Integrate (x-3) ²dx
Expanding the square and then integrating we get:
(x-3)²dx = x² - 6x + 9 dx
Integrating we get:
∫(x-3)²dx = ∫(x²-6x+9)dx = (1/3)x³ - 3x² + 9x + C
where C is the constant of integration.
To integrate (x-3)² dx, we can use the power rule of integration combined with the constant multiple rule.
Let's break down the expression:
(x-3)² = (x-3)(x-3) = x² - 3x - 3x + 9 = x² - 6x + 9
Now, let's integrate each term separately using the power rule of integration:
∫ x² dx = (1/3) x³ + C1
∫ -6x dx = -3x² + C2
∫ 9 dx = 9x + C3
Where C1, C2, and C3 are constants of integration.
Now, we can add up all the integrals:
∫ (x-3)² dx = ∫ (x² - 6x + 9) dx
= (1/3) x³ - 3x² + 9x + C
So, the indefinite integral of (x-3)² dx is:
(1/3) x³ - 3x² + 9x + C, where C is the constant of integration.