Answer
\[ = \frac{1}{{1 - p}}\,\,\,for\,\,p < 1\]
Work Step by Step
\[\begin{gathered}
\int_0^1 {f\,\left( x \right)} \,dx \hfill \\
\hfill \\
\,find\,\,value\,\,of\,\,p\,\,for\,\,which \hfill \\
\hfill \\
\int_0^1 {f\,\left( x \right)} \,dx{\text{ }}exists \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_0^1 {f\,\left( x \right)} \,dx\,\, = \mathop {\lim }\limits_{a \to 0} \,\,\left[ {\frac{{{x^{ - p + 1}}}}{{1 - p}}} \right]_a^1 \hfill \\
\hfill \\
use\,\,the\,\,ftc \hfill \\
\hfill \\
= \mathop {\lim }\limits_{a \to 0} \,\,\left[ {\frac{1}{{1 - p}} - \frac{{{a^{ - p + 1}}}}{{1 - p}}} \right] \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \frac{1}{{1 - p}}\,\,\,for\,\,p < 1 \hfill \\
\end{gathered} \]
