P1PS-FMM212-2014-02.pdf

   

a

(a)

a

 

2

x

=

a dx

0

e

(b)

L

= l im

4π

(4x2 − π2 ) +

loga (x) dx

1

π/2 π/ 2

sin x

e

dx

x

1 + cos(2x)

x→π/ π/2 2

(a)

 

a

x

ax

=

aa

ln a

 

;

loga xdx  =

a

−x

ln a

− 1 =  2 l n 2 − 2 − (1ln1 − 1)

ln a a

x ln x

ln a

− 1 = 2 ln2 − 1 =⇒ [aa = 2ln2]

     (∗)

(∗) (b) e L

= l im

4π

(4x

2

−π

2

2 π/2 π/

)+

2xe =

l im

2 x→π/ π/2

2e L

=

sin x

e

dx

x

=

1 + cos(2x)

2 x→π/ π/2

L

 

l im

x→π/ π/2 2

π

π

− esin x

−2 sin sin(2 (2x)

=

− cos(x)esin x − cos(4x)

0 0

=

e

2π

0 0

(a)

z  =

x+5

I  =

x+8

 

dx

(x + 5)(x + 8)

(b)

(a)

( a)

z  =

x+5

x

x+8

=

5 − 8z z−1

dx

x+5

=

=

3 (z − 1)2

−3z z−1

dz

;

x+8

=

−3 z−1

3

 

dx

(x + 5)(x + 8)

=

 

(z − 1)2 9z (z − 1)2

dz

1 = 3

 

1 z

dz

1 1 =  ln z + C  =  ln 3 3

=

1 1  − 3(x + 5) 3(x + 8)

  x+5 x+8

(b) 1 = (x + 5)(x + 8)

 

dx

(x + 5)(x + 8)

=

  

A x+5

 +

B x+8

1 1  − 3(x + 5) 3(x + 8)



dx

1 =  ln 3

  x+5 x+8

 + C 

 + C 

(a) I m

  √ 

xm

=

dx

ax + b

2 = a · (1 + 2m)

(b)



x

√ 

m

ax + b

− mb

  √ 

xm−1

dx

ax + b



I 1

(a) u  =  x

m

⇒ du =  mx m

  √    √ 

xm

ax + b xm

ax + b

  √    √ 

xm

ax + b xm

ax + b

1

−

dx  =

dx  =

dx  =

2xm

ax + b

−

a

2xm

√ 

ax

+b

− 2m

  √ 

xm

ax + b

xm

ax + b

dx  =

  √ 

xm

2 dx  = a(1 + 2m) ax + b

2m a

x

   

xm

x

m−1

2 xm

dx

a

m

√ 

−

ax + b

+ bdx

ax + b

ax + b

ax + b

ax

√ 

√ 

√ 

√ 

xm−1 (ax + b)

ax + b

a



⇒ dv  = √  dx

ax + b

  √ 

dx  =

2 xm

ax + b

− 2am

a

ax + b

√ 

a

√ 

√ 

  √ 

v  =

dx

a

dx + 2m

(1 + 2m)

2xm

2

−

2mb a

  √    √ 

xm−1

xm−1 ax + b

  √    √ 

xm−1

2mb

ax + b

− mb

xm−1 ax + b

dx



dx

(b) I 1

2 = 3a

 √  x

ax + b

− b · I 0



2 = 3a

 √  x

ax + b

2 √ 

−a

dx

ax + b

− 2mb a

a

dx

ax



+ b  + C 

dx