Respuesta :

LammettHash
The path starts at [tex]t=1[/tex] and ends at [tex]t=6[/tex], so we have an arc length of


[tex]\displaystyle\int_{\mathcal C}\mathrm dS=\int_{t=1}^{t=6}\|\mathbf c'(t)\|\,\mathrm dt[/tex]
[tex]=\displaystyle\int_1^6\sqrt{2^2+(2t)^2+\left(\frac1t\right)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_1^6\sqrt{4t^2+4+\frac1{t^2}}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_1^6\frac1t\sqrt{4t^4+4t^2+1}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_1^6\frac1t\sqrt{(2t^2+1)^2}\,\mathrm dt[/tex]
[tex]=\displaystyle\int_1^6\frac{2t^2+1}t\,\mathrm dt[/tex]
[tex]=\displaystyle\int_1^6\left(2t+\frac1t\right)\,\mathrm dt[/tex]
[tex]=t^2+\log t\bigg|_{t=1}^{t=6}[/tex]
[tex]=(6^2+\log6)-(1^2+\log1)=35+\log6[/tex]
shubhamchouhanvrVT

The integration will be equal to 35  +  Log6.

What is integration?

Integration is defined as the addition or summation of the small split parts to form a whole or a total part.

The solution will be given as:-

The path starts at t = 1 and ends at t = 6, so we have an arc length of

[tex]\int_cdS =\int_{t=1}^{t=6}c'(t)dt\\\\\\=\int_1^6\sqrt{2^2+(2t)^2+\dfrac{1}{t}^2dt}\\\\\\=\int_1^6\sqrt{4t^2+4+(\dfrac{1}{t^2})dt}\\\\\\=\int_1^6\dfrac{1}{t}\sqrt{(2t^2+1)^2dt}\\\\\\=\int_1^6(2t+\dfrac{1}{t})dt\\\\\\=t^2+Logt\\[/tex]

=  ( 6²  +  Log6  ) - (  1²  +  Log1)  =  35  +  Log6

Therefore integration will be equal to 35  +  Log6.

To know more about an integration follow

https://brainly.com/question/20156869

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