Math Problem Statement

Solve

Evaluate the following definite integrals: \( \int_{2}^{4} (x^2 + 3x + 1) \, dx - \int_{2}^{4} (2x^2 - 7x + 1) \, dx \)

Solution

To solve the problem, let's evaluate each definite integral separately and then subtract the results as specified.

Problem:

Evaluate: 24(x2+3x+1)dx24(2x27x+1)dx\int_{2}^{4} (x^2 + 3x + 1) \, dx - \int_{2}^{4} (2x^2 - 7x + 1) \, dx

Step 1: Evaluate 24(x2+3x+1)dx\int_{2}^{4} (x^2 + 3x + 1) \, dx

  1. Find the antiderivative of x2+3x+1x^2 + 3x + 1: (x2+3x+1)dx=x33+3x22+x\int (x^2 + 3x + 1) \, dx = \frac{x^3}{3} + \frac{3x^2}{2} + x

  2. Apply the limits of integration from 2 to 4: [x33+3x22+x]24\left[ \frac{x^3}{3} + \frac{3x^2}{2} + x \right]_{2}^{4}

    • At x=4x = 4: 433+3422+4=643+482+4=643+24+4=643+28=1483\frac{4^3}{3} + \frac{3 \cdot 4^2}{2} + 4 = \frac{64}{3} + \frac{48}{2} + 4 = \frac{64}{3} + 24 + 4 = \frac{64}{3} + 28 = \frac{148}{3}
    • At x=2x = 2: 233+3222+2=83+122+2=83+6+2=83+8=323\frac{2^3}{3} + \frac{3 \cdot 2^2}{2} + 2 = \frac{8}{3} + \frac{12}{2} + 2 = \frac{8}{3} + 6 + 2 = \frac{8}{3} + 8 = \frac{32}{3}
  3. Subtract the two results: 1483323=1163\frac{148}{3} - \frac{32}{3} = \frac{116}{3}

Step 2: Evaluate 24(2x27x+1)dx\int_{2}^{4} (2x^2 - 7x + 1) \, dx

  1. Find the antiderivative of 2x27x+12x^2 - 7x + 1: (2x27x+1)dx=2x337x22+x\int (2x^2 - 7x + 1) \, dx = \frac{2x^3}{3} - \frac{7x^2}{2} + x

  2. Apply the limits of integration from 2 to 4: [2x337x22+x]24\left[ \frac{2x^3}{3} - \frac{7x^2}{2} + x \right]_{2}^{4}

    • At x=4x = 4: 24337422+4=12831122+4=128356+4=128352=283\frac{2 \cdot 4^3}{3} - \frac{7 \cdot 4^2}{2} + 4 = \frac{128}{3} - \frac{112}{2} + 4 = \frac{128}{3} - 56 + 4 = \frac{128}{3} - 52 = \frac{-28}{3}
    • At x=2x = 2: 22337222+2=163282+2=16314+2=16312=203\frac{2 \cdot 2^3}{3} - \frac{7 \cdot 2^2}{2} + 2 = \frac{16}{3} - \frac{28}{2} + 2 = \frac{16}{3} - 14 + 2 = \frac{16}{3} - 12 = \frac{-20}{3}
  3. Subtract the two results: 283203=83\frac{-28}{3} - \frac{-20}{3} = \frac{-8}{3}

Step 3: Subtract the Integrals

Combine the results from Step 1 and Step 2: 1163(83)=1163+83=1243\frac{116}{3} - \left(\frac{-8}{3}\right) = \frac{116}{3} + \frac{8}{3} = \frac{124}{3}

Final Answer:

24(x2+3x+1)dx24(2x27x+1)dx=1243\int_{2}^{4} (x^2 + 3x + 1) \, dx - \int_{2}^{4} (2x^2 - 7x + 1) \, dx = \frac{124}{3}

Would you like more details on any part of the solution?


Here are some questions to further explore related topics:

  1. What is the process for finding antiderivatives for polynomial functions?
  2. How do definite integrals differ from indefinite integrals?
  3. Can definite integrals be interpreted in terms of area under a curve?
  4. How would you handle definite integrals with trigonometric functions?
  5. What are some common techniques for evaluating integrals without direct computation?

Tip: Always double-check limits of integration, as reversing them changes the sign of the result.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Polynomial Functions

Formulas

Antiderivative formula for polynomials \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
Definite integral evaluation using limits

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

College Calculus