Q.Integrate the following function: $\int (ax^2 + bx + c) dx$

The key idea is Power Rule Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$, applied term-by-term.

Step 1: Split the integral across the sum:

Step 2: Factor out constants and apply the power rule to each term:

$a \int x^2 \, dx = a \cdot \frac{x^{3}}{3}, \quad b \int x \, dx = b \cdot \frac{x^{2}}{2}, \quad c \int 1 \, dx = c x$ …

The key idea here is that integration is a linear operation — you can break a sum into separate integrals and factor out constants. So instead of tackling the whole expression at once, we handle each term individually.

The Power Rule for integration is the reverse of the derivative power rule. If you differentiate $x^{n+1}$, you get $(n+1)x^n$. So to undo that, when integrating $x^n$, you increase the exponent by 1 and divide by the new exponent. The constant $C$ appears because differentiation eliminates any constant term.

Let’s walk through it step by step.

1. Separate the integral using linearity. The integral of a sum is the sum of the integrals, and constants can be pulled out:

2. Apply the Power Rule to $x^2$.

   For $n = 2$, the rule says $\int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3}$.

   So $a \int x^2 \, dx = a \cdot \frac{x^3}{3} = \frac{a}{3}x^3$.

3. Apply the Power Rule to $x$.

   For $n = 1$, $\int x \, dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}$.

   So $b \int x \, dx = b \cdot \frac{x^2}{2} = \frac{b}{2}x^2$.

4. Integrate the constant term.

   The constant $c$ can be thought of as $c \cdot x^0$. Using the Power Rule with $n = 0$: $\int x^0 \, dx = \frac{x^{0+1}}{0+1} = x$.

   So $c \int 1 \, dx = c \cdot x = cx$.

5. Add the constant of integration. …

Method: Integrating a general polynomial

Integrate a polynomial term by term with the power rule; each coefficient rides along unchanged.

Steps

Step 1: Split by linearity.

$\int(ax^2+bx+c)\,dx=a\int x^2\,dx+b\int x\,dx+c\int 1\,dx$.

Step 2: Power-rule each term. …

Common Mistakes

Mistake 1: Integrating the constant $c$ as $c$ instead of $cx$.

Why it's wrong: $\int c\,dx=cx$ since $c=cx^0$ integrates to $cx^1$. Correct approach: the constant term contributes $cx$, not $c$.

Mistake 2: Omitting the constant of integration. …