Problem 47
Question
Consider the function \(f(x)=\frac{1-x}{x}\).
a. Are there numbers \(0
Step-by-Step Solution
Verified Answer
Answer: No, there are no values of \(a\) satisfying the given conditions in either range:
a. No values within the range \(01\) satisfy \(\int_{1 / a}^{a} f(x) d x=0\).
1Step 1: Calculate the integral of the function
To solve the given integral, find the antiderivative of the given function: \(f(x) = \frac{1-x}{x}\). To do this, simplify the fraction, if possible:
\(f(x) = \frac{1}{x} - 1\)
Now, find the antiderivative:
\[\int f(x) dx = \int\frac{1}{x}dx - \int 1 dx\]
Apply the power rule for integration:
\[F(x) = \ln|x| - x + C\]
Now we have the antiderivative, \(F(x) = \ln|x| - x + C\), and can proceed to solve the two parts of the exercise.
2Step 2: Part a: Solve for 0<a<1
First, write down the definite integral that needs to satisfy the condition in Part a:
\[\int_{1-a}^{1+a} f(x) d x = 0\]
Now, apply the fundamental theorem of calculus with the antiderivative we found in Step 1:
\[F(1+a)-F(1-a) = 0\]
\[F(1+a) = F(1-a)\]
Substitute the antiderivative into the equation:
\[\ln|1+a| - (1+a) = \ln|1-a| -(1-a)\]
Now, we want to solve for \(a\) in the range \((0,1)\), if possible. Observe that for \(0
3Step 3: Part b: Solve for a>1
Write down the definite integral that needs to satisfy the condition in Part b:
\[\int_{1 / a}^{a} f(x) d x = 0\]
Apply the fundamental theorem of calculus with the antiderivative we found in Step 1:
\[F(a)-F\left(\frac{1}{a}\right) = 0\]
Substitute the antiderivative into the equation:
\[\ln|a| - a = \ln\left|\frac{1}{a}\right| - \frac{1}{a}\]
Now we want to solve for \(a\) in the range \((1,\infty)\). Notice that when \(a>1\), the term \(\frac{1}{a}\) is smaller than \(a\). Because \(\ln(a)=\ln\left(\frac{1}{a}\right)\), the linear terms become equal when \(a=1\). Thus for \(a>1\), we have:
\[\ln|a| - a < \ln\left|\frac{1}{a}\right| - \frac{1}{a}\]
As a result, there are no numbers \(a>1\) such that \(\int_{1 / a}^{a} f(x) d x = 0\). So the answer to part b is: No, there are no numbers \(a>1\) such that \(\int_{1 / a}^{a} f(x) d x=0.\)
The final answers to the exercise are:
a. No, there are no numbers \(01\) such that \(\int_{1 / a}^{a} f(x) d x=0.\)
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