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Determine integral of 19,20,21,23,25,27

asked Jan 11, 2013 in CALCULUS

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19 ) ʃ ( cos x + 1/ 2 x )dx

ʃ ( cos x + 1/ 2 x )dx = ʃ cosx dx + ʃ 1 / 2 x dx                 ( ʃ (a+b )dx= ʃ adx +ʃ b dx )

=  sinx + 1/2 ʃ x dx                            ( ʃ x^n dx= x^n+1 / n+1 )

= sin x + 1 / 2 (x^2 / 2)

ʃ ( cos x + 1/ 2 x )dx = sinx + (x^2) / 4

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20)    ʃ (ex - 2 x2 )dx

Rewrite the above equation as

=   ʃ exdx - ʃ 2 x2dx

=   ex  -  2 ʃx2dx                (  ʃ exdx = ex )

=   ex -   2 x3/3               (  ʃxn dx = xn+1/n+1dx )

=  ex -   2 x3/3

The solution is    ʃ (ex - 2 x2 )dx  = ex -  2 x3/3

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23)0

ʃ (1/2(t^4) + 1/4(t^3) -t )dt

-2

0                                           0                   0               0

Given      ʃ (1/2(t^4) + 1/4(t^3) -t )dt =  ʃ 1/2t^4 dt +ʃ 1/4t^3dt -ʃ tdt

-2                                         -2                 -2             -2

0                    0              0

= 1 / 2 ʃ t^4dt +1 / 4 ʃ  t^3 dt - ʃ tdt

-2                   -2            -2

0                        0                0

= 1 /2 [ t^5 / 5 ] + 1 / 4 [ t^4 /4 ]   -    [t^2 / 2]

-2                      -2               -2

=  1 /2 ( 0^5 / 5 - ( -2^5 / 5 ) ) + 1 / 4  (0^4 / 4 -(-2^4 / 4 ) - (0^2 / 2 - (-2^2 / 2 )

=  1/ 2 (0 - (-32 / 5 ) ) +1 /4 (0 - (-16/ 4) - ( 0 - (-4 / 2 )

Product of two same sighs is possitive

0

ʃ (1/2(t^4) + 1/4(t^3) -t )dt = (1/2) 32/ 5 + (1 / 4 ) 16 / 4 + 4 /2

-2

0

ʃ (1/2(t^4) + 1/4(t^3) -t )dt = 16/5 + 1 +2

-2

0

ʃ (1/2(t^4) + 1/4(t^3) -t )dt =16 / 5 + 3 /1

-2

LCM of 5,1 is 5

0

ʃ (1/2(t^4) + 1/4(t^3) -t )dt = 16*1+3*5 / 5

-2

0

ʃ (1/2(t^4) + 1/4(t^3) -t )dt = 16+15 / 5

-2

0

ʃ (1/2(t^4) + 1/4(t^3) -t )dt  = 31 / 5

-2

Therefore the solution is 4.2.

3

ʃ ( x² - 3 ) dx

-2

Rewrite the above equation as

3            3

ʃ x² dx - ʃ 3 dx

-2            -2

3           3

[x³/3]  - [3x]            (   ʃxn dx = xn+1/n+1dx   ; ʃ dx = x )

-2          -2

Substitute 3 and -2 respecttively and subtract them

[(3)³/3] - [(-2)³/3] - [3(3) -3 (-2)]

[(27/3)-(-8/3)] - [9 +6]

[(27/3) + (8/3) - 15]

[(27 + 8 - 45)/3]

(35-45)/3

-10/3

3

The solution is  ʃ ( x² - 3 ) dx  =  -10/3

-2

2

ʃ (2x-3)  (4 x² +1)  dx

0

2                                    2                              2                        2                             2

ʃ (2x-3)  (4 x² +1)  dx =  ʃ ( 2x)* (4x^2 ) dx + ʃ (2x) * (1) dx + ʃ ( -3 )* (4x^2 ) dx+ ʃ (-3)* (1)dx

0                                    0                              0                       0                              0

2                      2                2                         2

=        ʃ ( 8x^3 ) dx + ʃ ( 2x)dx +  ʃ ( - 12x^2 ) dx + ʃ ( -3) dx                (   ʃxn dx = xn+1/n+1)

0                     0                  0                         0

2                  2                   2       2

= 8 (x^4 / 4) + 2 ( x^2 / 2 ) -12 (x^3 / 3 ) - 3(x)

0                  0                    0        0

= 8 ( (2)^4 / 4 - (0^4) / 4  ) + 2 ( (2^2 / 2 - 0^2 ) -12 ( 2^3 / 3 - 0^3 / 3 ) - 3 ( 2-0 )

= 8( 16 /4 - 0 ) -12 (8 / 3 ) +2 ( 4 / 2 ) -3 (2 )

=  8 (4)  -12 (8 / 3 ) + 2 (2) -6

= 32 -96 / 3 +4 -6

LCM of 3 ,1 is 3

= 32*3 -96 +4*3 -6*3 / 3

= 96-96+12-18 / 3

= -6 / 3

2

ʃ(2x-3 ) (4x^2 +1 )    = -2

0