Question

1)Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative. Remember to use ln |u| where appropriate.)

f(x) = (1/5)−(3/x)

-----> (x/5)-3lnx+C

2)Find f.
f ''(x) = 4 + 6x + 36x^2, f(0) = 2, f (1) = 10

----> 2x^2+x^3+3x^4+8x+2


A particle is moving with the given data. Find the position of the particle.
v(t) = 1.5(square root of(t)), s(4) = 17

----> s(t)=t^(3/2)+1

Answer 1

first: correct, also x/5)- ln(x^3)+C

second 2x^2+x^3+3x^4+8x+Cx + K
f(o)=2=k or K=2
f(1)=10=2+1+3+8+C +2 means C=10-16=-6
>>>f(x)=2x^2+x^3+3x^4+8x -6x + 2
f'= 4x + 3x^2 +12x^3 + 2
f"= 4+6x + 36x^2

Answer 2

how about the third one!

Answer 3

v = 3/2 √t
s = t^(3/2)+C
8+C = 17, so C=9

s(t) = t^(3/2) + 9

How did you get C=1?