- [[definite integral]], [[indefinite versus definite integral]] - [[integral formula for power functions]] # Idea An indefinite integral is also known as an antiderivative. It is a family of functions that reverses the process of [[differentiation]]. General form of the indefinite integral: $ \int f(x) \ d x=F(x)+C $ where $C$ can be any constant. This characteristic reflects the fact that the derivative of a constant is zero, which means adding a constant to a function does not change its derivative. If $f(x)$ is a function, its indefinite integral is written as: $\int f(x) \ dx$ This expression denotes the set of all functions $F(x)$ such that the derivative of $F(x)$ with respect to $x$ is $f(x)$. In other words, for any function $F(x)$ that satisfies this condition: $F'(x) = f(x)$ The integral includes an arbitrary constant $C$, known as the constant of integration, because the derivative of a constant is zero and does not affect the derivative of $F(x)$. ## Examples ### Example Consider the function $f(x) = x^2$. We want to find its indefinite integral: $ \int x^2 d x $ $ \int x^n d x=\frac{x^{n+1}}{n+1}+C $ Applying this to our example, where $n=2$ : $ \int x^2 d x=\frac{x^{2+1}}{2+1}+C=\frac{x^3}{3}+C $ Thus, the indefinite integral of $x^2$ is: $ \int x^2 d x=\frac{x^3}{3}+C $ This result means any function of the form $\frac{x^3}{3}+C$, where $C$ is any constant, is an antiderivative of $x^2$ because the derivative of $\frac{x^3}{3}+C$ with respect to $x$ returns $x^2$. ### Example Find the antiderivative. $ \int 2 x^4-6 x+2 d x $ Using the formula $\int f(x) d x=F(x)+C$ where $F^{\prime}(x)=f(x)$, we take the integral one term at a time and get $ \begin{aligned} & \int 2 x^4-6 x+2 d x=\frac{2 x^5}{5}-\frac{6 x^2}{2}+2 x+C \\ & \int 2 x^4-6 x+2 d x=\frac{2}{5} x^5-3 x^2+2 x+C \end{aligned} $ ### Example Evaluate the indefinite integral. $ \int 2 \pi x \ dx $ $ \frac{2 \pi x^{1+1}}{1+1}=\pi x^2 $ Then the integral is $ \int 2 \pi x d x=\pi x^2+C $ where $C$ is an arbitrary constant, meaning that any value for $C$ makes a valid antiderivative. ### Example $ \begin{aligned} & \int \frac{2 x^3-x^2+4}{x^2} d x \\ & \int 2 x-1+4 x^{-2} d x \\ & \frac{2}{2} x^2-x+\frac{4}{-1} x^{-1}+C \\ & x^2-x-\frac{4}{x}+C \\ & \frac{x^3}{x}-\frac{x^2}{x}-\frac{4}{x}+C \\ & \frac{x^3-x^2-4}{x}+C \end{aligned} $ ### Example Consider the function $f(x)=\sin (x)$. We need to find its indefinite integral: $ \int \sin (x) d x $ According to the fundamental rules of integration, the integral of $\sin (x)$ is: $ \int \sin (x) d x=-\cos (x)+C $ Here, $C$ represents the constant of integration. The result is derived from the fact that the derivative of $-\cos (x)$ is $\sin (x)$, satisfying the definition of the indefinite integral. Thus: $ \int \sin (x) d x=-\cos (x)+C $ This tells us that any function of the form $-\cos (x)+C$, where $C$ is any constant, is an antiderivative of $\sin (x)$. # References