91Ó°ÊÓ

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.

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 \(0<a<1\), the logarithm will be smaller on the left side than on the right side, and the linear part will be more significant as \(a\) approaches 1. Hence, there is no solution in this range. So the answer to part a is: No, there are no numbers \(0<a<1\) such that \(\int_{1-a}^{1+a} f(x) d x=0.\)

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 \(\int_{1 / a}^{a} f(x) d x=0.\)