Proving <msubsup> &#x222B;<!-- ∫ --> 0 1 </msubsup> sin

Wade Bullock 2022-07-01 Answered
Proving 0 1 sin π x x 2 + 1 d x = 2 π ( μ 1 2 + 1 ) = π 4 sin π μ 2 for certain μ 1 , μ 2 [ 0 , 1 ]
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Answers (1)

persstemc1
Answered 2022-07-02 Author has 18 answers
Hint: Notice that 0 1 sin π t d t = 2 π and 0 1 1 1 + t 2 d t = π 4 . The generalized mean value theorem states that for f,g continuous on an interval [a,b] and differentiable on (a,b), there exists c ( a , b ) such that
( f ( b ) f ( a ) ) g ( c ) = ( g ( b ) g ( a ) ) f ( c ) .
Apply this theorem for
f ( x ) = 0 x sin π t t 2 + 1 d t
and
g ( x ) = 0 x sin π t d t or g ( x ) = 0 x 1 t 2 + 1 d t .

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