Question #6978d

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Question #6978d

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Answer 1 · 2018-01-04T15:50:21

$\sqrt{3} x + C$ $sqrt(3)x+C$

Explanation:

$\int \sqrt{4 - \frac{x^{2}}{x^{2}}} d x$ $int sqrt(4-x^2/x^2)dx$ can be simplified to....

$\int \sqrt{4 - 1} d x$ $int sqrt(4-1)dx$

= $\int \sqrt{3} d x$ $int sqrt(3)dx$

Now, remember the rule that : $\int b x d x = b \int x d x$ $int bx dx=bintx dx$
Therfore, $\int \sqrt{3} d x = \sqrt{3} \int 1 d x$ $int sqrt(3)dx=sqrt3int1dx$
Using the anti-power rule, we know that: $\int 1 = \frac{x}{1} \Rightarrow x$ $int 1=(x)/1 =>x$
Now, we multiply this by $\sqrt{3}$ $sqrt3$ to get $\sqrt{3} x$ $sqrt3x$

Now, since constants do not affect the derivication process, we just turn $\sqrt{3} x$ $sqrt3x$ to $\sqrt{3} x + C$ $sqrt(3)x+C$

P.S. Don't forget the constant due to no limits!!!

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Question #6978d

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