Math Problem Statement

Solve the integrals using the Fundamental Theorem of Calculus as specified in the given problems: 1. \(\int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx\) 2. \(\int_{0}^{5} \left(3x - 2\right)^2 dx\) 3. \(\int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x dx\)

Solution

Let’s solve the integrals step by step as presented in the image. The instructions mention using the Fundamental Theorem of Calculus. Here’s the process for each problem:


Problem 8:

22(x54x3+9)dx\int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx

Solution:

  1. Integrate each term of the polynomial:

    • x5dx=x66\int x^5 dx = \frac{x^6}{6}
    • 4x3dx=4x44=x4\int -4x^3 dx = -\frac{4x^4}{4} = -x^4
    • 9dx=9x\int 9 dx = 9x

    The antiderivative is: F(x)=x66x4+9xF(x) = \frac{x^6}{6} - x^4 + 9x

  2. Apply the Fundamental Theorem of Calculus: 22(x54x3+9)dx=F(2)F(2)\int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx = F(2) - F(-2)

  3. Calculate F(2)F(2) and F(2)F(-2):

    • F(2)=26624+9(2)=64616+18=646+2=746=373F(2) = \frac{2^6}{6} - 2^4 + 9(2) = \frac{64}{6} - 16 + 18 = \frac{64}{6} + 2 = \frac{74}{6} = \frac{37}{3}
    • F(2)=(2)66(2)4+9(2)=6461618=64634=6462046=1406=703F(-2) = \frac{(-2)^6}{6} - (-2)^4 + 9(-2) = \frac{64}{6} - 16 - 18 = \frac{64}{6} - 34 = \frac{64}{6} - \frac{204}{6} = -\frac{140}{6} = -\frac{70}{3}
  4. Subtract: F(2)F(2)=373(703)=373+703=1073F(2) - F(-2) = \frac{37}{3} - \left(-\frac{70}{3}\right) = \frac{37}{3} + \frac{70}{3} = \frac{107}{3}

Answer: 22(x54x3+9)dx=1073\int_{-2}^{2} \left(x^5 - 4x^3 + 9\right) dx = \frac{107}{3}


Problem 9:

05(3x2)2dx\int_{0}^{5} \left(3x - 2\right)^2 dx

Solution:

  1. Expand the square: (3x2)2=9x212x+4(3x - 2)^2 = 9x^2 - 12x + 4

  2. Integrate each term:

    • 9x2dx=3x3\int 9x^2 dx = 3x^3
    • 12xdx=6x2\int -12x dx = -6x^2
    • 4dx=4x\int 4 dx = 4x

    The antiderivative is: F(x)=3x36x2+4xF(x) = 3x^3 - 6x^2 + 4x

  3. Apply the Fundamental Theorem of Calculus: 05(3x2)2dx=F(5)F(0)\int_{0}^{5} \left(3x - 2\right)^2 dx = F(5) - F(0)

  4. Calculate F(5)F(5) and F(0)F(0):

    • F(5)=3(5)36(5)2+4(5)=3(125)6(25)+20=375150+20=245F(5) = 3(5)^3 - 6(5)^2 + 4(5) = 3(125) - 6(25) + 20 = 375 - 150 + 20 = 245
    • F(0)=3(0)36(0)2+4(0)=0F(0) = 3(0)^3 - 6(0)^2 + 4(0) = 0
  5. Subtract: F(5)F(0)=2450=245F(5) - F(0) = 245 - 0 = 245

Answer: 05(3x2)2dx=245\int_{0}^{5} \left(3x - 2\right)^2 dx = 245


Problem 10:

12(1+x2)32xdx\int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x \, dx

Solution: This integral involves substitution:

  1. Let u=1+x2u = 1 + x^2. Then:

    • du=2xdxdu = 2x \, dx
    • When x=1x = -1, u=1+(1)2=2u = 1 + (-1)^2 = 2
    • When x=2x = 2, u=1+22=5u = 1 + 2^2 = 5
  2. Rewrite the integral: 12(1+x2)32xdx=25u3du\int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x \, dx = \int_{2}^{5} u^3 \, du

  3. Integrate:

    • u3du=u44\int u^3 \, du = \frac{u^4}{4}

    The antiderivative is: F(u)=u44F(u) = \frac{u^4}{4}

  4. Apply the Fundamental Theorem of Calculus: 25u3du=F(5)F(2)\int_{2}^{5} u^3 \, du = F(5) - F(2)

  5. Calculate F(5)F(5) and F(2)F(2):

    • F(5)=544=6254F(5) = \frac{5^4}{4} = \frac{625}{4}
    • F(2)=244=164=4F(2) = \frac{2^4}{4} = \frac{16}{4} = 4
  6. Subtract: F(5)F(2)=62544=6254164=6094F(5) - F(2) = \frac{625}{4} - 4 = \frac{625}{4} - \frac{16}{4} = \frac{609}{4}

Answer: 12(1+x2)32xdx=6094\int_{-1}^{2} \left(1 + x^2\right)^3 \cdot 2x \, dx = \frac{609}{4}


Final Answers:

  1. Problem 8: 1073\frac{107}{3}
  2. Problem 9: 245245
  3. Problem 10: 6094\frac{609}{4}

Would you like detailed explanations for any step? Let me know!

Here are 5 related questions to explore:

  1. How do you perform substitution in definite integrals?
  2. What is the Fundamental Theorem of Calculus and why is it essential?
  3. How do you handle integrals with symmetric limits (e.g., a-a to aa)?
  4. What is the role of expanding terms in polynomial integration?
  5. How can you confirm an integral solution graphically?

Tip: Always double-check bounds and substitutions when solving definite integrals!

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Math Problem Analysis

Mathematical Concepts

Definite integrals
Polynomial integration
Substitution in integrals

Formulas

\(\int x^n dx = \frac{x^{n+1}}{n+1} + C\)
Substitution: Let \(u = f(x)\), then \(du = f'(x)dx\)
Fundamental Theorem of Calculus: \(\int_a^b f(x)dx = F(b) - F(a)\)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grade 11-12 (or early college-level calculus)