Problem 76 - Thomas Calculus

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: Understanding the Problem

We need to find the area between the curve \(y=2^{1-x}\) and the \(x\)-axis within the interval \(-1 \leq x \leq 1\). This involves finding the area under the curve from \(x = -1\) to \(x = 1\).

Step 2: Setting Up the Integral

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

Step 3: Finding the Antiderivative

To solve the integral, we must first find the antiderivative of \(2^{1-x}\). Rewrite the function as \(2 \cdot 2^{-x}\) and use the substitution \(u = -x\), so \(du = -dx\). The integral becomes: \[ \int -2 \cdot 2^{u} \, du = -2 \cdot \frac{2^{u}}{\ln(2)} + C. \] Substituting back gives the antiderivative as \[-\frac{2^{1-x}}{\ln(2)} + C.\]

Step 4: Evaluating the Definite Integral

Now, we evaluate the definite integral from \(x = -1\) to \(x = 1\) using the antiderivative: \[ \left[ -\frac{2^{1-x}}{\ln(2)} \right]_{-1}^{1}. \] This gives: \(-\frac{2^{1-1}}{\ln(2)}\) evaluated at \(x = 1\) minus \(-\frac{2^{1+1}}{\ln(2)}\) evaluated at \(x = -1\).

Step 5: Calculating the Result

Let's compute the values: \(-\frac{2^{0}}{\ln(2)} = -\frac{1}{\ln(2)}\) and \(-\frac{2^{2}}{\ln(2)} = -\frac{4}{\ln(2)}\). The area is: \[ -\frac{1}{\ln(2)} - (-\frac{4}{\ln(2)}) = \frac{4}{\ln(2)} - \frac{1}{\ln(2)} = \frac{3}{\ln(2)}. \]

Step 6: Conclusion

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

Key Concepts

Definite Integrals

In integral calculus, definite integrals are a fundamental concept used to calculate the total accumulation of quantities, such as areas under curves. Unlike indefinite integrals, which provide a family of functions (antiderivatives), definite integrals yield a specific numerical value. This value represents the total accumulation of the function within a given interval. For example, when computing the area under the curve of the function \(y=2^{1-x}\) from \(x = -1\) to \(x = 1\), we use the definite integral notation:

• \(\int_{-1}^{1} 2^{1-x} \, dx\)

Once the antiderivative is found, the actual computation involves subtracting the value of the antiderivative at the lower limit of the interval from the value at the upper limit. By performing definite integration, we get a precise value, \(\frac{3}{\ln(2)}\), which represents the area between the curve and the x-axis over the specified interval.

Area Under a Curve

The area under a curve in the context of integral calculus often refers to the region bounded by the graph of a function, the x-axis, and vertical lines corresponding to the limits of integration. To find the area under the curve of a function like \(y=2^{1-x}\) over a particular interval, we can make use of integration. This involves calculating a definite integral. In more intuitive terms, this area is like cutting out a piece of paper shaped by the curve above the x-axis. The integral allows us to compute this area precisely, even if the curve is not a simple geometric shape. Calculating it means finding the sum of infinite tiny strips across the curve:

• Each strip is represented by a small rectangle underneath the curve, with height given by the function value and infinitesimal width \(dx\).
• This process captures the curve's continuous nature over the specified interval.

Thus, the area between the curve \(y = 2^{1-x}\) and the x-axis from \(x = -1\) to \(x = 1\) is determined to be \(\frac{3}{\ln(2)}\), as computed by integrating the function across the interval.

Antiderivatives

An antiderivative is a function that reverses the process of differentiation. For a given function \(f(x)\), an antiderivative \(F(x)\) satisfies the property that when differentiated, it returns the original function:

• \(F'(x) = f(x)\)

Finding antiderivatives is essential for solving definite integrals, as it allows us to express the total accumulation described by integration.In our example, to integrate \(2^{1-x}\), we first find its antiderivative. By rewriting the function and using substitution, we determine:

• \(-\frac{2^{1-x}}{\ln(2)} + C\)

The constant \(C\) is typically not needed for definite integrals, because the integration process, which involves evaluating bounds, effectively cancels it out. Ultimately, understanding antiderivatives helps us unlock the method to "undo" differentiation, illuminating the consistent area calculations under curves.