Problem 13 - Precalculus Mathematics for Calculus

Question

Use the definition of area as a limit to find the area of the region that lies under the curve. Check your answer by sketching the region and using geometry. $$y=3 x, \quad 0 \leq x \leq 5$$

Step-by-Step Solution

Answer

The area under the curve is 37.5 square units.

Step 1: Understand the Problem

We are given the linear function \( y = 3x \) and need to find the area under this curve from \( x = 0 \) to \( x = 5 \). This is essentially finding the area of a region under this line and above the x-axis within the given limits.

Step 2: Set Up the Integral

The area under the curve can be found using the integral of \( y = 3x \) from \( x = 0 \) to \( x = 5 \). Set up the integral:\[ A = \int_{0}^{5} 3x \, dx \]

Step 3: Integrate the Function

To find the antiderivative of \( 3x \), apply the power rule of integration. The antiderivative is \( \frac{3}{2}x^2 \). Evaluate the definite integral:\[ A = \left[ \frac{3}{2}x^2 \right]_{0}^{5} \]

Step 4: Evaluate the Definite Integral

Substitute the upper and lower limits into the antiderivative:\[ A = \frac{3}{2}(5)^2 - \frac{3}{2}(0)^2 \] \[ A = \frac{3}{2}(25) - 0 = \frac{75}{2} = 37.5 \].

Step 5: Verify Using Geometry

The region under the curve forms a right triangle with base \( b = 5 \) and height \( h = 15 \) (since \( 3 \times 5 = 15 \)). The area of a triangle is \( \frac{1}{2} \times b \times h \). \[ \text{Area} = \frac{1}{2} \times 5 \times 15 = 37.5 \] confirming our integral result.

Key Concepts

Limits in Calculus

In calculus, a limit is a fundamental concept that helps us understand how functions behave as they approach a certain point. When we're determining the area under a curve, limits play a crucial role. We use them to understand behavior as a value tends towards any specific point. Consider this: each tiny segment under a curve represents a rectangle's area. When added up, these segments can approximate the total area. As these segments become increasingly small (infinitesimally, to be precise), the sum of these rectangles approaches the exact area under the curve.
This process signifies the idea of a limit. By setting the individual segment widths to approach zero, we're moving towards the precise area calculation. This technique is foundational in calculus and is the building block for integral calculus. Here, the limit helps so that instead of approximating an area with finite widths, we compute it precisely.
Thus, the concept of limits allows calculus to transition from approximate measurements to exact solutions, a pivotal leap in mathematical accuracy.

Definite Integrals

Definite integrals are a core tool in calculus, used for calculating the area under a curve between two points, among other applications. When we talk about definite integrals, we are essentially discussing the integration of a function within a specific interval. In this exercise, the definite integral calculates the area under the line from 0 to 5.
The notation for a definite integral is \[ \int_{a}^{b} f(x) \, dx \]where \( a \) and \( b \) represent the limits of integration. These limits define the region of interest along the x-axis.

• The function \( f(x) \) is what we're integrating. In our case, this function was \( 3x \).
• The \( dx \) denotes that we're integrating concerning \( x \).

Evaluating the integral gives the actual area by substituting these limits into the integral's result. Then, you calculate the difference between the values at these limits.
This outcome is not just the sum of infinite slices under the curve but the exact area measured, thanks to the power of calculus.

Geometry Verification

Verifying the result using geometry is a straightforward method to confirm the accuracy of our integrals and calculations. Essentially, it's about seeing whether simpler geometric perspectives lead to the same outcomes. In this exercise, we showed that the triangle formed under the line from 0 to 5 is the focus.
Calculating geometrically, the base is 5, and the height is calculated by substituting the value of \( x \) at the end of the interval into the function (\( y = 3 \times 5 = 15 \)).

• Using the area formula for a triangle, \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \), gives us \( \frac{1}{2} \times 5 \times 15 = 37.5 \).
• This matches the area calculated through the definite integral, confirming our computation is correct.

Such verification strengthens our confidence in results derived using calculus, by showing they align with classical geometric principles, making it an essential check when solving similar problems.