The indefinite integral, or which call the "antiderivative", of f'(x)=6x-8x^3
is f(x) = 3x^2 - 2x^4 + C
where C is any constant.
Plug in x = 2 to that to see what C is.
3 = 3*4 - 2*16 + C
Determine the antiderivative of the following function given the initial condition: f'(x)=6x-8x^3 given that
f(2)=3
f'(x)=6x-8x^3
f(x)= 6x^2/2 - 8x^4/4 + C
Put in x=2 and solve for C
To determine the antiderivative of the function f'(x) = 6x - 8x^3, we can apply the power rule of integration.
The power rule states that the integral of x^n with respect to x is (x^(n+1))/(n+1) + C, where C is the constant of integration.
Using the power rule, we can integrate each term of f'(x) separately:
∫(6x - 8x^3) dx = (6 ∫ x dx) - (8 ∫ x^3 dx)
For the first term, the integral of x with respect to x is (x^2)/2.
For the second term, the integral of x^3 with respect to x is (x^4)/4.
Therefore, the antiderivative of f'(x) is:
f(x) = (6x^2)/2 - (8x^4)/4 + C
= 3x^2 - 2x^4 + C
Now, we need to determine the constant of integration, C, using the initial condition f(2) = 3.
Plugging in x = 2 into our expression for f(x), we get:
f(2) = 3(2)^2 - 2(2)^4 + C
= 3(4) - 2(16) + C
= 12 - 32 + C
= -20 + C
Since f(2) = 3, we can set -20 + C = 3 and solve for C:
-20 + C = 3
C = 3 + 20
C = 23
Therefore, the antiderivative of f'(x) = 6x - 8x^3, given that f(2) = 3, is f(x) = 3x^2 - 2x^4 + 23.