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Find the area of the region bounded by the hyperbola [tex]9x^2-4y^2 = 36[/tex] and the line [tex]x = 3[/tex].
I'm thinking that I have to integrate for x, so I'll have the sum of twice the area from [tex]2[/tex] to [tex]3[/tex].
The function will be [tex] + \sqrt {\frac {9x^2-36}{4}}[/tex]
hence, the integral will be[tex] 2\int_2^3 {\sqrt {\frac {9x^2-36}{4}}}dx [/tex]
I just wanted to know if my reasoning is right
Thanks in advance
I'm thinking that I have to integrate for x, so I'll have the sum of twice the area from [tex]2[/tex] to [tex]3[/tex].
The function will be [tex] + \sqrt {\frac {9x^2-36}{4}}[/tex]
hence, the integral will be[tex] 2\int_2^3 {\sqrt {\frac {9x^2-36}{4}}}dx [/tex]
I just wanted to know if my reasoning is right
Thanks in advance
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