consider the family of functions f(x) = 1 / x^p, where p is a real number. for what values of p does the…

Explanation:

Step1: Recall power - rule for integration

The antiderivative of $x^{-p}$ is $\frac{x^{-p + 1}}{-p+1}+C$ for $p\neq1$ and $\ln(x)$ for $p = 1$. We are dealing with the improper integral $\int_{0}^{1}x^{-p}dx=\lim_{a\rightarrow0^{+}}\int_{a}^{1}x^{-p}dx$.

Step2: Evaluate the integral for $p\neq1$

$\lim_{a\rightarrow0^{+}}\int_{a}^{1}x^{-p}dx=\lim_{a\rightarrow0^{+}}\left[\frac{x^{-p + 1}}{-p + 1}\right]{a}^{1}=\lim{a\rightarrow0^{+}}\left(\frac{1}{1 - p}-\frac{a^{1 - p}}{1 - p}\right)$. This limit exists when $1-p>0$ (i.e., $p<1$) since $\lim_{a\rightarrow0^{+}}a^{1 - p}=0$ for $1 - p>0$.

Step3: Evaluate the integral for $p = 1$

When $p = 1$, $\lim_{a\rightarrow0^{+}}\int_{a}^{1}x^{-1}dx=\lim_{a\rightarrow0^{+}}[\ln(x)]{a}^{1}=\lim{a\rightarrow0^{+}}(0-\ln(a))=\infty$.

Step4: Calculate the value of the integral for $p<1$

For $p<1$, $\int_{0}^{1}x^{-p}dx=\frac{1}{1 - p}-\lim_{a\rightarrow0^{+}}\frac{a^{1 - p}}{1 - p}=\frac{1}{1 - p}$.

Answer:

A. The integral exists for $p<1$. A. $\int_{0}^{1}f(x)dx=\frac{1}{1 - p}$