Problem 28 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The area under the curve is \( \frac{9}{2} \) square units.

Step 1: Understand the Problem

We need to find the area under the curve of the function \( y = 2 - x - x^2 \) from \( x = -2 \) to \( x = 1 \). This involves calculating a definite integral over the interval \([-2, 1]\).

Step 2: Set up the Definite Integral

The definite integral to find the area under the curve is given by \( \int_{-2}^{1} (2 - x - x^2) \, dx \). This integral will give us the total area between the curve and the x-axis from \( x = -2 \) to \( x = 1 \).

Step 3: Integrate the Function

Integrate the function \( 2 - x - x^2 \) with respect to \( x \). The integral of \( 2 \) is \( 2x \), the integral of \( -x \) is \( -\frac{x^2}{2} \), and the integral of \( -x^2 \) is \( -\frac{x^3}{3} \). Thus, the antiderivative is \( 2x - \frac{x^2}{2} - \frac{x^3}{3} \).

Step 4: Evaluate the Integral

Evaluate the antiderivative from \(-2\) to \(1\). First, substitute \( x = 1 \) into the antiderivative: \( 2(1) - \frac{(1)^2}{2} - \frac{(1)^3}{3} = 2 - \frac{1}{2} - \frac{1}{3} = \frac{12}{6} - \frac{3}{6} - \frac{2}{6} = \frac{7}{6} \). Then, substitute \( x = -2 \): \( 2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} = -4 - 2 + \frac{8}{3} \). Simplify: \( -6 + \frac{8}{3} = -\frac{18}{3} + \frac{8}{3} = -\frac{10}{3} \).

Step 5: Calculate the Area

Subtract the evaluated antiderivative at \(-2\) from the evaluated antiderivative at \(1\): \( \frac{7}{6} - ( -\frac{10}{3} ) = \frac{7}{6} + \frac{10}{3} \). To add, convert to a common denominator: \( \frac{10}{3} = \frac{20}{6} \). Then \( \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2} \).

Step 6: Conclusion

The area under the curve of \( y = 2 - x - x^2 \) from \( x = -2 \) to \( x = 1 \) is \( \frac{9}{2} \) or 4.5 square units.

Key Concepts

Area Under Curve

When we talk about the "Area Under Curve," we're essentially considering a slice of space captured between a function graph and the x-axis, within certain bounds or limits on the x-axis. Visualizing this can act as a simple bridge to understanding definite integrals—which help us find this area.

A definite integral, often represented as \( \int_{a}^{b} f(x) \, dx \), captures this area calculation. Here, \( f(x) \) is your function, and \( a \) and \( b \) highlight the interval over which you’re interested in measuring the area. This mathematical tool measures how much space is bounded by the curve over that interval, and it's especially useful for functions not simply lying above the x-axis.

Finding the area in our specific case means dealing with the curve created by \( y = 2 - x - x^2 \) from \( x = -2 \) to \( x = 1 \). You'll take into account the ways the curve dips below or rises above the x-axis, adding or subtracting these values as necessary via integration.

Integration Techniques

Integrating a function involves processes that help find the area under a curve, and different techniques exist for this. Understanding which technique to apply comes with practice and knowledge of the function involved. In our scenario of \( y = 2 - x - x^2 \), basic polynomial integration is utilized.

The integral of a constant, like 2, is quite straightforward; it's simply \( 2x \). For terms like \( -x \), the integration becomes \( -\frac{x^2}{2} \); we use the formula \( x^n \rightarrow \frac{x^{n+1}}{n+1} \). Lastly, for \( -x^2 \), we compute \( -\frac{x^3}{3} \).

• Understanding power rule basics is essential, as it helps tackle polynomial terms efficiently.
• Expressing integrated functions in reduced form helps when it comes to evaluating these at boundaries.

Evaluating these techniques correctly means keeping track of negative signs and staying clear of common pitfalls while finding the complete antiderivative.

Function Evaluation

Once you've identified the ways to integrate the given function, the process of "Function Evaluation" is used to finalize the solution of a definite integral. This involves calculating the antiderivative expression you got from integration, at both the upper boundary and the lower boundary you've set.

In the example \( y = 2 - x - x^2 \), we've integrated it to get \( 2x - \frac{x^2}{2} - \frac{x^3}{3} \). Now, you plug in the boundary values— here, \( x = 1 \) and \( x = -2 \) — and then subtract these results from each other.

• First, replace \( x \) with the upper limit of your integral.
• Similarly, put in the lower limit and calculate the output.
• Compare these to find the overall area render.

The key to accurate function evaluation involves careful arithmetic handling and ensuring the respect of order, particularly while handling negative areas below the axis.