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20 (x/sqrt(1+2x^2))dx

I got 1 as my final answer....but I'm not sure if thats correct. Even if it is right, would someone be able to go through this problem step by step?

 

also if anyone could help me with this following problem:

3π^2 π/2 cosx/sin^2(x)dx

 

thank you!!!

 Mar 25, 2019
edited by Guest  Mar 25, 2019
edited by Guest  Mar 25, 2019
edited by Guest  Mar 25, 2019
 #1
avatar+28159 
+3

Yes, 1 is the correct answer:

 

 

Are you sure you have the limits written correctly in your second integral?

 Mar 25, 2019
 #2
avatar+23128 
+2

integrating w substitution

\(\large{\int \limits_{0}^{2} \dfrac{x}{ \sqrt{1+2x^2} }dx}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\large{\int \limits_{x=0}^{x=2} \dfrac{x}{ \sqrt{1+2x^2} }dx}} \\ && \quad \boxed{ \text{substitute: } \\ u = 1+2x^2,\ du=4x\ dx,\\ x=0 \Rightarrow u=1+2\cdot 0^2=1,\\ x=2\Rightarrow u=1+2\cdot 2^2=9 } \\ &=& \int \limits_{u=1}^{u=9} \dfrac{x}{ \sqrt{u} }\dfrac{du}{4x} \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} \dfrac{1}{ \sqrt{u} }du \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} u^{-\frac{1}{2}} du \\\\ &=& \dfrac{1}{4} \left[ \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{4}\cdot \dfrac{2}{1} \left[ u^{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{2} \left( 9^{\frac{1}{2}}-1^{\frac{1}{2}} \right) \\\\ &=& \dfrac{1}{2} ( \sqrt{9} -1 ) \\\\ &=& \dfrac{1}{2} ( 3 -1 ) \\\\ &=& \dfrac{2}{2} \\\\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

 

laugh

 Mar 25, 2019

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