∫1在上 0在下 (2x^2+1)^3 dx=229/35

题目

∫1在上 0在下 (2x^2+1)^3 dx=229/35

答案

∫^1_0 (2x^2+1)^3 dx
=∫^1_0 [8x^6+12x^4+6x^2+1] dx
=[8x^7/7+12x^5/5+2x^3+x]|^1_0
=8/7+12/5+2+1
=229/35