Find the indefinite integral in two ways.
∫(2x-1)^2 dx
The first way I used was using the power rule and chain rule with substitution.
Let u = 2x - 1
du = 2 dx
(1/2)∫ u^2 du
(Applying power rule)
(1/2) * (u^3/3) + C
=(2x-1)^3/6 + C
What would be another way of finding the indefinite integral??
Just expand the polynomial and integrate all the terms:
4x^2-4x+1 dx
Another way of finding the indefinite integral of ∫(2x-1)^2 dx is by expanding the expression and then integrating each term separately.
Expanding the expression (2x-1)^2, we get:
(2x-1)^2 = (2x-1)(2x-1)
= 4x^2 - 2x - 2x + 1
= 4x^2 - 4x + 1
Now we can integrate each term separately:
∫4x^2 dx = (4/3)x^3 + C1 (using power rule)
∫-4x dx = -2x^2 + C2 (using power rule)
∫1 dx = x + C3 (since the integral of a constant is the constant times x)
Combining the integrals, we get:
∫(2x-1)^2 dx = (4/3)x^3 + C1 - 2x^2 + C2 + x + C3
Simplifying further, we can combine the constants:
= (4/3)x^3 - 2x^2 + x + C
So, the indefinite integral of ∫(2x-1)^2 dx can also be written as (4/3)x^3 - 2x^2 + x + C.