被積分関数は偶関数だから， S=2∫[0,√(2+√5)](-x⁴+4x²+1)dx =-2[x⁵/5-4x³/3-x]_[0,√(2+√5)] =-(2/15)[x(3x⁴-20x²-15)]_[0,√(2+√5)] =-(2/15)√(2+√5){3(2+√5)²-20(2+√5)-15} =-(2/15)√(2+√5){(27+12√5)-(40+20√5)-15} =(2/15)√(2+√5)(28+8√5) =(8/15)(7+2√5)√(2+√5)