Math Problem Statement

find the area of the shaded region

find the area of the shaded region

Solution

To find the area of the shaded region in the image, we need to analyze the curves that bound the region. The shaded region appears to be bounded by two parabolas.

Step 1: Identify the curves

The curves look like two parabolas, one opening upwards and the other downwards. Let's denote these parabolas by their general forms:

  • y=ax2+bx+cy = ax^2 + bx + c for the parabola opening upwards.
  • y=dx2+ex+fy = dx^2 + ex + f for the parabola opening downwards.

Step 2: Analyze the curves

From the symmetry and the shape, it appears that these parabolas might be of the form:

  • y=ax2y = ax^2 (upwards-opening parabola)
  • y=ax2+ky = -ax^2 + k (downwards-opening parabola)

where kk represents the vertical shift.

Step 3: Find the points of intersection

The intersection points between the two curves will provide the limits for integration. These points occur where:

ax2=ax2+kax^2 = -ax^2 + k

Simplifying this, we get:

2ax2=k    x2=k2a    x=±k2a2ax^2 = k \implies x^2 = \frac{k}{2a} \implies x = \pm \sqrt{\frac{k}{2a}}

These points give the bounds for xx.

Step 4: Set up the integral

The area between the curves can be calculated by integrating the difference between the upper curve and the lower curve:

Area=k2ak2a[(ax2+k)ax2]dx\text{Area} = \int_{-\sqrt{\frac{k}{2a}}}^{\sqrt{\frac{k}{2a}}} \left[(-ax^2 + k) - ax^2\right] \, dx

Simplifying:

Area=k2ak2a[2ax2+k]dx\text{Area} = \int_{-\sqrt{\frac{k}{2a}}}^{\sqrt{\frac{k}{2a}}} \left[-2ax^2 + k\right] \, dx

Step 5: Evaluate the integral

This can be split into two integrals:

Area=k2ak2akdx2ak2ak2ax2dx\text{Area} = \int_{-\sqrt{\frac{k}{2a}}}^{\sqrt{\frac{k}{2a}}} k \, dx - 2a \int_{-\sqrt{\frac{k}{2a}}}^{\sqrt{\frac{k}{2a}}} x^2 \, dx

The first integral is straightforward:

k2ak2akdx=k2k2a=2kk2a\int_{-\sqrt{\frac{k}{2a}}}^{\sqrt{\frac{k}{2a}}} k \, dx = k \cdot 2\sqrt{\frac{k}{2a}} = 2k\sqrt{\frac{k}{2a}}

The second integral, which is the integral of x2x^2, is:

k2ak2ax2dx=23(k2a)3/2\int_{-\sqrt{\frac{k}{2a}}}^{\sqrt{\frac{k}{2a}}} x^2 \, dx = \frac{2}{3} \left(\frac{k}{2a}\right)^{3/2}

Thus, the area is:

Area=2kk2a4a3(k2a)3/2\text{Area} = 2k\sqrt{\frac{k}{2a}} - \frac{4a}{3} \left(\frac{k}{2a}\right)^{3/2}

Conclusion

The exact expression for the area depends on the specific values of kk and aa, which can be derived from the original equations of the parabolas. To proceed further, I'd need the equations of the curves from which these areas are derived.

Let me know if you want to provide the equations of the curves or have any other questions!


Follow-up Questions:

  1. What are the equations of the curves shown in the graph?
  2. How would the area change if the parabolas were shifted horizontally?
  3. What happens to the area if one parabola is stretched vertically by a factor of 2?
  4. How does the integration method change if the parabolas intersect at different points?
  5. How can we verify the result of the area using numerical methods?

Tip: When dealing with areas between curves, always ensure you have correctly identified the upper and lower bounds of integration and the curves themselves.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Quadratic Functions
Area Between Curves

Formulas

y = ax^2
y = -ax^2 + k
Area = ∫ (upper curve - lower curve) dx

Theorems

Fundamental Theorem of Calculus
Area between two curves

Suitable Grade Level

Grades 11-12 (Advanced Calculus)