206.08.08.10 integral from --\infty

[karush]

\tiny{206.08.08.10 }

\begin{align*}
\displaystyle
I&= \int_{-\infty}^{0}\frac{dx}{(x+2)^{1/3}}\\
&=-\infty\\
\end{align*}

why does this go to $-\infty$

 

[Greg]

What have you tried? The improper integral does evaluate to $-\infty$.

 

[karush]

What have you tried? The improper integral does evaluate to $-\infty$.

my guess was just observation

how would $-\infty$ be plugged into the equation

and what would happen if the the upper limit would be greater than zero?

would the answer be the same no matter what the expression would be?

 

[skeeter]

\(\displaystyle I = \lim_{a \to -\infty} \int_a^{-10} \dfrac{dx}{(x+2)^{1/3}} + \lim_{b \to -2^-} \int_{-10}^b \dfrac{dx}{(x+2)^{1/3}} + \lim_{c \to -2^+} \int_c^0 \dfrac{dx}{(x+2)^{1/3}}\)

\(\displaystyle I = \dfrac{3}{2} \lim_{a \to -\infty} \bigg[(x+2)^{2/3}\bigg]_a^{-10} + \dfrac{3}{2} \lim_{b \to -2^-} \bigg[(x+2)^{2/3}\bigg]_{-10}^b + \dfrac{3}{2} \lim_{c \to -2^+} \bigg[(x+2)^{2/3}\bigg]_c^0
\)

\(\displaystyle I = \dfrac{3}{2} \lim_{a \to -\infty} \bigg[4 - (a+2)^{2/3} \bigg] + \dfrac{3}{2} \lim_{b \to -2^-} \bigg[(b+2)^{2/3}-4 \bigg] + \dfrac{3}{2} \lim_{c \to -2^+} \bigg[2^{2/3} - (c+2)^{2/3} \bigg]\)

\(\displaystyle I = \dfrac{3}{2} \lim_{a \to -\infty} \bigg[4 - (a+2)^{2/3} \bigg] + (-6) + \dfrac{3}{2^{1/3}}\)

... what happens with that first limit?

 

[karush]

as larger negative values are input the eq goes to $-\infty$