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How do you find the integral #int_0^13dx/(root3((1+2x)^2)# ?

Answer 1

By substitution, #int_0^{13}1/{root3{(1+2x)^2}}dx=3#

Let #u=1+2x#. By taking the derivative, #{du}/{dx}=2# By taking the reciprocal, #{dx}/{du}=1/2# By multiply by #du#, #dx={du}/2#

Since #x# goes from 0 to 13, #u# goes from 1 to 27. Now, we can rewrite the integral in terms of #u#. #int_0^{13}1/{root3{(1+2x)}}dx =int_1^{27}1/{root3{u^2}}{du}/2# by simplifying, #=1/2int_1^{27}u^{-2/3}du# by Power Rule, #=1/2[3u^{1/3}]_1^{27} =3/2[root3{27}-root3{1}]=3/2cdot 2=3#

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Answer 2

Isaiah Allen

To find the integral ∫₀¹₃ dx/√₃(1 + 2x)², you can use the substitution method. Let u = 1 + 2x. Then, du/dx = 2, which implies dx = du/2.

Substituting these into the integral, we have: ∫₀¹₃ dx/√₃(1 + 2x)² = ∫₀¹₃ (1/2) * du / √₃(u²)

This simplifies to: (1/2√₃) ∫₀¹₃ du/u

Now, integrate with respect to u: (1/2√₃) [ln|u|] from 1 to 7

Substitute back u = 1 + 2x: (1/2√₃) [ln|1 + 2x|] from 1 to 7

Evaluate at the upper and lower limits: (1/2√₃) [ln|1 + 14| - ln|1 + 2|]

Finally, simplify: (1/2√₃) [ln 15 - ln 3]

So, the integral evaluates to: (1/2√₃) [ln(15/3)] = (1/2√₃) ln(5)

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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.