Find f, f ''(x) = 20x3 + 12x2 + 10, f(0) = 3, f(1) = 5?

f"(x)=20x^3+12x^2+10, f(0)=3; f(1)=5

=>

f '(x)=5x^4+4x^3+10x+C1

f(x)=x^5+x^4+5x^2+C1x+C2

f(0)=3

=>

C2=3

Thus,

f(x)=x^5+x^4+5x^2+C1x+3

f(1)=5

=>

1+1+5+C1+3=5

=>

C1=-5

Thus

f(x)=x^5+x^4+5x^2-5x+3

is final.

f ''(x) = 20x3 + 12x2 + 10

f(x) = c_2 x + c_1 + x^5 + x^4 + 5 x^2

f(x) = x^5 + x^4 + 5 x^2 - 5 x + 3

f(0) = 3, f(1) = 5

The idea here is to integrate twice. This will give rise to two arbitrary constants. These constants can then be found via your two conditions.

Hopefully no one will spoil you the answer. That would be very irresponsible of them. I am still amazed that there are so many irresponsible people here! 

Presuming that is:

f"(x) = 20x³ + 12x² + 10

You want to get the anti-derivative of this twice.  Each time will add a constant term so we will have two new unknowns.  We now have:

f'(x) = 5x⁴ + 4x³ + 10x + c

And the next anti-derivative:

f(x) = x⁵ + x⁴ + 5x² + cx + k

You are now given two values of f(x):

f(0) = 3 and f(1) = 5

If we substitute those into the above equation we'll have a system of two equations and two unknowns that you can solve:

f(0) = 0⁵ + 0⁴ + 5(0)² + c(0) + k and f(1) = 1⁵ + 1⁴ + 5(1)² + c(1) + k

3 = 0 + 0 + 5(0) + 0 + k and 5 = 1 + 1 + 5(1) + c + k

3 = 0 + 0 + 0 + 0 + k and 5 = 1 + 1 + 5 + c + k

3 = k and 5 = 7 + c + k

3 = k and -2 = c + k

We have a value for k, now we can solve for c:

-2 = c + k

-2 = c + 3

-5 = c

Putting these back into your function, your answer is:

f(x) = x⁵ + x⁴ + 5x² + cx + k

f(x) = x⁵ + x⁴ + 5x² - 5x + 3

f ''(x) = 20x3 + 12x2 + 10

f ''(x) = 60x + 24x + 10

f ''(x) = 84x + 10

f' = 42x² + 10x + C

f = 14x³ + 5x² + Cx + K

f(0) = 3

f(1) = 5

3 = 14•0³ + 5•0² + C•0 + K

K = 3

f = 14x³ + 5x² + Cx + 3

5 = 14•1³ + 5•1² + C•1 + 3

5 = 14 + 5 + C + 3

C = –17

eq:

f = 14x³ + 5x² – 17x + 3