Math Problem Statement

Solution

Let's analyze the given problem. We need to evaluate the two integrals given the graph:

22f(x)dxand26f(x)dx\int_{-2}^{2} f(x) \, dx \quad \text{and} \quad \int_{2}^{6} f(x) \, dx

The graph shows the function y=f(x)y = f(x) with two shaded areas between x=2x = -2 and x=6x = 6.

  1. For 22f(x)dx\int_{-2}^{2} f(x) \, dx:

    • The area from x=2x = -2 to x=0x = 0 is below the x-axis.
    • The area from x=0x = 0 to x=2x = 2 is above the x-axis.
  2. For 26f(x)dx\int_{2}^{6} f(x) \, dx:

    • The entire area from x=2x = 2 to x=6x = 6 is above the x-axis.

Calculation of Areas

  1. From x=2x = -2 to x=2x = 2:

    • The area from x=2x = -2 to x=0x = 0 (below the x-axis) appears to be a semicircle with radius 2.

      • The area of a semicircle is 12πr2\frac{1}{2} \pi r^2.
      • Here, r=2r = 2, so the area = 12π(22)=2π\frac{1}{2} \pi (2^2) = 2\pi.
    • The area from x=0x = 0 to x=2x = 2 (above the x-axis) is also a semicircle with radius 2.

      • The area is again 2π2\pi.

    20f(x)dx=2π(negative because it’s below the x-axis)\int_{-2}^{0} f(x) \, dx = -2\pi \quad (\text{negative because it's below the x-axis}) 02f(x)dx=2π\int_{0}^{2} f(x) \, dx = 2\pi

    Therefore: 22f(x)dx=2π+2π=0\int_{-2}^{2} f(x) \, dx = -2\pi + 2\pi = 0

  2. From x=2x = 2 to x=6x = 6:

    • The area from x=2x = 2 to x=6x = 6 is the area of the quarter circle and rectangle combined.
      • The quarter circle has radius 2 (area = 14π(22)=π\frac{1}{4} \pi (2^2) = \pi).
      • The rectangle has dimensions 4 (length) by 2 (height) (area = 4×2=84 \times 2 = 8).

    Therefore: 26f(x)dx=π+8\int_{2}^{6} f(x) \, dx = \pi + 8

Final Answer

22f(x)dx=0\int_{-2}^{2} f(x) \, dx = 0 26f(x)dx=π+8\int_{2}^{6} f(x) \, dx = \pi + 8

Would you like more details or have any questions?

Relative Questions

  1. How do we determine the sign of the area under the curve for definite integrals?
  2. Why is the area under the x-axis considered negative for integrals?
  3. Can you explain the relationship between definite integrals and the net area under the curve?
  4. How would the integrals change if the areas under the curve were differently shaped regions?
  5. What is the significance of the Fundamental Theorem of Calculus in evaluating these integrals?
  6. How can you verify the results of these integrals using numerical methods?
  7. What other methods can be used to find the area under the curve besides integration?
  8. How do the properties of symmetry in the function affect the calculation of integrals?

Tip

When dealing with definite integrals involving areas under the curve, always carefully analyze the regions above and below the x-axis to determine the correct signs for the areas.

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Math Problem Analysis

Mathematical Concepts

Definite Integrals
Areas under Curves
Geometric Shapes

Formulas

Area of Semicircle
Area of Rectangle

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12