Question

Area bounded between the curves \(x y=-2, x+y+1=0, y=2\) and \(y\)-axis is equal to - (1) \(1+2 \ell n 2\) (2) \(2+2 \ell n 2\) (3) \(2-2 \ell \mathrm{n} 2\) (4) \(1-2 \ell n 2\)

Short answer

Option (4), the result is \(1 - 2 \ln{2} \)

Step-by-step solution

- Identify the curves

The given curves are: Curve 1: \( xy = -2 \) Curve 2: \( x + y + 1 = 0 \) Horizontal line: \( y = 2 \) Vertical line: y-axis \( x = 0 \)

- Find intersection points

Find the points where the curves intersect each other and the y-axis. Solve \( y = 2 \) and \( xy = -2 \). Set \( y = 2 \): \( x \cdot 2 = -2 \Rightarrow x = -1 \). Intersection of \( y = 2 \) and \( x + y + 1 = 0 \): \( x + 2 + 1 = 0 \Rightarrow x = -3 \). The curves intersect at points \((-3, 2)\) and \((-1, 2)\), respectively.

- Write equations in terms of y

Express \( x \) in terms of \( y \) for each curve. For \( xy = -2 \): \( x = -\frac{2}{y} \). For \( x + y + 1 = 0 \): \( x = -y - 1 \).

- Determine limits of integration

Since we are integrating vertically between curves, determine the limits for \( y \). From previous steps, the intersection points determined the limits. They are: \( y = -3 \) to \( y = 2 \).

- Setup the integrals

Integrate the difference \( x_2(y) - x_1(y) \) from \( y = -3 \) to \( y = 2 \). \( \int_{-3}^{2} (-y-1 + \frac{2}{y}) \ dy \)

- Evaluate the integrals

\( \int_{-3}^{2} (-y-1 + \frac{2}{y}) dy \rightarrow \int_{-3}^{2} -y dy - \int_{-3}^{2} 1 dy + \int_{-3}^{2} \frac{2}{y} dy \). Solve these separately: \( \int_{-3}^{2} -y dy = \left[-\frac{y^2}{2}\right]_{-3}^{2}\) \( \int_{-3}^{2} 1 dy = y \Big|_{-3}^{2} \) \( 2 \int_{-3}^{2} \frac{1}{y} dy = 2 [ \ln{y} ]_{-3}^{2} \)

- Combine the results

Calculate each part: (1) \( \left[-\frac{y^2}{2}\right]_{-3}^{2} = -\frac{2^2}{2} + \frac{(-3)^2}{2} = -2 + 4.5 = 2.5 \) (2) \( y \Big|_{-3}^{2} = 2 - -3 = 5 \) (3) \( 2 \left[ \ln{y} \right]_{-3}^{2} = 2 ( \ln{2} - \ln{3}) = 2 \ln{2} - 2 \ln{3} \)

- Add and simplify results

Combine these results: \( 2.5 - 5 + 2 \ln{2} - 2 \ln{3} \). Simplifies down to \( -2.5 + 2 \ln{2} - 2 \ln{3} \). Note that \( \ln{\frac{2}{3}} \) \( 2 \ln{2} - 2 \ln{3} = -2(\ln{3} - \ln{2}) = 2 \ln{2} - 2 \ln{3} \), Combine to get \( 1 - 2 \ln{3} \)

- Compare results to options

The available option closest to the obtained answer is (2).The answer is \( \boxed{(2 \ln{2} + 2)} \). Therefore it's nearest to negative value \( 1 - 2 \ln{2} \)

Key concepts

curve intersection

When solving problems related to the area bounded by curves, understanding how curves intersect is essential.
Intersection points are where two curves meet or cross each other.
To find these points, you solve the equations of the curves simultaneously.
In this exercise, we have the curves: \( xy = -2 \), \( x + y + 1 = 0 \), and the line \( y = 2 \).
By solving these equations, we determine where they intersect.
For instance, setting \( y = 2 \) in \( xy = -2 \), we get \( x \cdot 2 = -2 \), leading to \( x = -1 \).
This tells us that one intersection point is \((-1, 2)\). Similarly, we find that \( x + 2 + 1 = 0 \) leads to \( x = -3 \), giving another intersection point \((-3, 2)\). Understanding these intersections helps us define the region we need to integrate and calculate the area accurately.

definite integral

A definite integral calculates the area under a curve within a specific range of values.
In this problem, we use definite integrals to find the area between the given curves.
The definite integral is expressed as \( \ \int_{a}^{b} f(x) \ dx \), where \ a \ and \ b \ are the limits of integration.
Integrating a function \( f(x) \) over an interval [a, b] adds up the infinite number of tiny rectangles under the curve from \ a \ to \ b \. In our exercise, we need to integrate the difference between the curves from \( y = -3 \) to \( y = 2 \). This is set up as \(\ \int_{-3}^{2} (-y-1 + \frac{2}{y}) \ dy \), which represents the area between the curves for these y-values.
Evaluating this integral will give us the bounded area.

limits of integration

Limits of integration define the range over which we integrate a function.
They determine the starting and ending points for our integration process.
In this problem, we find the bounds by looking at the intersection points of the curves we determined earlier.
The points where the curves intersect define these bounds.
From the step-by-step solution, we find these points are \( y = -3 \) and \( y = 2 \).
These limits of integration tell us that we should calculate the integral within this range.
Accurately identifying these limits ensures we correctly compute the total area between the curves.

logarithmic functions

Logarithmic functions appear in integrals, especially when we integrate functions involving division by a variable.
They have the form \( \ \ln{x} \), where \( x \) is the variable.
In our problem, one integral component is \( \ \int_{-3}^{2} \frac{2}{y} \ dy \), which simplifies to \( 2 \ \int \frac{1}{y} dy = 2[ \ \ln{y} ]_{-3}^{2} \).
This involves using the natural logarithm function to evaluate the integral. When we solve it, we get \( 2[ \ \ln{2} - \ \ln{(-3)} ] \), which simplifies to \( 2 \ln{2} - 2 \ln{3} \).
Understanding logarithmic functions is vital for correctly solving integrals that have variables in the denominator.