- #1

- 1

- 0

x = sqrt(t)

y = 6t - 2

Interval from 0 to 5 inclusive.

Whenever I do this, I get a long answer with big numbers in the numerator all divided by 48. Can someone walk me through the steps? THanks.

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- Thread starter wh00ps
- Start date

- #1

- 1

- 0

x = sqrt(t)

y = 6t - 2

Interval from 0 to 5 inclusive.

Whenever I do this, I get a long answer with big numbers in the numerator all divided by 48. Can someone walk me through the steps? THanks.

- #2

mathman

Science Advisor

- 7,942

- 496

- #3

- 2

- 0

[tex]L=\int\limits_0^{\sqrt 5 } {\sqrt {1 + 144x^2 } dx} [/tex] ,after [tex]x = \frac{1}{{12}}\tan \theta [/tex] substution

[tex]\frac{1}{{12}}\int {\sec ^3 \theta d\theta } = \frac{1}{{24}}(\ln \left| {\tan \theta + \sec \theta } \right| + \tan \theta \sec \theta )

\][/tex]by using integration by parts

[tex]\[

\left. {\frac{1}{{24}}(\ln \left| {12x + \sqrt {1 + 144x^2 } } \right| + 12x\sqrt {1 + 144x^2 } )} \right|_0^{\sqrt 5 }

\]

[/tex]

[tex]\[

L = \frac{1}{{24}}(\ln \left| {12\sqrt 5 + \sqrt {721} } \right| + 12\sqrt 5 \sqrt {721} )

\][/tex]

approximately L=30.19 units

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