How do you integrate #f(x)=(2-2x+2x^2-2x^3-3x^4)(1+x-2x^2-3x^3)# using the product rule?

Question

Scarlett Allison · Accepted Answer

#(2-2x+2x^2-2x^3-3x^4) (1 - 4x - 9x^2) + (-2 + 4x - 6x^2 - 12x^3) (1 + x -2x^2 -3x^3)#

Remember the product rule is f g'+f' g, so: #(2-2x+2x^2-2x^3-3x^4) (1 - 4x - 9x^2) + (-2 + 4x - 6x^2 - 12x^3) (1 + x -2x^2 -3x^3)#

You are taking the derivative of f and that what is g. When you take the derivative of #(2-2x+2x^2-2x^3-3x^4)# you are multiplying by the exponent. Step by step for #(2-2x+2x^2-2x^3-3x^4)# is #2^0#, which is 0. the 2x is #2x^1#, where you multiply by the 1, where you get the 2, and take away the x. If you go down the line, for each and everyone one it will fall the exact same. #3x^4# become #12x^3#. Remember you are multiplying by the exponent and then subtracting it by 1.

If someone were to give you find the derivative of #(2x^5-2x^2-5) (3x^2+2)#, what would you write?

one step at a time, beginning at the beginning.

Formula, then: f g' + f' g

2, find the derivatives: #(2x^5-2x^2-5)# becomes #(10x^4-4x)# #(3x^2+2)# becomes (6x)

3, Now put in what you have: #(2x^5-2x^2-5)# #(6x)# + #(10x^4-4x)##(3x^2+2)#

Really simple, huh?

Elijah Abram · Answer

undefined To integrate the function $ f(x) = (2 - 2x + 2x^2 - 2x^3 - 3x^4)(1 + x - 2x^2 - 3x^3) $ using the product rule, follow these steps:

1. Expand the product $ f(x) $.
2. Use the power rule to integrate each term.
3. Apply the sum rule to find the total integral.

The integral of $ f(x) $ will be the sum of the integrals of each expanded term.