10.8k views
3 votes
[4] 7. Find f if
f"(x) = x^2+ sin x
f'(0) = 2
f(0) =4

1 Answer

4 votes

Answer:


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
User Zufar Muhamadeev
by
7.7k points

No related questions found