Is the indefinite integral of x with respect to x equal to x^2/2? Justify the answer.

Step-by-step solution

1. 1

   Apply the power rule for antiderivatives. For any exponent $n\ne -1$,

   $\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$

   Here $x=x^{1}$, so $n=1$ and

   $\int x\,dx=\frac{x^{1+1}}{1+1}+C=\frac{x^{2}}{2}+C$

2. 2

   Verify the candidate by differentiating back. Integration is only ever checked one way — differentiate the answer:

   $\frac{d}{dx}\!\left(\frac{x^{2}}{2}\right)=\frac{2x}{2}=x\ \checkmark$

   So $\dfrac{x^{2}}{2}$ is an antiderivative of $x$. The stated formula is not wrong so much as incomplete.

3. 3

   See why $+C$ cannot be dropped. Differentiation kills constants, so for every real $C$,

   $\frac{d}{dx}\!\left(\frac{x^{2}}{2}+C\right)=x$

   For instance $\tfrac{x^{2}}{2}$, $\tfrac{x^{2}}{2}+7$ and $\tfrac{x^{2}}{2}-\tfrac{1}{3}$ all differentiate to $x$. Writing a single one of them as "the" integral silently discards infinitely many equally valid answers.

4. 4

   Know that $C$ captures all of them. The mean value theorem guarantees that two antiderivatives of the same function on an interval differ by a constant, so the family $\tfrac{x^{2}}{2}+C$ is not just some antiderivatives — it is every antiderivative of $x$. That is exactly what the notation $\int x\,dx$ denotes.

5. 5

   Note when the constant genuinely disappears. In a definite integral the constant cancels:

   $\int_{0}^{3}x\,dx=\left[\frac{x^{2}}{2}+C\right]_{0}^{3}=\left(\frac92+C\right)-\left(0+C\right)=\frac92$

   So you must write $+C$ for an indefinite integral, and you may safely ignore it for a definite one.