Anti Derivative of
F''(x) = 4 + 6x + 24x^2, f(0) = 3, f(1)=10
Not sure how to find the value of the constant for F'(x).
It is not clear if you want F'(x) or whatever you call f(x).
F'(x) = integral of F"(x)
= 4x + 3x^2 + 8x^3 + C
where C is any constant
F(x) = 2x^2 + x^3 + 2x^4 + Cx + C'
where C' is a different constant.
You could choose C and C' to make F(0) = 3 and F(1) = 10, but the problem does not ask for F(x)
F(0) = 3 = C
F(1) = 2 + 1 + 2 + 3 + C' = 10
C' = 2
F(x) = 2x^2 + x^3 + 2x^4 + 3x + 2
To find the antiderivative of F''(x), we first need to integrate the given function. Here's how you can solve it step-by-step:
Step 1: Integrate F''(x) with respect to x to find F'(x):
To integrate F''(x) = 4 + 6x + 24x^2, we integrate each term separately. The integral of a constant is the constant times x, the integral of x raised to the power of n is x raised to the power of n+1 divided by n+1.
So, integrating 4 with respect to x gives you 4x. Integrating 6x gives you 6x^2/2 = 3x^2. Integrating 24x^2 gives you 24x^3/3 = 8x^3.
Therefore, F'(x) = 4x + 3x^2 + 8x^3 + C₁, where C₁ is the constant of integration.
Step 2: Find the constant of integration, C₁:
To determine the value of C₁, we can use the given information. In this case, we are given that f(0) = 3.
Substituting x = 0 into F'(x), we have:
3 = 4(0) + 3(0)^2 + 8(0)^3 + C₁
3 = 0 + 0 + 0 + C₁
3 = C₁
Therefore, C₁ = 3.
Step 3: Find F(x) by integrating F'(x):
To find F(x), we integrate F'(x):
F(x) = ∫[F'(x)] dx
F(x) = ∫[(4x + 3x^2 + 8x^3 + 3)] dx
F(x) = 2x^2 + x^3 + 2x^4 + 3x + C₂, where C₂ is the constant of integration.
Step 4: Find the constant of integration, C₂:
To determine the value of C₂, we use the given information that f(1) = 10.
Substituting x = 1 into F(x), we have:
10 = 2(1)^2 + (1)^3 + 2(1)^4 + 3(1) + C₂
10 = 2 + 1 + 2 + 3 + C₂
10 = 8 + C₂
C₂ = 10 - 8
C₂ = 2
Therefore, the antiderivative of F''(x) = 4 + 6x + 24x^2, with f(0) = 3 and f(1) = 10, is F(x) = 2x^2 + x^3 + 2x^4 + 3x + 2.