4.1 Antiderivatives: Beginning of Integrations

As name suggested, the antiderivative of a function f is a function F such that derivative of F will give f i.e, F'(x) = f(x) The identified function F is known as prime antiderivative of f. It is well-known that the derivative of a constant c is zero. Thus, the general antiderivative of f is F+ c.

The antiderivative is also known as Indefinite Integration of f . The indefinite integration is denoted as \[\int f(x) dx = F(x) + c.\]

How to find the antiderivative of a function f?

Example: Find the antiderivative of f(x) = x^2 .
To find the antiderivative of a given function, we have to search a function whose derivative is the given function. Here is the steps:

Step 1 : The function whose derivative will contain x^2 is x^3
Step 2 : But derivative of x^3 will give 3 x^2
Step 3: Thus, the function whose derivative will be exactly x^2 is \frac{x^3}{3}.
Step 4 : Finally, the antiderivative of f(x) = x^2 is F(x) = \frac{x^3}{3} + c .
Generalization: For any constant \(m\), \[f(x)=x^m \implies F(x)=\frac{x^{m+1}}{m+1}+c.\]

Example: Find the antiderivative of f(x) = \sin(x).
Here is the thought process:

Step 1 : The function whose derivative will contain \sin(x) is \cos(x)
Step 2: But derivative of \cos(x) will give -\sin(x)
Step 3: Thus, the function whose derivative will be exactly \sin(x) is \(- \cos(x) \)
Step 4 : Finally, the antiderivative of f(x) = \sin(x) is F(x) = - \cos(x) + c .
Generalization: For any constant \(m\), \[f(x)=\sin(m x) \implies F(x)= – \frac{\cos(m x)}{m}+c.\]

Example: Find the antiderivative of f(x) = e^{2x}.
Here is the thought process:

Step 1 : The function whose derivative will contain e^{2x} is \( e^{2x} \)
Step 2: But derivative of e^{2x} will give 2e^{2x}
Step 3: Thus, the function whose derivative will be exactly e^{2x} is \(\frac{e^{2x}}{2} \)
Step 4 : Finally, the antiderivative of \(f(x) = e^{2x}\) is \(F(x) = \frac{e^{2x}}{2} + c. \)
Generalization: For any constant \(m\), \[f(x)= e^{mx}\implies F(x)=\frac{e^{2x}}{2} +c.\]

Using the thought process of the antiderivatives as explained in the above examples, following table can be generated for the antiderivatives.
Function \(f(x)\) Antiderivative \(F(x)\)
\(f(x) = x^m\) \(F(x)= \frac{x^{m+1}}{m+1}+c, \; m \neq – 1 \)
\(f(x) = e^{m x}\) \(F(x)= \frac{e^{m x}}{m}+c\)
\(f(x) = \sin(m x)\) \(F(x)= – \frac{\cos(m x)}{m}+c\)
\(f(x) = \cos(m x)\) \(F(x)= \frac{\sin(m x)}{m}+c\)
\(f(x) = \sec^2(m x)\) \(F(x)= \frac{\tan(m x)}{m}+c\)
\(f(x) = \tan(mx)\sec(m x)\) \(F(x)= \frac{\sec(m x)}{m}+c\)
\(f(x) = \csc^2(m x)\) \(F(x)= -\frac{\cot(m x)}{m}+c\)
\(f(x) = \cot(mx)\csc(m x)\) \(F(x)= – \frac{\csc(m x)}{m}+c\)
\(f(x) = \frac{1}{x}\) \(F(x)= \ln\left(|x|\right) +c\)
\(f(x) = \frac{1}{1+x^2}\) \(F(x)= \tan^{-1}(x) +c\)
\(f(x) = \frac{1}{\sqrt{1-x^2}}\) \(F(x)= \sin^{-1}(x) +c\)
\(f(x) = a^x\) \(F(x)= \frac{a^x}{ln(a)} +c\)

You can watch the Video lecture below. For more video lecture of Calculus 1 topics please click here

The Antiderivatives are required to solve the integration in next section and calculus 2

Affiliated