Results Summary

If you’re searching for “integration of x 2 2,” here are some possible interpretations and their solutions:

$$\boxed{\int \frac{x – 2}{2} \, dx = \frac{x^2}{4} – x + C}$$

$$\boxed{\int (x – 2)^2 \, dx = \frac{x^3}{3} – 2x^2 + 4x + C}$$

$$\boxed{\int \frac{x^2}{2} \, dx = \frac{x^3}{6} + C}$$

$$\boxed{\int \frac{x}{2} + 2 \, dx = \frac{x^2}{4} + 2x + C}$$

Introduction

The “integration of x 2 2” could have multiple meanings. Here are four possible options with detailed solutions for each.

Option 1: \( \displaystyle \int \frac{x – 2}{2} \, dx \)

Let’s separate the terms:

\[\begin{align*} = \frac{1}{2} \int x \, dx – \frac{1}{2} \int 2 \, dx \end{align*}\]

Now integrate each term:

\[\begin{align*} = \frac{1}{2} \cdot \frac{x^2}{2} – \frac{1}{2} \cdot 2x + C \end{align*}\]

Simplify:

\[\begin{align*} = \frac{x^2}{4} – x + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int \frac{x – 2}{2} \, dx = \frac{x^2}{4} – x + C \end{align*}\]

Option 2: \( \displaystyle \int (x – 2)^2 \, dx \)

Let’s expand \( (x – 2)^2 \):

\[\begin{align*} = \int (x^2 – 4x + 4) \, dx \end{align*}\]

Now integrate each term:

\[\begin{align*} = \frac{x^3}{3} – 2x^2 + 4x + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int (x – 2)^2 \, dx = \frac{x^3}{3} – 2x^2 + 4x + C \end{align*}\]

Option 3: \( \displaystyle \int \frac{x^2}{2} \, dx \)

Here, we factor out the constant \( \frac{1}{2} \):

\[\begin{align*} = \frac{1}{2} \int x^2 \, dx \end{align*}\]

Using the power rule:

\[\begin{align*} = \frac{1}{2} \cdot \frac{x^3}{3} + C = \frac{x^3}{6} + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int \frac{x^2}{2} \, dx = \frac{x^3}{6} + C \end{align*}\]

Option 4: \( \displaystyle \int \frac{x}{2} + 2 \, dx \)

We first separate the terms:

\[\begin{align*} = \int \frac{x}{2} \, dx + \int 2 \, dx \end{align*}\]

Now integrate each term:

\[\begin{align*} = \frac{x^2}{4} + 2x + C \end{align*}\]

Thus, the result is:

\[\begin{align*} \int \frac{x}{2} + 2 \, dx = \frac{x^2}{4} + 2x + C \end{align*}\]