Integrate term by term: \(\int 3x^2 dx = x^3\), \(\int 2x dx = x^2\), so \(\int_0^1 (3x^2 + 2x) dx = [x^3 + x^2]_0^1 = (1 + 1) - (0 + 0) = 2\).

["Understanding Integration: A Step-by-Term Breakdown with (\int_0^1 (3x^2 + 2x) , dx = 2)", "Integration is a fundamental concept in calculus, essential for solving problems across science, engineering, and mathematics. One of the clearest ways to grasp integration is by integrating term by term—breaking down complex functions into simpler parts for easier computation. In this article, we’ll walk through the integration of (\int 3x^2 , dx = x^3), (\int 2x , dx = x^2), and combine them to solve (\int_0^1 (3x^2 + 2x) , dx = 2). Let’s dive in.", "---", "### The Term-by-Term Integration Process", "Calculus teaches us that the integral of a sum is the sum of integrals—this is known as the linearity property of integration. For expressions like (\int (u(x) + v(x)) , dx), we can write:", "[\n\int u(x) + v(x) , dx = \int u(x) , dx + \int v(x) , dx\n]", "This principle makes integration drastically simpler, especially when dealing with polynomial functions composed of multiple terms.", "---", "### Step 1: Integrate Each Term Separately", "Consider the function inside the definite integral:\n[\n\int_0^1 (3x^2 + 2x) , dx\n]", "By the linearity property, we integrate each term individually:", "[\n\int_0^1 3x^2 , dx + \int_0^1 2x , dx\n]", "#### Integral of (3x^2)", "We know from basic integration rules that:", "[\n\int x^n , dx = \frac{x^{n+1}}{n+1} + C \quad \ ext{for } n <br/>\neq -1\n]", "Applying this to (3x^2):", "[\n\int 3x^2 , dx = 3 \cdot \frac{x^{3}}{3} = x^3\n]", "Evaluating from 0 to 1:", "[\n[x^3]_0^1 = (1)^3 - (0)^3 = 1 - 0 = 1\n]", "#### Integral of (2x)", "Similarly:", "[\n\int 2x , dx = 2 \cdot \frac{x^{2}}{2} = x^2\n]", "Evaluating from 0 to 1:", "[\n[x^2]_0^1 = (1)^2 - (0)^2 = 1 - 0 = 1\n]", "---", "### Step 2: Combine the Results", "Now add the results of the two integrals:", "[\n\int_0^1 (3x^2 + 2x) , dx = \int_0^1 3x^2 , dx + \int_0^1 2x , dx = 1 + 1 = 2\n]", "---", "### Final Evaluation via Antiderivative Notation", "Alternatively, we express the result using definite integral antiderivatives:", "[\n\int_0^1 (3x^2 + 2x) , dx = \left[ x^3 + x^2 \right]_0^1\n]", "Evaluate the antiderivative at the upper and lower bounds:", "[\n(1^3 + 1^2) - (0^3 + 0^2) = (1 + 1) - (0 + 0) = 2 - 0 = 2\n]", "---", "### Conclusion: Why Integrate Term by Term?", "Breaking down (\int (3x^2 + 2x) , dx) into (\int 3x^2 , dx + \int 2x , dx) allows for straightforward integration using standard rules. This method reduces complexity and minimizes errors, forming the backbone for solving polynomial and more advanced functions in calculus.", "When computing definite integrals such as (\int_0^1), evaluating the antiderivative at bounds gives a precise, efficient result—here, confirming that:", "[\n\int_0^1 (3x^2 + 2x) , dx = 2\n]", "Understanding this term-by-term approach not only strengthens foundational calculus skills but also empowers learners to tackle integrals confidently in real-world applications.", "---", "Key Takeaways:\n- Use linearity of integration: (\int (u + v) , dx = \int u + \int v)\n- Apply standard power rule: (\int x^n , dx = \frac{x^{n+1}}{n+1})\n- Evaluate definite integrals by subtracting antiderivatives at bounds\n- Term-by-term integration simplifies complex expressions efficiently", "Whether you’re a student, educator, or enthusiast, mastering term-by-term integration—like solving (\int_0^1 (3x^2 + 2x) , dx = 2)—opens doors to deeper mathematical understanding."]