Q.Find the anti derivative $F$ of $f$ defined by $f(x) = 4x^3 - 6$, where $F(0) = 3$.

Concept: Initial Value Problem — we integrate the derivative and then use the given condition to solve for the constant.

Step 1: Integrate $f(x)$ to find the general antiderivative:

Step 2: Use the initial condition $F(0) = 3$: …

The problem gives us a function $f(x)=4x^3-6$ and asks for its antiderivative $F$ — that is, a function whose derivative is $f$. But there’s a catch: antiderivatives are not unique. Because the derivative of any constant is zero, if $F(x)$ is an antiderivative, then $F(x)+C$ is also one for any constant $C$.

To pin down exactly which antiderivative we want, we’re given an initial condition: $F(0)=3$. This is an Initial Value Problem (IVP): find the function whose derivative is known and which passes through a specific point. The constant $C$ is determined by plugging in that point.

1. Find the general antiderivative Integrate $f(x)$ term by term:

Using the power rule $\int x^n\,dx = \frac{x^{n+1}}{n+1}$ for $n \neq -1$:

and

So the general antiderivative is:

where $C$ is an arbitrary constant.

2. Apply the initial condition We know $F(0)=3$. Substitute $x=0$ into $F(x)$:

Setting this equal to $3$ gives $C = 3$.

3. Write the particular solution …

Method: Solving an Initial Value Problem (Finding $C$)

Use this whenever you are asked for a specific antiderivative that passes through a given point $F(x_0)=y_0$, not just the general family.

Steps

Step 1: Integrate to get the general antiderivative.

Integrate $f(x)$ term-by-term, keeping the arbitrary constant:

The $+C$ is essential here — it is the unknown the condition will pin down.

Step 2: Substitute the initial condition. …

Common Mistakes

Mistake 1: Dropping $+C$ during integration.

Why it's wrong: without $C$ there is nothing for the initial condition to determine, and the whole point of the IVP is lost. Correct approach: always keep $+C$ until the condition fixes it.

Mistake 2: Applying the condition before finishing the integration.

Why it's wrong: the condition constrains the antiderivative $F$, so $F$ must be found first. Correct approach: integrate fully, then substitute $x=x_0$. …