Problem 29 - Calculus and its applications

Question

Find the area under the graph of each function over the given interval. $$ y=5-x^{2} ; \quad[-1,2] $$

Step-by-Step Solution

Answer

The area under the curve is 12.

Step 1: Set Up the Integral

To find the area under the curve of the function \( y = 5 - x^2 \) over the interval \([-1, 2]\), we need to set up the definite integral:\[\int_{-1}^{2} (5 - x^2) \, dx.\]

Step 2: Integrate the Function

Find the antiderivative of the function \(5 - x^2\). The antiderivative of \(5\) is \(5x\), and the antiderivative of \(-x^2\) is \(-\frac{x^3}{3}\). Thus, the antiderivative of \(5 - x^2\) is:\[5x - \frac{x^3}{3} + C.\]

Step 3: Evaluate the Integral at the Upper Limit

Substitute \(x = 2\) into the antiderivative \(5x - \frac{x^3}{3}\):\[\left. 5x - \frac{x^3}{3} \right|_{x=2} = 5(2) - \frac{2^3}{3} = 10 - \frac{8}{3} = 10 - 2.67 = 7.33.\]

Step 4: Evaluate the Integral at the Lower Limit

Substitute \(x = -1\) into the antiderivative \(5x - \frac{x^3}{3}\):\[\left. 5x - \frac{x^3}{3} \right|_{x=-1} = 5(-1) - \frac{(-1)^3}{3} = -5 - (-\frac{1}{3}) = -5 + 0.33 = -4.67.\]

Step 5: Subtract to Find the Area

The area under the curve from \(-1\) to \(2\) is the difference between the value of the antiderivative at the upper limit and the lower limit:\[7.33 - (-4.67) = 7.33 + 4.67 = 12.\]

Key Concepts

Antiderivative

An antiderivative is essentially the opposite of taking a derivative. While derivatives tell us the rate of change of a function, antiderivatives allow us to go backwards and find an original function from its rate of change. When you have a function like \( f(x) = 5 - x^2 \), finding its antiderivative gives us a new function whose derivative is \( 5 - x^2 \). Evaluating powers of \( x \) or simple constants, for example:

• The antiderivative of a constant \( 5 \) is \( 5x \).
• The antiderivative of \( -x^2 \) involves increasing the power of \( x \) by one and dividing by the new power. So, it turns into \( -\frac{x^3}{3} \).

To represent the most general form of the antiderivative, we add \( C \), which accounts for any constant that could get "lost" during differentiation. In this context of definite integrals, \( C \) will cancel out, so it typically doesn't affect our calculations for area.

Calculating Area Under a Curve

Calculating the area under a curve is a common problem in calculus, and definite integrals help solve this by providing a precise way to find such areas. For the function \( y = 5 - x^2 \), we needed to find the area over a specific interval, from \(-1\) to \(2\). The definite integral is denoted as \( \int_{-1}^{2} (5 - x^2) \, dx \), and it represents the sum of an infinite number of infinitesimally small rectangles under the curve.
For example:

• Visualize splitting the region between the curve and the \( x \)-axis into skinny vertical strips.
• The width of these strips is infinitely small, while each strip's height matches the function's value at that point.

This idea is captured through integration, allowing us to calculate areas even with complex curves accurately. In this exercise, the definite integral helps us to find the exact area without underestimation or overestimation, which might occur when using basic geometric methods.

Definite Integration Steps

To solve a definite integral, such as finding the area under the curve of \( y = 5 - x^2 \) from \(-1\) to \(2\), you need to follow specific steps:1. **Set Up the Integral:** Start by identifying the function and the interval. Here, it's \( \int_{-1}^{2} (5 - x^2) \, dx \).2. **Find the Antiderivative:** Determine the antiderivative, which in this case is \( 5x - \frac{x^3}{3} \).3. **Evaluate at the Upper Limit:** Plug the upper limit \( x = 2 \) into the antiderivative: \[ 5(2) - \frac{2^3}{3} = 10 - \frac{8}{3} \] Simplifying this gives \( 7.33 \).
4. **Evaluate at the Lower Limit:** Do the same for the lower limit \( x = -1 \): \[ 5(-1) - \frac{(-1)^3}{3} = -5 + \frac{1}{3} \] This simplifies to \( -4.67 \).5. **Calculate the Area:** Subtract to find the total area under the curve: \[ 7.33 - (-4.67) = 7.33 + 4.67 = 12 \] These steps illustrate the methodical process of definite integration, providing the exact total area beneath a curve between specified intervals. By following this procedure, one can consistently achieve accurate results.