Question

第三题。大一高等数学定积分。

Answer 1

3. f(x)+(sinx)^4 = ∫<0,π/4>f(2x)dx
= (1/2)∫<0,π/4>f(2x)d(2x) = (1/2)∫<0,π/2>f(u)du
记∫<0,π/2>f(x)dx = A, 则 f(x)+(sinx)^4 = A/2，
两边在 [0,π/2] 上对 x 积分，得
A + ∫<0,π/2>(sinx)^4dx = (A/2)(π/2)
得 A = -4/(4-π)∫<0,π/2>(sinx)^4dx
= -1/(4-π)∫<0,π/2>(1-cos2x)^2dx
= -1/(4-π)∫<0,π/2>[1-2cos2x+(cos2x)^2]dx
= -1/(4-π)∫<0,π/2>[3/2-2cos2x+(1/2)cos4x]dx
= -1/(4-π)[3x/2-sin2x+(1/8)sin2x]<0,π/2> = -3π/[4(4-π)].
即 ∫<0,π/2>f(x)dx = -3π/[4(4-π)].