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16 - x^2
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(16-x^2)-1 x^2-16 x=4 and x=-4
Use the limit process for integral {1,3] to evaluate
I don't know how this website works, I'm sorry.
\(\large\color{black}{x^2-16\color{red}{=0} }\) I am correcting your little mistake with red.
I mean the other way, \(\large\color{black}{ 16-x^2=0 }\) but that doesn't matter, because x-solutions will either way be the same.
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yes, \(\large\color{black}{ x=\pm4 }\) is correct.
I'm doing calculus haha, I need to use the limit process to find the area of the region [1,3]
What is the exact question?
you want \[\Large \int\limits_{1}^{3} 16-x^2~~dx\]right?
yes!
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