$f(x)=x^a$ Definite Integral

Question

Consider $f(x)=x^a $

Now $\int_0^1 x^a = 1/(1+a)$ gives the area bounded by the function, $x $ axis, $x=0$ and $x=1$.

Now consider $a<-1$

On LHS the function is positive for all $0<x<1$

Still the area bounded by the function comes out to be negative? How?

This formula is given in my book and is proved by considering that differentiation of $\frac {x^{1+a}}{a+1} $ is $ x^a $

What am I doing wrong?

Please Explain.

Answer 1

The formula $$\int_0^1 x^adx=\frac{1}{1+a}$$ is based on the assumption that $a>-1$. Otherwise, the integral diverges, $$\int_0^1x^adx=\infty,\quad\text{for }a\leq -1.$$

Answer 2

For $\;a<-1\;$ as you want, the function $\;x^a\;$ isn't defined at $\;x=0\;$ nor bounded on a right nieghborhood of zero, and then integral $\;\int_0^1 x^a\;dx\;$ becomes an improper non-convergent integral