Problem 25 - Calculus and its applications

Question

Find the area of the region bounded by the graphs of the given equations. $$ y=5, y=\sqrt{x}, x=0 $$

Step-by-Step Solution

Answer

The area of the bounded region is \( \frac{125}{3} \) square units.

Step 1: Identify the Boundaries and Region

The area we need to find is bounded by three curves: the horizontal line \( y = 5 \), the curve \( y = \sqrt{x} \), and the vertical line \( x = 0 \). The line \( y = 5 \) intersects \( y = \sqrt{x} \) at the points where \( \sqrt{x} = 5 \). Solving this, we find \( x = 25 \). The region of interest is thus bound between \( x = 0 \) and \( x = 25 \).

Step 2: Set Up the Integral for Area

The area between the curve \( y = \sqrt{x} \) and the line \( y = 5 \) can be described by the integral of their difference. This area A can be written as: \[ A = \int_{0}^{25} (5 - \sqrt{x}) \, dx. \]

Step 3: Evaluate the Integral

We now need to calculate the integral \[ A = \int_{0}^{25} 5 \, dx - \int_{0}^{25} \sqrt{x} \, dx. \] The first integral is straightforward: \[ \int_{0}^{25} 5 \, dx = 5[x]_{0}^{25} = 5 \times (25 - 0) = 125. \]But the second integral: \[ \int_{0}^{25} \sqrt{x} \, dx = \int_{0}^{25} x^{1/2} \, dx = \left[ \frac{2}{3}x^{3/2} \right]_{0}^{25} = \frac{2}{3}[25^{3/2} - 0^{3/2}]. \]

Step 4: Simplify and Calculate

We must simplify \[ \frac{2}{3} \times 25^{3/2} = \frac{2}{3} \times (25 \times 5) = \frac{2}{3} \times 125 = \frac{250}{3}. \]Now, subtract the resolved integrals to find the area:\[ A = 125 - \frac{250}{3} = \frac{375}{3} - \frac{250}{3} = \frac{125}{3}. \]

Step 5: Conclude the Area Calculation

Thus, the area of the region bounded by the equations \( y = 5 \), \( y = \sqrt{x} \), and \( x = 0 \) is \( \frac{125}{3} \) square units. This represents the total space between the horizontal line and the square root curve.

Key Concepts

Integral Calculus

Integral Calculus is a fundamental part of calculus focused on the concept of integration. While Calculus is broadly divided into two sections, differential and integral calculus, the latter deals with the accumulation of quantities like areas under curves, total distances, and continuous sums.
Within Integral Calculus, we employ integrals to determine areas bounded by curves, evaluate total accumulation, and solve real-world problems involving rates of change.

• One fundamental theorem in Integral Calculus directs that the definite integral of a function over an interval provides the net area between the function and the horizontal axis on that interval.
• There are two types of integrals: definite and indefinite. Indefinite integrals yield a general form of antiderivatives, while definite integrals compute specific numerical values like areas.

By understanding how to set up and solve integrals, students can tackle various applied math problems that include finding curves' areas, volumes of solids, and more. This makes integral calculus a powerful tool to translate real-world situations into mathematical solutions.

Area Under a Curve

When we talk about finding the area under a curve, we mean the space between the curve of a graph and the x-axis, typically within specified bounds.
To visualize, imagine the graph of a function laid out on a coordinate plane. The area between the curve and the x-axis can be filled with infinitely small rectangles. Summing these rectangles' areas gives us the total area under the curve.

• This concept is crucial in areas ranging from Physics, helping to determine work done over an interval, to Economics, calculating consumer and producer surplus.
• In practical calculus terms, finding this area involves setting up a definite integral with the function of the curve as the integrand and the specified interval as the limits.

Mastering this concept allows one to graphically and quantitatively understand changes and areas related to cumulative quantities, easing the complexities of accumulation and distribution problems.

Definite Integrals

Definite Integrals are a cornerstone of integration techniques and are used to calculate the exact area under a curve between two points along the x-axis. They are represented as \(\int_{a}^{b} f(x) \, dx\) where \(a\) and \(b\) are the lower and upper bounds, respectively, and \(f(x)\) is the integrand.
Through this process, we determine the net area between the curve and the x-axis from \(x = a\) to \(x = b\).

• Unlike an indefinite integral, which provides a general form of antiderivative, definite integrals produce a real number that represents accumulated value in the context of the area under a curve or total accumulation in practical applications.
• The evaluation of a definite integral simplifies to utilizing the antiderivative and applying the bounds to find a numerical result.

When solving definite integrals, students often encounter fundamental challenges but mastering them serves as a gateway to understanding volume, area, and many phenomena described by Continuous Mathematics.