Please help. See description, Please click on the question to read it accurately

Question

d

— (∫(x²-3x+1)dx)=

dt

Answer 1

First, we take the integral of x² - 3x + 1  with respect to x.

 

∫(x² - 3x + 1)dx

 

 

The rule for integrating a term with an exponent is

 

∫Cxⁿ dx = C∫[(x^{n + 1}) / (nⁿ⁺¹)]

 

So

 

∫(x² - 3x + 1)dx = ∫x²dx - 3∫xdx + ∫1dx

 

                         = (1/3)x³ + (3/2)x² + x + C

 

 

The solution to this problem is

 

d/dt[ (1/3)x³ + (3/2)x² + x + C]

 

We cannot take the derivative of this because the notation tells us to take the derivative with respect to t.  The variable in the function here is x.  Two different variables.  This means that we are done here.

 

Answer 2

Is x a function of t?