Problem 1 - Essential Calculus

Question

In Exercises \(1-4\), set up the definite integral that gives the area of the region. $$ \begin{array}{l} f(x)=x^{2}-6 x \\ g(x)=0 \end{array} $$

Step-by-Step Solution

Answer

The definite integral that gives the area of the region between the functions is \[\int_{0}^{6}|x^2 - 6x|dx.\]

Step 1: Find the Intersection Points

The intersection points of the two functions can be found by setting the two functions equal to each other and solving for \(x\).\n\(f(x) = g(x)\) gives us \(x^2 - 6x = 0\). Solving for \(x\), we find \(x = 0, 6\).

Step 2: Setup the Definite Integral

The definite integral is set up with the intersection points as the boundaries and the absolute difference of the functions as the integrand.\nHence, the integral will be \[\int_{0}^{6}|f(x)-g(x)|dx\]. Since \(g(x)=0\), the absolute value of \(f(x)-g(x)\) is simply \(|f(x)|\), so the integral becomes \[\int_{0}^{6}|x^2 - 6x|dx.\]

Step 3: Write the Final Result

We have now successfully set up the definite integral that gives the area of the region between the curves \(f(x)\) and \(g(x)\):\nThe final answer is \[\int_{0}^{6}|x^2 - 6x|dx.\]

Key Concepts

Intersection Points

To understand intersection points, think of them as the places where two curves meet or cross on a graph. In the context of finding the area between curves, discovering these points is a crucial first step. These are the x-values where both functions have the same y-value.

• For the functions given: \(f(x) = x^2 - 6x\) and \(g(x) = 0\), we set them equal to find intersection points: \(x^2 - 6x = 0\).
• Solving the equation involves factoring out an \(x\): \(x(x - 6) = 0\).
• This gives the solutions \(x = 0\) and \(x = 6\), which are the intersection points.

These values help us define the limits of integration for calculating the area between the curves. Knowing how to find intersection points is essential whenever you are dealing with multiple functions in calculus.

Area between Curves

To find the area between two curves, we need to calculate the definite integral between their intersection points. The definite integral allows us to sum up an infinite number of infinitesimally small areas to find the total region between the curves.

• The area between two functions \(f(x)\) and \(g(x)\) from \(x=a\) to \(x=b\) is given by \(\int_{a}^{b} |f(x) - g(x)| \, dx\).
• Here, \(|f(x) - g(x)|\) ensures we consider only positive areas, regardless of which function is on top.
• In our problem, since \(g(x) = 0\), the area simplifies to \(\int_{0}^{6} |x^2 - 6x| \, dx\).

By setting up and computing this integral, we find the exact area between \(f(x)\) and the x-axis over the interval [0, 6]. This process is a common application of definite integrals in calculus.

Absolute Value

Using absolute value in integrals helps manage situations where a curve crosses the axis. The absolute value ensures that the area calculated is always positive since area cannot be negative.

• When dealing with \(f(x) = x^2 - 6x\), using \(|f(x)|\) ensures that dips below the x-axis count positively towards the area.
• Graphically, absolute value reflects any portion of the curve below the x-axis across the x-axis to be above it, maintaining positivity for area.
• In the integral \(\int_{0}^{6} |x^2 - 6x| \, dx\), the absolute value is crucial since \(x^2 - 6x\) changes sign within the interval [0, 6].

This concept ensures the area computation is accurate and reflects the true geometric space between the curve and the axis. Using absolute values is a fundamental technique in integration when working with functions that can dip below the x-axis.