1.设L为从(0,0)点沿曲线y=x^2到点(1,1),则∫L2xydx+(y^2+1)dy等于?2.设f（x）在（0,+无穷）上连续,且∫(0,x^2(1+x))f(t)dt=x,则f(2)等于什么

Question

1.设L为从(0,0)点沿曲线y=x^2到点(1,1),则∫L2xydx+(y^2+1)dy等于?
2.设f（x）在（0,+无穷）上连续,且∫(0,x^2(1+x))f(t)dt=x,则f(2) 等于什么

Answer 1

1、设L是从(0,0)点沿曲线y = x²到点(1,1),则∫_L (2xy) dx + (y² + 1) dy =
dy = 2xdx
∫_L (2xy) dx + (y² + 1) dy
= ∫(0~1) 2x(x²) dx + (x⁴ + 1) 2xdx
= ∫(0~1) (2x³ + 2x⁵ + 2x) dx
= 2 * x⁴/4 + 2 * x⁶/6 + x² |(0~1)
= 1/2 + 1/3 + 1
= 11/6
2、设f(x)在(0,+∞)上连续,且∫(0~x²(1 + x)) f(t) dt = x,则f(2) =
∫(0~x²(1 + x)) f(t) dt = x,两边求导
[d/dx x²(1 + x)] * f[x²(1 + x)] = 1
(2x + 3x²) * f[x²(1 + x)] = 1
令x = 1,
(2 + 3)f(1 + 1) = 1
==> f(2) = 1/5