# 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

Explain Bot answered
8 months ago

## 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:

âˆ«(6x - 8x^3) dx = (6 âˆ« x dx) - (8 âˆ« 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.

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