Problem 3 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=4 x, y=0, x=1$$

Step-by-Step Solution

Answer

The area bounded by the curves is 2 square units.

Step 1: Understand the Curves and Bounds

The given curves are \( y = 4x \), \( y = 0 \), and the line \( x = 1 \). \( y = 4x \) is a straight line through the origin with slope 4, and \( y = 0 \) is the x-axis. The vertical line \( x = 1 \) is the boundary perpendicular to the x-axis. We need to find the area of the region above \( y = 0 \), below \( y = 4x \), and to the left of \( x = 1 \).

Step 2: Set Up the Integral

The area between the curves \( y = 4x \) and \( y = 0 \) from \( x = 0 \) to \( x = 1 \) can be found by integrating the function \( y = 4x \) with respect to \( x \). So, we set up the integral as \( \int_0^1 4x \, dx \).

Step 3: Compute the Integral

To solve the integral \( \int_0^1 4x \, dx \), first, find the antiderivative of \( 4x \), which is \( 2x^2 \). Then evaluate it from 0 to 1: \[ 2x^2 \Big|_0^1 = 2(1)^2 - 2(0)^2 = 2 - 0 = 2. \]

Step 4: Conclude the Solution

The computed result of the integral \( 2 \) represents the area bounded by the given curves. This tells us that the region enclosed by the line \( y = 4x \), the x-axis \( y = 0 \), and the vertical line \( x = 1 \) has an area of 2 square units.

Key Concepts

Area under a curve

In calculus, the area under a curve is an important concept related to integration. When we talk about the "area under a curve," we are referring to the space between the curve and the x-axis, within a certain interval. For example, consider the problem where we need to find the area bounded by the line \( y = 4x \), the x-axis (\( y = 0 \)), and the line \( x = 1 \). Here, we calculate the space under the line \( y = 4x \), from \( x = 0 \) to \( x = 1 \). Finding this area helps us understand how the values of the function add up over a segment.

• The curve \( y = 4x \) is linear, making the area a simple geometric shape.
• Visualizing this "under the curve" area makes it clear why integrals are used for such tasks — they capture how much "space" the function covers.

Definite integral

A definite integral is a fundamental tool in calculus used to find the exact area under a curve between two points. In this exercise, the task is to determine the area under the line \( y = 4x \), bounded by \( x = 0 \) and \( x = 1 \). The definite integral is represented by the notation \( \int_a^b f(x) \, dx \), where \( f(x) \) is the function, and \( a \) and \( b \) are the bounds.

• The integral \( \int_0^1 4x \, dx \) captures all accumulated values of \( 4x \) from \( x = 0 \) to \( x = 1 \).
• The evaluation of a definite integral involves computing the antiderivative at the upper and lower bounds and subtracting the results.

This process provides us with an exact value, which tells how much 'area' is contained below the function and above the x-axis, within the specified range.

Antiderivative

Understanding the concept of an antiderivative is crucial for solving integral calculus problems. An antiderivative of a function is another function whose derivative is the original function. If you can find the antiderivative, you can evaluate integrals. In our exercise, the function \( y = 4x \) has an antiderivative \( F(x) = 2x^2 \).

• The antiderivative allows us to compute the integral \( \int 4x \, dx \) more easily.
• Finding \( 2x^2 \) means that \( \frac{d}{dx}(2x^2) = 4x \), confirming that it is indeed the antiderivative of \( 4x \).

By evaluating \( F(x) \) at the bounds \( x = 1 \) and \( x = 0 \), we calculate the built-up area, indicating the total value accumulating under the function over that interval.

Mathematical problem-solving

Problem-solving in mathematics is about breaking down complex questions into manageable steps. In integral calculus, this involves using concepts like antiderivatives and definite integrals to find areas under curves or solve other problems. The exercise here is an example of this process: finding the area bounded by specific curves.

• Step 1: Identify the region by establishing boundaries and the functions involved.
• Step 2: Setup the definite integral to model the problem.
• Step 3: Compute the antiderivative, then evaluate it over the defined interval.
• Step 4: Interpret the solution in terms of the original question.

Each of these steps ensures that the problem is solved accurately, transforming an abstract question into a solved reality through a logical sequence. Effective mathematical problem-solving demands a clear understanding of each concept and careful application to the task at hand.