Preguntas del examen de cálculo Ap

=π∫(0,5)(4 4sinx sin^2x)dx

=π(4x- 4cosx)|(0,5) π/2∫(0,5)(1-cos2x)dx

=4π[(5-0)-(cos5-cos0)] π/2x|( 0,5)-π/4∫(0,5)cos2xd(2x)

= 20π-4πcos 5 4π π/2(5-0)-π/4 sen 2x |(0,5 )

= 26.5π-4π×0.2837π/4(sen 10-sen 0)

=26.5π-1.1348π-π/4×(-0.5440)

=25.3652π 0.136π

=25.5012π

=80.07

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