1 tan x+C

6. Determina las siguientes integrales por el método que consideres más conveniente:

a)

∫



x

1





3

x

dx

= x=t6 dx =6t5_{d t}

∫ √

t6 1+3

√

t66t 5_{dt =}

∫

6 t8 1+t2dt

⏟

t8_{: (t}2_+1)=t6 −t4 +t2_{−1 (Resto: 1)} =6

_∫

(

t6_−t4_+t2_{−1+ 1} 1+t2

)

dt = =6 7t 7₋6 5t 5₊6 3t 3_{−6t +6 arctan(t )+C=}6 7 6

√

x7₋6 5 6

√

x5₊₂

_√

_x−66

√

x +6 arctan

√

6x +C

b)

∫

ln



1

x



dx

=

∫

ln

(

x−1

)

dx=

_∫

−ln x dx = u =ln(x) → u '= 1_x v '=−1 → v =−x −x ln x +

_∫

1dx=x−ln x +C=x (1−ln∣x∣)+C

c)

_∫



x



1 ln



x



1



dx

= u =ln(x+ 1) → u ' = 1_{x +1} v '=√x +1 → v =2 3(x +1) 3 2 2 3(x+1) 3 2_{ln(x +1)−}2 3

∫

√

x +1dx= 2 3(x+1) 3 2_{ln∣x +1∣−}4 9(x + 1) 3 2_+C

d)

∫

dx

x



1



ln

2

x



= t =ln x dt =dx x

∫

_1+tdt2=arctan(t )+C=arctan(ln∣x∣)+ C

e)

_∫

cos(ln x)dx

∫

_⏟

cos (ln x )dx I = u=cos(lnx ) → u '=− 1_xsen(ln x) v '=1 → v= x x cos(ln x)+

_∫

sen(ln x)dx = u =sen(lnx ) → u '= 1_xcos(ln x) v '=1 → v=x

=x cos(ln x)+ x sen(ln x)−

_∫

_⏟

cos(ln x)

I

→ 2I=x cos(ln x)+ x sen(ln x)+C → I =x

2

(

cos(ln∣x∣)+ sen (ln∣x∣)

)

f)

_∫

_

_{1 – x}

2

_dx

₌

x=sent → dx =cost dt

√1− x2

=√1−sen2_{t =√cos}2_{t =cost}

∫

cos2_{t dt}

⏟

I = u =cost → u '=−sent v ' =cost → v =sen t

sent cost +

_∫

sen2_{t dt =}

=sen t cost +

_∫

(

1−cos2_t

₎

_{dt =sent cost +}

_∫

_{dt −I → 2I =sent cost +t +C} I=1 2sen t cost + t 2+C= 1 2x cos(arcsen x)+ arcsen x 2 +C= x

√

1− x2 2 + arcsen x 2 +C

g)

∫

_

_x

x

_

₁

dx

= t =√x +1 → dt= 1₂ dx √x +1 x =t2₋₁

∫

2(t2_{−1)dt =}2 3t 3_{−2t +C=}2 3(x +1) 3 2₋₂

_√

_{x +1+C}

h)

_∫

dx

x



x



1

t=√x +1 → dt =1= 2 dx √x +1 x=t2₋₁

∫

_t2₋₁2 dt=

∫

(

A t +1+ B t−1

)

dt =

∫

A(t −1)+B(t +1) (t +1)(t −1) dt x=1 → 2 B=2= x =−1 → −2 A=2 =

∫

(

1 t−1− 1t +1

)

dt=ln(t−1)−ln(t +1)+C=ln

∣

√

x +1−1

√

x+ 1+1

∣

+C

i)

_∫

x

5

_e

−x3

dx

₌

_∫

_x3_e−x3 x2_{d x} = t =x3 _{→ dt =3 x}2_dx

∫

1 3te −t_dt = u= 1 3t → u ' =13 v ' =e−t _{→ v=−e}−t −1 3t e −t +1 3

∫

e −t_dt= =−1 3t e −t₋1 3e −t_+C=−1 3

(

1+ x 3

₎

_e−x3 +C

j)

∫

e

x

e

2x



3 e

x



2

dx

= t=ex → dt =ex dx

∫

1 t2_{+3t +2}dt=

∫

(

A t +1+ B t +2

)

dt =

∫

A(t +2)+B(t +1) (t +1)(t +2) dt x =−1 → A=1= x=−2 → −B=1 =

∫

(

1 t +1− 1 t +2

)

dt =ln(t +1)−ln(t +2)+C=ln

(

ex₊₁ ex₊₂

)

+C

7. Calcula la función f(x) que cumple que f''(x) = 6x+1, f(0)=1 y f(1)=0.

f ' (x)=

_∫

f ' ' (x )dx =3 x2 +x +C f (x )=

_∫

f ' ( x)dx= x3 +1 2x 2 +C x +D →

{

x =0 → f (0)=D=1 x =1 → f (1)=1+1 2+C+1=0 → f (x )= x3₊1 2x 2₋5 2x+ 1

8. Encuentra la familia de funciones que tienen como segunda derivada f''(x)=4x sabiendo que

tienen un máximo relativo en x=-1.

Determina cuál de las funciones de la familia pasa por el punto



1,

−

5

6



.

f ' (x )=

_∫

f ' ' (x)dx =2x2_+C f (x )=

∫

f ' (x )dx=2 3x 3 +C x +D →

{

f ' (−1)=0 →_Máximo 2+C=0 Familia: f ( x )=2 3 x 3 −2 x +D → Por el punto

(

1,− 5 6

)

f ( x)= 2 3x 3 −2 x +1 2

9. Busca una primitiva F(x) de la función f(x)=2x – 4 que verifique que

F



x

≥

0

∀

x

∈[

0, 4

]

.

F ( x)=

_∫

(2 x – 4)dx= x2_{– 4 x+C Si hacemos C=4 tendremos: F ( x )= x}2_{−4 x + 4=(x−2)}2_{≥0∀ x ∈ℝ}

10.Halla f(x) sabiendo que:

f ' '



x

=

cos

x

2

f '



2 

=

0

f(0)=1

f ' (x)=

_∫

cos

(

x 2

)

dx=2sen

(

x 2

)

+C f (x )=

_∫

[

2sen

(

x 2

)

+C

]

dx=−4cos

(

x 2

)

+Cx+D →

{

f ' (2 π)=0 → C=0 f (0)=1 → −4+D=1 → f ( x)=−4cos

(

x 2

)

+5

11.Calcula el área encerrada entre la gráfica de las funciones

f x=x , g x=2 x , hx=x

2

_.

A=

_∫

0 1

(

g(x )−f ( x)

)

dx +

_∫

1 2

(

g(x )−h(x)

)

dx =

_∫

0 1 (2 x−x )dx +

_∫

1 2

(

2x −x2

)

dx= =

[

1 2x 2

]

0 1 +

[

x2−1 3x 3

]

1 2 =1 2+4− 8 3−1+ 1 3= 7 6u.d.s.

12.Integrales definidas:

a)

∫

0 

sen2x dx

=

[

−1 2cos 2x

]

0 π =−1 2cos2 π+ 1 2cos0=0

b)

∫

1 2

x

2

ln xdx

u =lnx → u '= 1= x v '=x2 _{→ v =}1 3x 3

[

1 3x 3 ln x

]

1 2 −

∫

1 2 x2 3dx= 8 3ln 2−0−

[

x3 9

]

1 2 =8 3ln 2− 8 9+ 1 9= 24 ln(2)− 7 9

c)

_∫

−1 3

∣2 x

−

4∣dx

= ∣2 x− 4∣=

{

4−2 x si x ≤2 2 x− 4 si x >2

∫

−1 2 (4−2 x)dx +

_∫

2 3 (2 x−4)dx =

[

4x −x2

_]

−1 2 +

[

x2_{−4 x}

_]

2 3 =4+5−3+4=10

d)

_∫

−1 0

x



1



x dx

u =x → u ' =1= v ' =√1+ x → v= 2 3( 1+x ) 3 2

[

2 3x (1+x ) 3 2

]

−1 0 −

∫

1 2 2 3(1+x ) 3 2_dx=0−

[

4 15(1+x ) 5 2

]

−1 0 =− 4 15

e)

_∫

−3 3

∣

x

−

1∣dx

= ∣x −1∣=

{

1− x si x≤1_{x−1 si x>1}

∫

−3 1 (1−x)dx +

_∫

1 3 (x−1)dx=

[

x −1 2x 2

]

−3 1 +

[

1 2x 2_−x

]

1 3 =1 2+ 15 2+ 3 2+ 1 2=10

f)

∫

0 1



x dx

=

[

2 3x 3 2

]

0 1 =2 3−0= 2 3

g)

∫

1 e

ln x

x

dx

= t =lnx → d t = 1 xdx x =1→t =0 x =e→t =1

∫

0 1 t dt =

[

1 2t 2

]

0 1 =1 2−0= 1 2

h)

_∫

0 2

x



x

2

_

₁

dx

= t =x2_{+1 → d t =2 x dx} x=0→t =1 x =2→t =5

∫

1 5 1 2

√

tdt=

[√

t

]

1 5 =

√

5−1

i)

∫

1 e e

2 ln x dx

= u =lnx → u '= 1 x v '=2 → v=2 x

[

2x ln x

]

₁ e e −

∫

1 e e 2dx=2e+2 e−

[

2x

]

1 e e =2e+2 e−2e+ 2 e= 4 e

j)

_∫

−3 3

x

2

x

2



1

dx

función= par 2

∫

0 √3 x2 x2₊₁dx=2

∫

0 √3 x2₊₁₋₁ x2₊₁ dx=2

∫

0 √3

(

1− 1 1+x2

)

dx=2

[

x −arctan x

]

0√ 3 =2

₍

√

3− π 3

)

13.Dadas la parábola

y

=

x

2

₋

₁

_{y la recta y=5 – x, representa y calcula el área de los}

siguientes recintos:

a) Recinto acotado limitado por ambas curvas.

Intersección de funciones: x2_{−1=5−x → x}2_{+x−6=0 → x} 1=−3 x2=2 A=

_∫

−3 2

(

(5−x )−( x2₋₁₎

₎

_dx=

∫

−3 2 (−x2_{−x +6)dx=}

[

−1 3x 3₋1 2x 2_{+6 x}

]

−3 2 =22 3 −

(

− 27 2

)

= 125 6 u.d.s.

b) Recinto acotado limitado por la parábola a la izquierda, la recta a la

derecha y el eje OX por debajo.

A=

_∫

1 2 (x2_{−1)dx +}

∫

2 5 (5− x)dx=

[

1 3x 3_−x

]

1 2 +

[

5x −1 2x 2

]

2 5 =2 3+ 2 3+ 25 2−8= 35 6 u.d.s.

c) Recinto acotado limitado por el eje OY a la izquierda, la parábola a la

derecha, la recta por encima y el eje OX por debajo.

A=

_∫

0 1 (5−x )dx +

_∫

1 2

[

(5− x)−(x2₋₁₎

_]

_dx=

[

5 x−1 2x 2

]

0 1 +

[

−1 3x 3₋1 2x 2_+6x

]

1 2 =9 2+ 22 3− 31 6= 20 3 u.d.s.

14. Calcula el área encerrada entre las gráficas de las siguientes funciones en cada caso:

a)

f



x

=

x

2

_{– 5 g}



x

=−

x

2



5

Intersección de funciones: x2_−5=−x2_{+5 → 2x}2_{−10=0 → x=±}

_√

₅ A=

_∫

−√5 √5

(

(5− x2_)−(x2₋₅₎

₎

_{dx =} Función par 2

_∫

0 √5 (10− 2x2_)dx=2

[

10 x−2 3x 3

]

0 √5 =20

√

5−4

√

125 3 u.d.s.

b)

f



x

=

4 – x

2

_g



x

=

8 – 2 x

2 Intersección de funciones: 4−x2_=8−2x2 _{→ x}2_{=4 → x =±2} A=

_∫

−2 2

(

(8−2 x2_)−(4−x2₎

₎

_{dx =} Función par 2

_∫

0 2 (4−x2_)dx=2

[

4 x−1 3x 3

]

0 2 =32 3 u.d.s.

c)

f ( x)= x( x−3) g( x)=2 x−4

Intersección de funciones: x (x−3)=2 x− 4 → x2_{−5 x +4=0 → x} 1=1 x2=4 A=

_∫

1 4

(

(2x−4)−x (x−3)

)

dx=

_∫

1 4

(

−x2_{+5 x−4}

₎

_dx=

[

−1 3x 3₊5 2x 2_{−4 x}

]

1 4 =9 2 u.d.s.

d)

f



x

=

6

x

g



x

=

7

−

x

Intersección de funciones:6 x=7−x → x 2_{−7 x+6=0 → x} 1=1 x2=6 A=

_∫

1 6

(

7−x−6 x

)

dx=

[

7x − 1 2x 2_{−6ln x}

]

1 6 =35 2−6ln6 u.d.s.

e)

f



x

=

sen x g



x

=

cos x

con 0

≤

x

≤

2 

Intersección de funciones:sen x=cos x → tanx=1 → x₁= π 4

⏟

45º x₂=5 π 4

⏟

235º A=

_∫

π 4 5 π 4

(sen x −cosx ) dx=

[

−cos x−sen x

]

π 4 5 π 4₌₂

_√

_{2 u.d.s.}

f)

f



x

=



x

2

g



x

=

∣1

−

x∣

Intersección de funciones:

√

x 2=∣1−x∣ → x 2=(1− x) 2 _{→ 2 x}2_{−5 x +2=0 → x} 1= 1 2 x2=2 A=

∫

1 2 2

(

_√

x 2−∣(1− x)∣

)

dx ∣1−x∣=

{

1− x si x ≤1_{x −1 si x>1}=

∫

1 2 1

(

_√

x 2−(1− x)

)

dx+

∫

1 2

(

_√

x 2−(x−1)

)

dx= =

[

2x

√

x 3

√

2 −x + 1 2x 2

]

1 2 1 +

[

2 x

√

x 3

√

2 − 1 2x 2_+x

]

1 2 =13 24 u.d.s.

g)

f



x

=

x

3

₋

_{2x g}

_

_x

_=

_x

2 Intersección de funciones: x3_{−2 x= x}2 _{→ x}3_−x2_{−2 x=0 → x} 1=−1 x2=0 x3=2 A=

_∫

−1 0

(

(x3_{−2x )−x}2

₎

_dx+

∫

0 2

(

x2_−(x3_{−2 x)}

₎

_dx=

[

1 4x 4₋1 3x 3_−x2

]

−1 0 +

[

−1 4x 4₊1 3x 3_+x2

]

0 2 =37 12 u.d.s.

15.Halla el área del recinto acotado limitado por las gráficas de las funciones

f



x

=

e

x2

_,

_g

_

_x

_=

_e

−x

_{y el eje OY}

Intersección:ex+2_=e−x _{→ x +2=−x → x =−1 A=}

∫

−1 0

(

ex +2_−e−x

₎

_dx=

_[

_ex +2_+e−x

_]

−1 0 =e2_{+1− 2e u.d.s.}

16.Considera el área de la región encerrada entre la parábola

y

=

x

2

_{y la recta y=1.}

Esta región se divide mediante una recta horizontal y=a.

Determina a que hace que dicha región quede dividida en mitades de igual superficie.

Intersección con y=1: x2_{=1 → x=±1 Intersección con y=a : x}2_{=a → x =±a} ATOTAL=

∫

−1 1

(

1−x2

₎

_dx=

[

x −1 3x 3

]

−1 1 =4

3u.d.s. Área bajo y=a: A(a)=₋

∫

_√a

√a (a −x2_)dx=

[

ax−1 3x 3

]

−√a √a =2a

√

a−2 3(

√

a) 3 =4 3a 3 2 4 3a 3 2 ₌ 1 2ATOTAL 2 3 → a 3 2₌1 2 → a= 1 3

√

4