This problem has me stumped: Find the exact area of the region: Bounded by y=(1)/(sqrt(x^2+9)), y=0, x=0, x=3 I'm a little rusty on trigonometric substitutions
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Originally Posted by Nitz456 This problem has me stumped: Find the exact area of the region: Bounded by y=(1)/(sqrt(x^2+9)), y=0, x=0, x=3 I'm a little rusty on trigonometric substitutions Are you wanting: 0 = int(1/(sqrt(x^2 + 9))dx), from x = 0...3, in which case this is not possible. The integral is ln(sqrt(2) + 1), and this will never equal 0.
Originally Posted by AfterShock Are you wanting: 0 = int(1/(sqrt(x^2 + 9))dx), from x = 0...3, in which case this is not possible. The integral is ln(sqrt(2) + 1), and this will never equal 0. What I'm looking for, I'm quite sure, is the area of the integral of the equation i was given, for 0..3. i think the y=0 is just meaning that it doesnt want the integral below the x-axis.
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