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How do you integrate #{x/(sqrt(4+4x^2))} dx# from 0 to 2?

Answer 1

Christopher Agresta

The answer is #=1/2(sqrt5-1)#

We need #intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Let #u=4+4x^2#

#du=8xdx#,

#xdx=1/8du#

Therefore,

#int(xdx)/sqrt(4+4x^2)=1/8int(du)/sqrtu=1/8u^(-1/2+1)/(1/2)#

#=1/4sqrtu=1/4sqrt(4+4x^2)=1/2sqrt(1+x^2)#

so,

#int_0^2(xdx)/sqrt(4+4x^2)=[1/2sqrt(1+x^2)]_0^2#

#=(sqrt5/2)-(1/2)#

#=1/2(sqrt5-1)#

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Answer 2

Aurora Alvarado

To integrate ( \frac{x}{\sqrt{4+4x^2}} ) with respect to (x) from 0 to 2, you can use the substitution method. Let ( u = 4 + 4x^2 ). Then, ( du = 8x , dx ). Rearranging, we get ( \frac{1}{8} du = x , dx ). Substituting (u) and (du), the integral becomes ( \int \frac{1}{8} \frac{1}{\sqrt{u}} , du ) from 4 to 20. This simplifies to ( \frac{1}{8} \int u^{-1/2} , du ) from 4 to 20. Integrating ( u^{-1/2} ) yields ( 2u^{1/2} ). Evaluating this from 4 to 20 and then multiplying by ( \frac{1}{8} ) gives ( \frac{1}{8} (2\sqrt{20} - 2\sqrt{4}) ), which simplifies to ( \frac{1}{4} (\sqrt{20} - 2) ). Hence, the definite integral from 0 to 2 is ( \frac{1}{4} (\sqrt{20} - 2) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.