evaluate the integral

Question

∫x(t+4)^2dt

  0

 

I hope you can understand this x is on top 0 is on bottom

Answer 1

Antidifferentiate (t+4)² to get (1/3)(t+4)³ + C

 

Then ∫(t+4)²dt (from 0 to x) = (1/3)(x+4)³ - (1/3)(4)³

 

                                                    = (1/3)[(x+4)³ - 64]

 

 

Answer 2

Expand the (x + 4)² term and work your way from there. 

 

∫(t + 4)² dt =

 

∫(t² + 8t + 16) dt

 

 

We can break this up as the sum of integrals.  Don't forget to put constants in front of the integral sign.

 

∫t² dt + 8∫t dt + 16∫ dt =

 

(1/3)t³ + 8(1/2)t² + 16t =

 

(1/3)t³ + 4t² + 16t

 

 

Now we just plug in the bounds to evaluate the definite integral.   The bounds are  t=0  to  t=x.  Note that setting t=0 makes the integral zero.

 

 

[(1/3)x³  + 4x² + 16x] - 0 = 

 

(1/3)x³ + 4x² + 16x

 

 

 

 

 

 

Answer 3

 ok

 

thanks for specifying that:) 

 

now integral of (t+4)² is (t+4)³  divided by 3  

 

apply the limits now to get( (x+4)³  -3³  )/3