Problem 3 - Calculus and its applications

Question

Find the area under the graph of \(g\) over [-2,3] . $$ g(x)=\left\\{\begin{array}{ll} x^{2}+4, & \text { for } \quad x \leq 0 \\ 4-x, & \text { for } \quad x>0 \end{array}\right. $$

Step-by-Step Solution

Answer

The total area under the curve over \([-2,3]\) is \(\frac{109}{6}\).

Step 1: Understand the Function Pieces

The function \( g(x) \) is defined piecewise. For \( x \leq 0 \), it is \( g(x) = x^2 + 4 \), and for \( x > 0 \), it is \( g(x) = 4 - x \). We need to find the area under these curves from \(-2\) to \(3\).

Step 2: Calculate Area for \( x \leq 0 \)

Consider \( g(x) = x^2 + 4 \) for \( x \leq 0 \). We need the area from \(-2\) to \(0\). Compute the integral:\[ \int_{-2}^{0} (x^2 + 4) \, dx. \] The antiderivative of \( x^2 + 4 \) is \( \frac{x^3}{3} + 4x \). Evaluate it from \(-2\) to \(0\): \[ \left[ \frac{x^3}{3} + 4x \right]_{-2}^{0} = \left( \frac{0^3}{3} + 4 \cdot 0 \right) - \left( \frac{(-2)^3}{3} + 4 \cdot (-2) \right) = 0 - \left( \frac{-8}{3} - 8 \right) = \frac{8}{3} + 8 = \frac{32}{3}. \]

Step 3: Calculate Area for \( x > 0 \)

Consider \( g(x) = 4 - x \) for \( x > 0 \). We need the area from \(0\) to \(3\). Compute the integral:\[ \int_{0}^{3} (4 - x) \, dx. \] The antiderivative of \( 4 - x \) is \( 4x - \frac{x^2}{2} \). Evaluate it from \(0\) to \(3\): \[ \left[ 4x - \frac{x^2}{2} \right]_{0}^{3} = \left( 4 \cdot 3 - \frac{3^2}{2} \right) - \left( 4 \cdot 0 - \frac{0^2}{2} \right) = 12 - \frac{9}{2} = \frac{24}{2} - \frac{9}{2} = \frac{15}{2}. \]

Step 4: Add Areas Together

Add the results from the previous steps to determine the total area under \( g(x) \) from \(-2\) to \(3\). We have \( \frac{32}{3} \) from \(-2\) to \(0\) and \( \frac{15}{2} \) from \(0\) to \(3\). Convert \( \frac{32}{3} \) and \( \frac{15}{2} \) to a common denominator (6) before adding: \[ \frac{32}{3} = \frac{64}{6}, \quad \frac{15}{2} = \frac{45}{6}. \] Thus, the total area is: \( \frac{64}{6} + \frac{45}{6} = \frac{109}{6} \).

Key Concepts

Piecewise Functions

Piecewise functions are an essential concept in mathematics, especially when dealing with real-world problems where a function behaves differently under different conditions. The function is defined piece by piece, or segment by segment.
In the exercise provided, the function \(g(x)\) is an example of a piecewise function:

• For \(x \leq 0\), the function behaves as \(g(x) = x^2 + 4\), which is a quadratic function.
• For \(x > 0\), the function shifts to a linear function, \(g(x) = 4 - x\).

Each piece of the function is applicable only within its defined interval. To find quantities like area under the curve, you must analyze each piece individually. This makes it crucial to understand the specific segment you are working with for accurate calculations.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are used to find a function whose derivative is the given function. Calculating antiderivatives is a critical step when determining the area under a curve.
When dealing with the piecewise function \(g(x)\):

• For \(x^2 + 4\), the antiderivative is \(\frac{x^3}{3} + 4x\).
• For \(4 - x\), the antiderivative is \(4x - \frac{x^2}{2}\).

These antiderivatives help compute the definite integrals needed to find exact areas. When you evaluate these antiderivatives within specific bounds, you get a numeric value representing the area under the curve for that interval. This process is crucial to solving area problems in calculus.

Area under a Curve

Finding the area under a curve is a fundamental application of integration in calculus, often used to determine accumulated quantities.
In this exercise, the total area under the graph of \(g(x)\) is found by adding the areas under each piece of the piecewise function:

• Calculate the integral from \(-2\) to \(0\) for \(x^2 + 4\), yielding \(\frac{32}{3}\).
• Calculate the integral from \(0\) to \(3\) for \(4 - x\), resulting in \(\frac{15}{2}\).

After finding these separate areas, the next step is to add them, ensuring the denominators are consistent. The combined area from \(-2\) to \(3\) is \(\frac{109}{6}\).
This illustrates how definite integration is applied to piecewise functions to find the area under a curve, an important tool in many fields like physics and engineering.