Question

[4] 7. Find f if
f"(x) = x^2+ sin x
f'(0) = 2
f(0) =4

Answer 1

`f(x) = (x^4)/(12) - sin(x) + 3x + 4`

General Formulas and Concepts:

Calculus

• Antiderivatives
• Integration Constant C
• [Int Rule] Reverse Power Rule:
  `\int {x^n} \, dx = (x^(n+1))/(n+1) + C`
• Integration Property 1:
  `\int {cf(x)} \, dx = c\int {f(x)} \, dx`
• Integration Property 2:
  `\int {f(x)+g(x)} \, dx = \int {f(x)} \, dx + \int {g(x)} \, dx`

Explanation:

Step 1: Define

f"(x) = x² + sin(x)

Condition f'(0) = 2

Condition f(0) = 4

Step 2: Integrate Pt. 1

1. Set up:
   `f'(x) = \int {f`
2. Substitute:
   `f'(x) = \int [{x^2 + sin(x)}] \, dx`
3. Rewrite [Int Property 2]:
   `f'(x) = \int {x^2} \, dx + \int {sin(x)} \, dx`
4. Integrate [Reverse Power Rule/Trig]:
   `f'(x) = (x^3)/(3) - cos(x) + C`

Step 3: Find f'(x)

Use the given condition to find the differential equation.

1. Substitute:
   `f'(0) = (0^3)/(3) - cos(0) + C`
2. Substitute:
   `2 = (0^3)/(3) - cos(0) + C`
3. Evaluate:
   `2 = 0 - 1 + C`
4. Solve:
   `3 = C`
5. Define:
   `f'(x) = (x^3)/(3) - cos(x) + 3`

Step 4: Integrate Pt. 2

1. Set up:
   `f(x) = \int {f'(x)} \, dx`
2. Substitute:
   `f(x) = \int [{(x^3)/(3) - cos(x) + 3}] \, dx`
3. Rewrite [Int Property 2]:
   `f(x) = \int {(x^3)/(3) } \, dx + \int {-cos(x)} \, dx + \int {3} \, dx`
4. Rewrite [Int Property 1]:
   `f(x) = (1)/(3) \int {x^3} \, dx - \int {cos(x)} \, dx + \int {3} \, dx`
5. Integrate {Reverse Power Rule/Trig]:
   `f(x) = (1)/(3)((x^4)/(4) ) - sin(x) + 3x + C`
6. Simplify:
   `f(x) = (x^4)/(12) - sin(x) + 3x + C`

Step 5: Find f(x)

Use the given condition to find the equation.

1. Substitute:
   `f(0) = (0^4)/(12) - sin(0) + 3(0) + C`
2. Substitute:
   `4 = (0^4)/(12) - sin(0) + 3(0) + C`
3. Evaluate:
   `4 = 0 - 0 + 0+ C`
4. Solve:
   `4 = C`
5. Define:
   `f(x) = (x^4)/(12) - sin(x) + 3x + 4`