Plane Figure $S_1$ Definition
The inequalities $|x-1| \le 2$ define the range for the x-coordinate:
- $-2 \le x-1 \le 2$
- Adding 1 to all parts gives: $-1 \le x \le 3$.
The inequalities $|y+2| \le 3$ define the range for the y-coordinate:
- $-3 \le y+2 \le 3$
- Subtracting 2 from all parts gives: $-5 \le y \le 1$.
Therefore, $S_1$ represents a rectangle with vertices at $(-1, -5)$, $(3, -5)$, $(3, 1)$, and $(-1, 1)$. The width of this rectangle is $3 - (-1) = 4$ and the height is $1 - (-5) = 6$.
$S_1$ Area Calculation
The area of rectangle $S_1$ is calculated as:
$ Area(S_1) = \text{width} \times \text{height} = 4 \times 6 = 24 $
Plane Figure $S_2$ Definition
The figure $S_2$ is defined by the following inequalities:
- $x - y \ge -2$, which is equivalent to $y \le x + 2$.
- $y \ge 1$.
- $x \le 3$.
These inequalities define a triangular region. The vertices are found by solving the systems of equations for the boundary lines:
- Intersection of $y=1$ and $x=3$: This gives the vertex $(3, 1)$.
- Intersection of $y=1$ and $y=x+2$: Substituting $y=1$ gives $1 = x+2$, so $x = -1$. This gives the vertex $(-1, 1)$.
- Intersection of $x=3$ and $y=x+2$: Substituting $x=3$ gives $y = 3+2 = 5$. This gives the vertex $(3, 5)$.
$S_2$ is a triangle with vertices $(-1, 1)$, $(3, 1)$, and $(3, 5)$.
$S_2$ Area Calculation
The base of the triangle lies along the line $y=1$, extending from $x=-1$ to $x=3$. The length of the base is $3 - (-1) = 4$. The height of the triangle is the perpendicular distance from the base ($y=1$) to the third vertex $(3, 5)$, which is $5 - 1 = 4$.
The area of triangle $S_2$ is:
$ Area(S_2) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 $
Union Area Calculation: $S = S_1 \cup S_2$
The set $S$ is the union of $S_1$ and $S_2$, denoted as $S = S_1 \cup S_2$. The area of the union is calculated using the principle of inclusion-exclusion:
$ Area(S) = Area(S_1) + Area(S_2) - Area(S_1 \cap S_2) $
We need to determine the area of the intersection $S_1 \cap S_2$.
Conditions for $S_1$: $-1 \le x \le 3$ and $-5 \le y \le 1$.
Conditions for $S_2$: $y \ge 1$, $x \le 3$, and $y \le x+2$.
For a point to be in the intersection $S_1 \cap S_2$, it must satisfy all conditions:
- Comparing $y$ bounds: $S_1$ requires $y \le 1$ and $S_2$ requires $y \ge 1$. The only possibility is $y=1$.
- Comparing $x$ bounds: $S_1$ requires $-1 \le x \le 3$ and $S_2$ requires $x \le 3$. Combined, this means $-1 \le x \le 3$.
- Checking the final condition $y \le x+2$: With $y=1$, this becomes $1 \le x+2$, which simplifies to $x \ge -1$. This condition is already satisfied by $-1 \le x \le 3$.
The intersection $S_1 \cap S_2$ consists of points where $y=1$ and $-1 \le x \le 3$. This is a line segment, and its area is 0.
$ Area(S_1 \cap S_2) = 0 $
Final Area Calculation for $S$
Substitute the calculated areas into the union formula:
$ Area(S) = Area(S_1) + Area(S_2) - Area(S_1 \cap S_2) $
$ Area(S) = 24 + 8 - 0 $
$ Area(S) = 32 $
The total area of the union $S$ is 32.
Important Questions from Application Of Definite Integral (Area)
The equation of a closed curve in two-dimensional polar coordinates is given by $r = \frac{2}{\sqrt{\pi}}(1 - \sin \theta)$. The area enclosed by the curve is ______ (answer in integer).
The area bounded by the curves, $y = \sqrt{x}$, and $y = 8x^2$ is _______________(rounded off to 3 decimal places)
Two straight lines pass through the origin $(x_0, y_0) = (0,0)$. One of them passes through the point $(x_1, y_1) = (1,3)$ and the other passes through the point $(x_2, y_2) = (1,2)$.
What is the area enclosed between the straight lines in the interval $[0, 1]$ on the x-axis?
Consider the equation for a curve, $y = f(x) = x^2 + x$.
The area enclosed by the curve, the x -axis ($y = 0$ line); the vertical lines passing through $x = 1$ and $x = 2$ is _________ (rounded off to 2 decimal places)- The area of the region (rounded off to one decimal place) enclosed between the curves $y = x$ and $y = 3\sqrt{x}$ and between the lines $x = 0$ and $x = 1$ is ________ units.
