Ap微积分考试真题

=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 sin 2x |(0，5)

= 26.5π-4π×0.2837π/4(sin 10-sin 0)

=26.5π-1.1348π-π/4×(-0.5440)

=25.3652π+0.136π

=25.5012π

=80.07

选择e