How to find the arc length of the curve y=ln x over the interval [1,2]?

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Sdunkenf9cu · Accepted Answer

y    =          ln      x            y    ′    =          1      x      Arc length is given by:      L    =                  ∫        1        2                            1        +                  1                      x            2                                      d      x      Rearrange:      L    =                  ∫        1        2                                                  x            2                    +          1                    x              d      x      Multiply numerator and denominator by                               x          2                +        1            :      L    =                  ∫        1        2                                      x          2                +        1                    x                                            x              2                        +            1                                      d      x      Integration is distributive:      L    =                  ∫        1        2                    x                                    x            2                    +          1                            d      x        +                  ∫        1        2                    1              x                                            x              2                        +            1                                      d      x      Apply the substitution       x    =          tan      θ      :      L    =                            [                                                    x                2                            +              1                                ]                1        2              +    ∫                  sec        θ                    tan        θ              d    θ  Simplify:      L    =          5        -          2        +    ∫          csc      θ        d    θ  Integrate directly:      L    =          5        -          2        -          [      ln              |                  csc          θ                +                  cot          θ                |            ]      Rewrite in terms of             tan      θ       and             sec      θ      :      L    =          5        -          2        -          [      ln              |                              1            +                          sec              θ                                            tan            θ                          |            ]      Reverse the substitution:      L    =          5        -          2        -                            [                      ln                          (                                                1                  +                                                                                    x                        2                                            +                      1                                                                      x                            )                                ]                1        2            Insert the limits of integration:      L    =          5        -          2        -          ln              (                              1            +                          5                                            2                          (              1              +                              2                            )                                      )            Rearrange for clarity:      L    =          5        -          2        +          ln      2        -          ln              (                              1            +                          5                                            1            +                          2                                      )

daneetuhxxtj · Accepted Answer

The arc length of a function       y    =          f              (        x        )             over the interval [a,b] is given by       L    =                  ∫        a        b                            1        +                              (                                          d                y                                            d                x                                      )                    2                            d      x      .      y    =          ln      x        →                  d        y                    d        x              =          1      x            ∴    L    =                  ∫        1        2                            1        +                              (                          1              x                        )                    2                            d      x            L    =                  ∫        1        2                            1        +                  1                      x            2                                      d      x      I will use this solution       ∫                  1        +                  1                      x            2                                      d      x        =                            x          2                +        1              +          1      2        ln          |                                                                  x                2                            +              1                                -          1                                                                    x                2                            +              1                                +          1                    |        +    C   courtesy of mason m to find that      L    =                            x          2                +        1              +          1      2        ln          |                                                                  x                2                            +              1                                -          1                                                                    x                2                            +              1                                +          1                    |                      ]        1        2            Hence,       ∴    L    =          5        +          1      2              ln              (                                            5                        -            1                                              5                        +            1                          )              -          2        -          1      2              ln              (                                            2                        -            1                                              2                        +            1                          )              ≈    1.222

How do you find the arc length of the curve y=ln x over the interval [1, 2]?

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