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+1 vote

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

 

asked Jan 11, 2013 in CALCULUS by angel12 Scholar

6 Answers

+1 vote

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

answered Jan 11, 2013 by ricky Pupil
+1 vote

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

    

      I hope this will help you!!!!!

      If you are satisified then select my answer as the best!!! thank you!!!

answered Jan 11, 2013 by Johncena Apprentice
+2 votes

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                                     

                                   

                                  

                                  

                                  

                                   

                                   

 

answered Jan 11, 2013 by ricky Pupil

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Therefore the solution is 4.2.

+2 votes

                 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

 

I hope this will help you!!!!!

answered Jan 11, 2013 by Johncena Apprentice
+2 votes

                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                                                      

                             

                                                          

 

                            

 

answered Jan 12, 2013 by ricky Pupil
0 votes

The integral expression is

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answered Jun 30, 2014 by casacop Expert

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