How do you integrate #f(x)=(-3+3x+3x^2-3x^3)(3-3x-x^2+3x^3+2x^4)# using the product rule?

1 Answer
Jan 1, 2018

Answer:

#-3/4x^8-3/7x^7+3x^6+9/5x^5-15/2x^4+x^3+9x^2-9x+C#

Explanation:

There is no product rule for integrals - the product rule only applies to derivatives.

In this case, we have to just expand the entire thing:
#(-3+3x+3x^2-3x^3)(3-3x-x^2+3x^3+2x^4)=#

#=-9+9x+3x^2-9x^3-6x^4+9x-9x^2-3x^3+9x^4+6x^5+9x^2-9x^3-3x^4+9x^5+6x^6-9x^3+9x^4+3x^5-9x^6-6x^7#

Combining like terms, we get:
#-6x^7-3x^6+18x^5+9x^4-30x^3+3x^2+18x-9#

We can then integrate using the reverse power rule, which says:
#int\ x^n\ dx=x^(n+1)/(n+1)+C#

In our case, we get:
#int\ -6x^7-3x^6+18x^5+9x^4-30x^3+3x^2+18x-9\ dx=#

#=-6/8x^8-3/7x^7+18/6x^6+9/5x^5-30/4x^4+3/3x^3+18/2x^2-9x+C=#

Simplifying, we get:
#=-3/4x^8-3/7x^7+3x^6+9/5x^5-15/2x^4+x^3+9x^2-9x+C#