微分與不定積分求解?

1. f ' (x)=5(2x-1)^4
∫f ' (x)dx=∫5(2x-1)^4dx
f(x)=∫(5/2)(2x-1)^4d(2x-1) (因為d(2x-1)/dx=2 => dx=(1/2)d(2x-1))
f(x)=(1/2)(2x-1)^5+C (C是常數)
f(x)通過(1,3)
3=(1/2)(2-1)^5+C
C=3-1/2
C=5/2
所以, f(x)=(1/2)(2x-1)^5+5/2
f(x)=(1/2)[(2x-1)^5+5]

2. f ' (x)=x√(1-x^2)
∫f ' (x)dx=∫x√(1-x^2)dx
f(x)=∫(-1/2)√(1-x^2)d(1-x^2) (因為d(1-x^2)/dx=-2x => xdx=(-1/2)d(1-x^2))
f(x)=∫[(-1/2)(1-x^2)^(1/2)]d(1-x^2)
f(x)=(-1/2)(2/3)(1-x^2)^(3/2)]+C (C是常數)
f(x)=(-1/3)(1-x^2)^(3/2)+C
f(x)通過(0,4/3)
4/3=(-1/3)(1-0^2)^(3/2)+C
4/3=(-1/3)+C
C=(4+1)/3=5/3
所以,f(x)=(-1/3)(1-x^2)^(3/2)+5/3
f(x)=(-1/3)[(1-x^2)^(3/2)-5]