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About LinkBackscan someone help me solve
integral from 0 to 4 of
(x)/ (square root of x^2+9) dx
Have you made any attempt at it yourself? This is a pretty straight-forward problem and im sure you've covered problems similar to this in school or uniOriginally Posted by cummings15
If ur really stuck, Hint: substitute u = x^2+9 and hence get a substitution for dx
No, that's not quite correct, you should have
Finding the anti-derivative is the main problem, you have two choices as regards the limits of the integral: you can transform the limits along with the integrand:what do i do with the 0 to 4 part?
or you can do a back-substitution before you apply the old limits:
![\int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(4)}\frac{du}{\sqrt{u}} \, du = \Big[\sqrt{u}\Big]_{u=u(0)}^{u(4)} = \Big[\sqrt{x^2+9}\Big]_{x=0}^4=\ldots = 2](http://latex.codecogs.com/png.latex?\int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(4)}\frac{du}{\sqrt{u}} \, du = \Big[\sqrt{u}\Big]_{u=u(0)}^{u(4)} = \Big[\sqrt{x^2+9}\Big]_{x=0}^4=\ldots = 2)
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