Find the area of the shaded region. WebAssign Plot у 4h 3 y=V x 2 1 X1, 1) X= 27 1 у х х 5 10 15 20 25
The Correct Answer and Explanation is:
It looks like you’re asking to find the area of a shaded region based on a WebAssign plot or graph. However, the information you provided is a bit unclear because the description seems incomplete or jumbled (for example, “у 4h 3 y=V x 2 1 X1, 1) X= 27 1 у х х 5 10 15 20 25”).
To help you accurately, I need to clarify a few points:
- What are the exact curves or boundaries of the shaded region?
- Are you given any functions or equations for the curves?
- Are there any limits of integration (x-values or y-values) provided?
- Is the shaded region between curves or under a curve?
Here’s a typical example to explain how to find the area of a shaded region bounded by curves:
Suppose the plot shows the curve y=xy = \sqrt{x} and the line y=3y = 3, with vertical boundaries at x=1x = 1 and x=9x = 9, and the shaded region is between these curves and these vertical lines.
Step-by-step method to find the area:
- Identify the curves and boundaries:
- Lower curve: y=xy = \sqrt{x}
- Upper curve: y=3y = 3
- Limits: x=1x = 1 and x=9x = 9
- Set up the integral for the area:
The area AA between two curves from x=ax=a to x=bx=b is A=∫ab(upper curve−lower curve)dxA = \int_a^b \left( \text{upper curve} – \text{lower curve} \right) dx
Here, A=∫19(3−x) dxA = \int_1^9 (3 – \sqrt{x}) \, dx
- Calculate the integral:
∫193 dx=3x∣19=3(9)−3(1)=27−3=24\int_1^9 3 \, dx = 3x \bigg|_1^9 = 3(9) – 3(1) = 27 – 3 = 24 ∫19x dx=∫19×1/2 dx=23×3/2∣19=23(93/2−13/2)=23(27−1)=23×26=523\int_1^9 \sqrt{x} \, dx = \int_1^9 x^{1/2} \, dx = \frac{2}{3} x^{3/2} \bigg|_1^9 = \frac{2}{3} (9^{3/2} – 1^{3/2}) = \frac{2}{3} (27 – 1) = \frac{2}{3} \times 26 = \frac{52}{3}
- Find the total area:
A=24−523=723−523=203≈6.67A = 24 – \frac{52}{3} = \frac{72}{3} – \frac{52}{3} = \frac{20}{3} \approx 6.67
Explanation:
The shaded region’s area is found by integrating the vertical difference between the upper and lower curves over the interval where the shading occurs. We subtract the lower curve from the upper curve because the area is vertical: at each xx, the height of the shaded region is the vertical distance between the two curves.
In this example, y=3y = 3 is a horizontal line above y=xy = \sqrt{x} in the interval from x=1x=1 to x=9x=9. By integrating the difference, we effectively sum up the tiny vertical slices (rectangles) whose widths are infinitesimally small, covering the entire interval.
Integration converts these slices into a continuous sum, which gives the total shaded area. This process works for any curves or lines enclosing a region, provided you know their equations and boundaries.
If you can provide:
- The exact functions or equations,
- The limits of the shaded region