Question

Compute the area of the figure bounded by a portion of the straight line \(y=x\) and segments of the straight lines \(\mathrm{y}=0\) and \(\mathrm{x}=3\).

Short answer

Answer: The area of the figure bounded by the given lines is 4.5 square units.

Step-by-step solution

Identify the Figure and Intersection Points

First, let's find the intersection points of the lines. The line \(y=x\) intersects the x-axis at \((0,0)\), and the line \(y=0\) is the x-axis itself. The vertical line \(x=3\) intersects the line \(y=x\) at the point \((3,3)\). Therefore, the figure is a triangle with vertices at points \((0,0)\), \((3,0)\), and \((3,3)\).

Determine the Base and Height of the Triangle

Now that we have identified the figure as a triangle, we can determine its base and height. The base can be represented by the line segment between \((0,0)\) and \((3,0)\), which has a length of 3 units. The height can be represented by the line segment between \((3,0)\) and \((3,3)\), which has a length of 3 units as well.

Compute the Area of the Triangle

We can now compute the area of the triangle using the formula for the area of a triangle, which is: \(Area = \frac{1}{2} * Base * Height\) Plug in the values for the base and height: \(Area = \frac{1}{2} * 3 * 3\)

Simplify and Find the Area

Simplify the expression to obtain the area of the triangle: \(Area = \frac{1}{2} * 9 = 4.5\) The area of the figure bounded by the given lines is 4.5 square units.

Key concepts

Area of a Triangle

Understanding the area of a triangle is fundamental to geometry and integral calculus. The area can be thought of as the amount of space a triangle occupies.

For any triangle, the area can be calculated by taking the product of its base and height and then dividing that product by two. We express this with the formula:
\[Area = \frac{1}{2} \times Base \times Height\]
This equation represents one of the simplest ways to find a triangle's area. In the context of the given problem, we're dealing with a right-angled triangle, making the calculation even more straightforward since the base and height correspond to the legs of the triangle.

Intersection Points

Finding intersection points is critical when analyzing graphs of linear equations. The point where two or more lines intersect is an 'intersection point'. These points are significant because they can be the vertices of geometric figures we're interested in, like triangles or polygons.

When dealing with linear equations, we often set the equations equal to each other to find their points of intersection. For example, if two lines in a Cartesian plane intersect each other, the x and y coordinates at which they do so can be found by solving the equations simultaneously.

Integral Calculus for IIT JEE

Integral calculus is a highly focused topic for the Indian Institutes of Technology Joint Entrance Examination (IIT JEE). It encompasses a broad range of concepts, methods, and applications.

Integral calculus covers techniques of integration, definite and indefinite integrals, and the application of these concepts to compute areas under curves, volumes of solids of revolution, and much more. For IIT JEE aspirants, mastering integral calculus, including understanding areas and the applications of integrals, is crucial for solving complex problems.

Linear Equations and Graphs

Linear equations form straight lines when graphed on a coordinate plane. The standard form of a linear equation is \(y = mx + b\), where \(m\) is the slope, and \(b\) is the y-intercept. Graphs of linear equations can be utilized to visualize problems, find intersection points, and solve for variables.

When dealing with shapes bounded by linear equations, as in our exercise, understanding how to graph these equations is essential. The intercepts on the axes and the slope of the line give us crucial information about the position and orientation of the line in relation to other geometric features in the plane.