Use the fundamental theorem of calculus:Find F ′ ( x ), where F ( x ) = ∫ 0 x 3 sin 3 ⁡ t d t

Question

Use the fundamental theorem of calculus:Find       F    ′    (  x  ), where   F  (  x  )  =      ∫    0                  x        3                  sin    3    ⁡  t    d  t

Semaj Stark · Accepted Answer

F    ′    (  x  )  is                     d        F                    d        x            . By the first part of the fundamental theorem of calculus, you know that      ∫          0                      x        3                  sin          3        ⁡  (  t  )   d  t  =  G  (      x    3    )  −  G  (  0  )  ,where   G is an antiderivative of       sin          3      ; that is,       G    ′    (  x  )  =      sin          3        ⁡  (  x  ). Take the derivative of this quantity and you get                    d        F                    d        x              =            d              d        x                  [          ∫              0                              x          3                            sin              3              ⁡    (    t    )     d    t    ]    =            d              d        x                  [    G    (          x      3        )    −    G    (    0    )    ]    =      G    ′    (      x    3    )  (  3      x    2    )  −  0  =      G    ′    (      x    3    )  (  3      x    2    )  .I made use of the chain rule for derivatives and that   G  (  0  ) is a constant and thus has derivative   0. By definition,       G    ′    (  x  )  =      sin          3        ⁡  (  x  ) and thus your answer is   3      x    2        sin          3        ⁡  (      x    3    ).

Use the fundamental theorem of calculus: Find F ′ </msup> ( x )

Answered question

Answer & Explanation

New Questions in Integral Calculus