Find the function f whose 2nd derivative is f ' ' (x) = x^3 + 2x^2 + 4 and which has the boundary values f (0) = 1, and f (1) = 3

Question

Find the function f whose 2nd derivative is f ' ' (x) = x^3 + 2x^2 + 4 and which has the boundary values f (0) = 1, and f (1) = 3

Answer 1

Given:

f''(x) = x³ + 2x² + 4

To find f(x), we can integrate twice. Lets first integrate f''(x) to get f'(x):

f'(x) = ∫(x³ + 2x² + 4) dx

= x⁴/4 + 2x³/3 + 4x + C₁

We can integrate again to get f(x):

f(x) = ∫ (x⁴/4 + 2x³/3 + 4x + C₁) dx

= x⁵/20 + x⁴/6 + 2x² + C₁x + C₂

Now, we can use the boundry values. Since f(0) = 1, we can subsitute 0 for x and 1 for f(x) to get our C₁ and C₂:

1 = 0⁵/20 + 0⁴/6 + 2(0)² + C₁(0) + C₂

C₂ = 1

We can now use f(1) = 3:

3 = 1⁵/20 + 1⁴/6 + 2(1)² + C₁(1) + 1

3 = 1/20 + 1/6 + 2 + C₁ + 1

3 = 1/20 + 1/6 + 3 + C₁

0 = 1/20 + 1/6 + C₁

C₁ = - 13/60

We can plug in our C₁ into the equation:

f(x) = x⁵/20 + x⁴/6 + 2x² - (13/60)x + 1

Answer 2

Answer 3

Just integrate twice and use the iniitial (boundary) conditions to evaluate the 2 constants of integration.