Chapter 7: Problem 76
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.
Short Answer
Expert verified
The area is \(\frac{3}{\ln(2)}\).
Step by step solution
01
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 \(x = 1\).
02
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 \[ \int_{-1}^{1} 2^{1-x} \, dx. \]
03
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 \[-\frac{2^{1-x}}{\ln(2)} + C.\]
04
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 \(x = -1\).
05
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: \[ -\frac{1}{\ln(2)} - (-\frac{4}{\ln(2)}) = \frac{4}{\ln(2)} - \frac{1}{\ln(2)} = \frac{3}{\ln(2)}. \]
06
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 \(\frac{3}{\ln(2)}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \(\int_{-1}^{1} 2^{1-x} \, dx\)
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.
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:
- \(F'(x) = f(x)\)
- \(-\frac{2^{1-x}}{\ln(2)} + C\)