find area bounded by 2(x^2+1)^(1/2), x-axis, x = 1
- Thread starter kika09
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Is this what you mean?.
You didn't say it was bounded by the y-axis, though.
\(\displaystyle \L\\2\int_{0}^{1}\sqrt{x^{2}+1}dx\)
To integrate, one way is to use \(\displaystyle \L\\x=tan({\theta}), \;\ dx=sec^{2}({\theta})d{\theta}\)
Don't forget to change your limits of integration.
This gives \(\displaystyle \L\\\int_{0}^{\frac{\pi}{4}}sec^{3}({\theta})d{\theta}\)
Now, use \(\displaystyle \L\\\int{sec^{n}(x)}dx=\frac{sec^{n-2}(x)tan(x)}{n-1}+\frac{n-2}{n-1}\int{sec^{n-2}(x)}dx\)
I used x instead of theta in the formula, for sake of typing. That's neither here nor there. Use what you need to.
You didn't say it was bounded by the y-axis, though.
\(\displaystyle \L\\2\int_{0}^{1}\sqrt{x^{2}+1}dx\)
To integrate, one way is to use \(\displaystyle \L\\x=tan({\theta}), \;\ dx=sec^{2}({\theta})d{\theta}\)
Don't forget to change your limits of integration.
This gives \(\displaystyle \L\\\int_{0}^{\frac{\pi}{4}}sec^{3}({\theta})d{\theta}\)
Now, use \(\displaystyle \L\\\int{sec^{n}(x)}dx=\frac{sec^{n-2}(x)tan(x)}{n-1}+\frac{n-2}{n-1}\int{sec^{n-2}(x)}dx\)
I used x instead of theta in the formula, for sake of typing. That's neither here nor there. Use what you need to.
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You are going to have to try a little harder. You are not seeing that the region you are describing is NOT bounded. If you have a picture in front of you, and it is bounded, then you are not describing it sufficiently. Please provide the entire problem statement. galactus gave it a try, including the limiting y-axis. Since you have declined this description, perhaps you can check for typos and take another shot at the ENTIRE problem statement. If you are not understanding the problem, you should not try to decide which parts of the problem are needed. Just provide ALL of it.
pka
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Look at the graph above.kika09 said:
Is there a region bounded by the x axis, and the line x=1?
No there is not. So there must be a typo somewhere, right?
Please review what you posted. Is there an error, maybe a sign error somewhere in it?
You do know that textbooks contain errors!
