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Mathematics 17 Online
OpenStudy (anonymous):

URGENT PLEASE HELP! ATTACHED CALC 2 QUESTIONS. please please please hw due tonight!

OpenStudy (anonymous):

OpenStudy (anonymous):

need help with #2

hero (hero):

Did you figure it out yet?

OpenStudy (anonymous):

noo i need help!

hero (hero):

\[\lim_{a \rightarrow -2} \int\limits_{a}^{0}\left( \frac{1}{\sqrt{x+2}} \right) dx\]

hero (hero):

Sketching the graph couldn't possibly be that difficult

hero (hero):

Especally if you write the function as \[ y = \frac{1}{\sqrt{x+2}} \]

OpenStudy (anonymous):

can you show me?

hero (hero):

Show you what exactly?

hero (hero):

I just gave you two hints.

OpenStudy (anonymous):

the area of s

OpenStudy (anonymous):

i have like no idea what to do

hero (hero):

Find the integral of the function first. Let me know what you get.

OpenStudy (anonymous):

im getting stuck, i am really lost on this

OpenStudy (anonymous):

if you can show me step by step i could really use the help

hero (hero):

What if I re-wrote it like this: \[\lim_{a \rightarrow -2} \int\limits_{a}^{0}\left( (\sqrt{x+2})^{-1}\right) dx\] Would you know how to integrate it now?

hero (hero):

By the way, we cannot use -2 as the lower limit because \[\frac{1}{\sqrt{-2+2}} = \frac{1}{0} = \text{undefined}\]

OpenStudy (anonymous):

can you show me?

hero (hero):

Okay, what if I re-wrote it like this: \[\lim_{a \rightarrow -2} \int\limits_{a}^{0}\left( ({x+2})^{\frac{1}{2}}\right)^{-1} dx\] Does that help?

hero (hero):

I already showed you why We cannot use -2 as the lower limit. But if we take the integral of the function, we can then apply the limits of the function and find the area

OpenStudy (anonymous):

The integral is?

OpenStudy (anonymous):

Please can you show me the answer i have to hand this in . Please please please

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