Question

Find an antiderivative F(x) with F′(x) = f(x) = 6 + 24x^3 + 18x^5 and F(1)=0.

Answer 1

Answer:

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$.

Step-by-step explanation:

Antiderivative F(x)

This is the integral of $F^{\prime}(x)$

So

F′(x) = f(x) = 6 + 24x^3 + 18x^5

Then:

$F(x) = \int (6 + 24x^3 + 18x^5) dx$

$F(x) = 6x + \frac{24x^4}{4} + \frac{18x^6}{6} + K$

$F(x) = 6x + 6x^4 + 3x^6 + K$

F(1)=0

$F(X) = 0$ when $x = 1$. We use this to find K.

$F(x) = 6x + 6x^4 + 3x^6 + K$

$0 = 6 + 6 + 3 + K$

$K = -15$

Thus

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$.