How to find the arc length of the curve f ( x ) = x 3 6 + 1 2 x over the interval [1,3]?

Question

How to find the arc length of the curve             f              (        x        )              =                  x        3            6        +          1              2        x             over the interval [1,3]?

abuhishcy · Accepted Answer

The arc length of a curve on the interval [a, b] is given by evaluating                     ∫        a        b                            1        +                              (                                          d                y                                            d                x                                      )                    2                            d      x      . The derivative of f&#039;(x), given by the power rule, is      f    ′          (      x      )        =          1      2              x      2        -          1              2                  x          2                      =                            x          4                -        1                    2                  x          2                    Substitute this into the above formula.                      ∫        1        3                            1        +                              (                                                            x                  4                                -                1                                            2                                  x                  2                                                      )                    2                            d      x      Expand.                      ∫        1        3                            1        +                                            x              8                        -            2                          x              4                        +            1                                4                          x              4                                                  d      x      Put on a common denominator.                      ∫        1        3                                                                x              8                        +            2                          x              4                        +            1                                4                          x              4                                                d          x                    Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form                     (        a        x        )            2      .                      ∫        1        3                                                                (                              x                4                            +              1              )                        2                                              (              2                              x                2                            )                        2                                      d      x      Eliminate the square root using                     (                  a          2                )                              1          2                      =    a                      ∫        1        3                                      x          4                +        1                    2                  x          2                            d      x      Factor out a             1      2       and put it in front of the integral.             1      2                      ∫        1        3                                      x          4                +        1                    x        2                    d      x      Separate into different fractions.              1      2                      ∫        1        3                            x        4                    x        2              +          1              x        2                    d      x      Simplify using                     a        n                    a        m              =          a              n        -        m             and             1              a        n              =          a              -        n            .              1      2                      ∫        1        3                    x      2        +          x              -        2                    d      x      Integrate using       ∫          x      n              d      x        =                  x                  n          +          1                            n        +        1            , with       n    ∈    ℝ    ,    n    ≠    -    1  .              1      2                                [                      1            3                                x            3                    -                      1            x                    ]                1        3            Evaluate using the second fundamental theorem of calculus, which states that for                     ∫        a        b              F          (      x      )        =          f              (        b        )              -          f              (        a        )            , if f(x) is continuous on [a, b] and where       f    ′          (      x      )        =    F          (      x      )      .            1      2              (              1        3                              (          3          )                3            -              1        3            -              (                  1          3                                      (            1            )                    3                -                  1          1                )            )      Combine fractions and simplify.              1      2              (      9      -              1        3            -              1        3            +      1      )                  1      2              (      10      -              2        3            )            5    -          1      3                  14      3      Hence, the arc length is             14      3       units.

How do you find the arc length of the curve f(x)=(x^3)/6+1/(2x) over the interval [1, 3]?

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