What is the integral of [(2x-y)^3]dx?
Sorry I mean [(2x-y)^2]dx
since x is the variable of integration, y acts just like any other constant, so it's just
(1/8)(2x-y)^4 + C
sorry - I mean
(1/6)(2x-y)^3 + C
To find the integral of the expression [(2x-y)^3]dx, you can use the power rule for integrals. The power rule states that the integral of x^n with respect to x is (x^(n+1))/(n+1), where n is any real number except -1.
In this case, we have [(2x-y)^3]dx, which can be expanded using the binomial theorem as (2x-y)(2x-y)(2x-y). This requires repeated application of the power rule.
Let's go step by step:
1. Start by expanding (2x-y)(2x-y)(2x-y):
(2x-y)(2x-y)(2x-y) = (2x-y)^2 * (2x-y)
= (4x^2 - 4xy + y^2) * (2x-y)
2. Next, simplify further by distributing:
(4x^2 - 4xy + y^2) * (2x-y) = 8x^3 - 8x^2y + 2xy^2 - 4x^2y + 4xy^2 - y^3
= 8x^3 - 12x^2y + 6xy^2 - y^3
3. Now, we can integrate each term one by one using the power rule:
∫(8x^3 - 12x^2y + 6xy^2 - y^3) dx = (8/4)x^4 - (12/3)x^3y + (6/2)x^2y^2 - (1/1)y^3 + C
Simplifying the result, we get:
2x^4 - 4x^3y + 3x^2y^2 - y^3 + C
Therefore, the integral of [(2x-y)^3]dx is 2x^4 - 4x^3y + 3x^2y^2 - y^3 + C, where C represents the constant of integration.