Problem 50 - Applied Calculus

Question

Find the area bounded by the given curves. \(y=x^{2}-6 x\) and \(y=0\)

Step-by-Step Solution

Answer

The area bounded by the curves is 36 square units.

Step 1: Find Intersection Points

To find the area bounded by the curve and the x-axis, we'll first find where the curve intersects with the line \(y=0\). Set \(y = x^2 - 6x = 0\) and solve for \(x\): \[x(x-6)=0\]This gives the solutions \(x = 0\) and \(x = 6\).

Step 2: Set Up the Integral

The area bounded by the curve \(y = x^2 - 6x\) and the line \(y = 0\) is the integral of \(x^2-6x\) with respect to \(x\) from 0 to 6. The integral expression is:\[\int_0^6 (x^2 - 6x) \, dx\]

Step 3: Integrate the Function

Calculate the integral:\[\int (x^2 - 6x) \, dx = \int x^2 \, dx - \int 6x \, dx\]This gives:\[\frac{x^3}{3} - 3x^2 + C\] where \(C\) is the constant of integration.

Step 4: Evaluate the Integral

Evaluate the antiderivative from \(x = 0\) to \(x = 6\):\[\left[\frac{x^3}{3} - 3x^2\right]_0^6 = \left[\frac{6^3}{3} - 3(6^2)\right] - \left[\frac{0^3}{3} - 3(0^2)\right]\]

Step 5: Simplify the Calculation

Calculate each term:\[\frac{6^3}{3} = \frac{216}{3} = 72\]\[3 \times 6^2 = 3 \times 36 = 108\]Subtract these values:\[72 - 108 = -36\]

Step 6: Calculate the Absolute Area

Since area cannot be negative, take the absolute value:\[| -36 | = 36\]

Key Concepts

Integral Calculus

Integral calculus is a fundamental area of mathematics that helps us understand various concepts, such as area under curves, volume, and accumulation processes. It is one of the two primary branches of calculus, the other being differential calculus, which focuses on rates of change.
In integral calculus, we use integrals to find the accumulation of quantities. Especially when dealing with functions and curves, integrals enable us to compute the total area beneath a curve bounded by specific limits on the x-axis. The calculus of integrals deals with both indefinite integrals and definite integrals.

• Indefinite integrals give us a general form for antiderivatives and include an arbitrary constant, typically denoted as 'C'.
• Definite integrals calculate the exact area under the curve between two specific points on the x-axis, providing a numerical value.

Calculating integrals is akin to finding the reverse of derivatives, where we take a function and determine what original function could give such a derivative when differentiated. In the given exercise, we find the area under certain curves using definite integrals.

Definite Integrals

Definite integrals are essential for calculating the exact area bounded by a curve and the axes over a given interval. Unlike indefinite integrals, which result in a family of functions, definite integrals provide a specific numeric result.
In mathematical notation, a definite integral of a function f(x) from a to b is expressed as \[ \int_a^b f(x) \, dx\]This represents the area under the curve y = f(x) from x = a to x = b. It is computed by evaluating the difference between the antiderivative at these endpoints.
Consider the exercise where we seek the area under the curve \(y = x^2 - 6x\) from x = 0 to x = 6. The integration process involves finding an antiderivative of the function and evaluating it at the limits of integration.

• The first step is to calculate the integral of \(x^2 - 6x\) which yields \(\frac{x^3}{3} - 3x^2 + C\).
• Next, we find the antiderivative's value at both x = 0 and x = 6.
• The result at x = 0 is zero since all terms multiply to zero.
• At x = 6, substituting gives \(72 - 108\), leading to a negative area.

The final step is taking the absolute value of the result to get the area, which is 36 square units, since area cannot be negative.

Intersection Points of Curves

When dealing with curves, we often need to determine the points where they intersect. These points are where two functions have the same value of y for specific x values. In our exercise, it was necessary to find where the curve intersects with the x-axis.
For the curve \(y = x^2 - 6x\), finding intersection points with the line \(y = 0\) required solving the equation \(x^2 - 6x = 0\).

• We factored the quadratic expression to \(x(x-6) = 0\).
• Setting each factor to zero revealed the solutions \(x = 0\) and \(x = 6\), these are the x-values where the curve intersects the axis.

Determining intersection points is a crucial step in setting limits for definite integrals. By figuring out where the curve crosses zero, or another curve, we establish boundaries for integration. This process ensures we consider only the relevant section of the curve to find the area bounded by it. Such calculations are helpful in a variety of mathematical applications, from solving geometrical problems to analyzing real-world phenomena.