Integration
Integration transforms calculus students from basic derivative calculations into powerful problem-solvers who can find areas, volumes, and accumulation rates. Most students struggle with the power rule's "add 1 to exponent, divide by new exponent" pattern until they practice with structured problems ranging from simple x³ terms to definite integrals with bounds.
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Why it matters
Integration solves real-world accumulation problems across multiple disciplines. Engineers use definite integrals to calculate the 47,500 gallons of water flowing through a pipe over 8 hours when flow rates vary. Economists apply integration to find total profit when marginal profit functions change, such as determining that ∫(12x - 0.3x²)dx from 0 to 20 yields $1600 in accumulated profit. Physics students calculate displacement from velocity functions, finding that an object moving at v(t) = 2t + 5 travels 85 meters in the first 6 seconds. Business analysts use integration to model continuous growth, determining that a company growing at rate r(t) = 150e^(0.05t) will add 3,247 customers over 12 months. These applications demonstrate why mastery of basic integration techniques directly enables advanced problem-solving in STEM careers.
How to solve integration
Integration
- Integration is the reverse of differentiation.
- Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1).
- Definite integral: evaluate at upper and lower bounds, subtract.
- The definite integral gives the area under the curve.
Example: ∫x² dx = x³/3 + C. ∫₁² x² dx = 83 − 13 = 73.
Worked examples
Find the integral: ∫ 2 x3 dx
Answer: x4/2 + C
- Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) → ∫ 2 x^3 dx = 2·x^4/4 — Increase the exponent by 1 (to 4) and divide by the new exponent.
- Simplify and add constant → x^4/2 + C — Always add the constant of integration C for indefinite integrals.
Find the integral: ∫ (2 x2 + 2 x - 5) dx
Answer: 2 x3/3 + x2 - 5 x + C
- Write out the rule → ∫xⁿ dx = xⁿ⁺¹/(n+1) — The power rule for integration: raise the exponent by 1 and divide by the new exponent.
- Integrate the first term: ∫ 2 x^2 dx → 2 x^3/3 — Exponent 2 becomes 3, divide by 3: 2x³/3 = 2 x^3/3.
- Integrate the second term: ∫ 2 x dx → x^2 — Exponent 1 becomes 2, divide by 2: 2x²/2 = x^2.
- Integrate the constant: ∫ -5 dx → - 5 x — The integral of a constant k is kx.
- Combine and add C → 2 x^3/3 + x^2 - 5 x + C — Add all terms together. Always include the integration constant C.
Find the integral: ∫ 4 sin(x) dx
Answer: - 4 cos(x) + C
- Apply the rule: ∫sin(x) dx = −cos(x) → - 4 cos(x) + C — The constant 4 is carried through the integration.
Common mistakes
- ✗Students forget to add 1 to the exponent before dividing, writing ∫x³dx = x³/3 instead of x⁴/4, missing the fundamental power rule step.
- ✗When integrating constants, students often write ∫5dx = 5 instead of 5x, forgetting that constants become linear terms through integration.
- ✗Students frequently omit the constant C in indefinite integrals, writing ∫2x dx = x² instead of x² + C, losing points on formal assessments.
- ✗For definite integrals, students subtract bounds incorrectly, computing [x²]₁³ as 1² - 3² = -8 instead of 3² - 1² = 8, reversing the subtraction order.
Practice on your own
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