Problem 138 - Thomas Calculus in SI Units

Question

Find the area of the region between the curve \(y=2^{1-x}\) and the interval \(-1 \leq x \leq 1\) of the \(x\) -axis.

Step-by-Step Solution

Answer

The area is \(\frac{3}{\ln 2}\).

Step 1: Identify the Area to be Calculated

The area to be calculated is between the curve \(y = 2^{1-x}\) and the x-axis from \(x = -1\) to \(x = 1\). We need to integrate the function within these bounds.

Step 2: Set Up the Definite Integral

To find the area between the curve and the x-axis, set up the definite integral of \(y = 2^{1-x}\) from \(-1\) to \(1\). The integral is: \[ \int_{-1}^{1} 2^{1-x} \, dx \]

Step 3: Integrate the Function

To integrate \(2^{1-x}\), let \(u = 1-x\), then \(du = -dx\). The limits of integration change to \(u = 2\) when \(x = -1\) and \(u = 0\) when \(x = 1\). The integral becomes: \[ -\int_{2}^{0} 2^{u} \, du \]

Step 4: Calculate the Antiderivative

The antiderivative of \(2^{u}\) is \(\frac{2^{u}}{\ln 2}\). So, evaluate: \[ -\int 2^{u} \, du = -\left[ \frac{2^{u}}{\ln 2} \right]_{2}^{0} \]

Step 5: Evaluate the Definite Integral

Substitute the limits into the antiderivative: \[ -\left( \frac{2^{0}}{\ln 2} - \frac{2^{2}}{\ln 2} \right) = -\left( \frac{1}{\ln 2} - \frac{4}{\ln 2} \right) = - \left( -\frac{3}{\ln 2} \right) = \frac{3}{\ln 2} \]

Step 6: Calculate the Final Result

Thus, the area between the curve \(y = 2^{1-x}\) and the x-axis from \(x = -1\) to \(x = 1\) is \(\frac{3}{\ln 2}\).

Key Concepts

Understanding the Area Under Curve

When tasked with finding the area under a curve, you're essentially looking at the amount of space between a graph of a function and the x-axis over a given interval. For the function in question, the area lies between the curve of the function \(y = 2^{1-x}\) and the x-axis, from \(x = -1\) to \(x = 1\). To calculate this, you integrate the function over the specified interval. This gives you the total accumulation of the function's values between these bounds. In simple terms, integration sums up tiny slices of the area to provide the total area under the curve.

Exploring Integration Techniques

Integration is a powerful mathematical tool that lets us find areas, among other things. For the specific function \(y = 2^{1-x}\), direct integration requires a clever technique. We use substitution to simplify it. Begin by letting \(u = 1-x\). Consequently, \(du = -dx\). By substituting \(u\) into the integral, we transform our problem:

• The integral of \(2^{1-x}\) becomes \(-\int 2^{u} \, du\).

This step, known as substitution, makes the integration process straightforward. The advantage here is it turns potentially complex integrals into easier ones, enabling the calculation of the antiderivative using more basic formulas.

Change of Limits in Integration

One essential part of using substitution in integrals is adjusting the limits of integration. Initially, the function \(2^{1-x}\) was integrated from \(x = -1\) to \(x = 1\). Once we substitute \(u = 1-x\), these limits transform as well:

• When \(x = -1\), \(u = 2\).
• When \(x = 1\), \(u = 0\).

Now, the integral \(-\int_{2}^{0} 2^{u} \, du\) captures the same area under the curve. This change simplifies our calculations and allows us to proceed smoothly with finding the antiderivative and solving the integral, ensuring we accurately evaluate the area between the defined limits.