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Antidifferentiation

Antidifferentiation is the reverse operation of differentiation. You need it to solve problems involving area under a curve, total accumulated change, and reversing rates of change. By mastering antidifferentiation, you'll be able to reconstruct original functions and solve real-world problems in physics, engineering, and beyond.

If you have the derivative of a function \(f'(x)\), you can find the original function \(f(x)\) using integration or antidifferentiation.

Here, the original function is called the antiderivative.

Isaac Newton
Isaac Newton co-invented calculus which includes differentiation and antidifferentiation (integration). This portrait of Newton at age 46 was done by Godfrey Kneller in 1689.
Image via Wikimedia Commons, licensed in the public domain.

Video tutorial – integrating functions

Watch this video to learn about how to integrate functions or find antiderivatives.

Antiderivatives and indefinite integrals by Khan Academy, licensed under CC BY-NC-SA 3.0

We know how to take derivatives of functions. If I apply the derivative operator to x squared, I get 2x. Now, if I also apply the derivative operator to x squared plus 1, I also get 2x.

If I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. Derivative, with respect to x of pi of a constant, is just 0. Derivative, with respect to x of 1, is just a constant, is just 0. So once again, this is just going to be equal to 2x.

In general, the derivative, with respect to x of x squared plus any constant, is going to be equal to 2x. The derivative of x squared, with respect to x, is 2x. Derivative of a constant, with respect to x, a constant does not change with respect to x, so it's just equal to 0. So you have– You apply the derivative operator to any of these expressions and you get 2x.

Now, let's go the other way around. Let's think about the antiderivative. And one way to think about it is we're doing the opposite of the derivative operator. The derivative operator, you get an expression and you find it's derivative.

Now, what we want to do, is given some expression, we want to find what it could be the derivative of. So if someone were to tell– or give you 2x– if someone were to say 2x– let me write this. So if someone were to ask you what is 2x the derivative of? They're essentially asking you for the antiderivative. And so you could say, well, 2x is the derivative of x squared. But you could also say 2x is the derivative of x squared plus 1. You could also say that 2x is the derivative of x squared plus pi, I think you get the general idea.

So if you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. So this is what you would consider the antiderivative of 2x.

Now, that's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. And the convention here is to use kind of a strange looking notation, is to use a big elongated s looking thing like that, and a dx around the function that you're trying to take the antiderivative of.

So in this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c.

Now, you might be saying, why do we use this type of crazy notation. It'll become more obvious when we study the definite integral and areas under curves and taking sums of rectangles in order to approximate the area of the curve.

Here, it really should just be viewed as a notation for antiderivative. And this notation right over here, this whole expression, is called the indefinite integral of 2x, which is another way of just saying the antiderivative of 2x.

Antiderivatives

In general, an antiderivative of \(f'(x)\) is given by \(f(x)+c\) where \(c\) is a constant. For example:

  • If \(f(x)=\dfrac{x^{3}}{3}\), then \(f'(x)=x^{2}\), so \(\dfrac{x^{3}}{3}\) is an antiderivative of \(x^{2}\).
  • If \(f(x)=\dfrac{x^{3}}{3}+1\), then \(f'(x)=x^{2}\), so \(\dfrac{x^{3}}{3}+1\) is an antiderivative of \(x^{2}\).
  • If \(f(x)=\dfrac{x^{3}}{3}+2\), then \(f'(x)=x^{2}\), so \(\dfrac{x^{3}}{3}+2\) is an antiderivative of \(x^{2}\).

You can see that adding a constant to \(\dfrac{x^{3}}{3}\) does not change the fact it is an antiderivative of \(x^{2}\). This is because the derivative of a constant is zero. We often write \(c\in\mathbb{R}\), which means that \(c\) is a real number.

Finding antiderivatives

Antidifferentiation is more complicated than differentiation. But there are some rules to help us. One of the most important is the power rule which says:

If \(f'(x)=x^{n}\), \(n\neq-1\), the antiderivative \(f(x)=\dfrac{1}{n+1}x^{n+1}+c\), where \(c\) is a constant.

If \(n=-1\), \(\dfrac{1}{n+1}\) would be \(\dfrac{1}{0}\) which has no meaning. Apart from this restriction, \(n\) can be any number.

Integrals

We express antiderivatives using intervals. For example, \(\dfrac{1}{3}x^{3}\) is the antiderivative of \(y=x^{2}\). Using intervals, we would use the mathematical notation:
\[ \int x^{2}dx=\frac{1}{3}x^{3}+c,\,c\in\mathbb{R}\]

We read this as "the indefinite integral of \(x^{2}\) with respect to \(x\) is equal to \(\dfrac{1}{3}x^{3}+c\)". The \(\int\) sign means indefinite integral and \(dx\) means with respect to \(x\). The words "indefinite integral" are the same as "antiderivative".

Example 1 – finding antiderivatives

Given \(\dfrac{dy}{dx}=x\), find the antiderivative.

Here, \(n=1\) because \(x=x^{1}\). We add \(1\) to the power of \(1\), divide by the new power, and add a constant \(c\).
\[\begin{align*} y & = \frac{x^{1+1}}{1+1}+c,\,c\in\mathbb{R}\\
& = \frac{x^{2}}{2}+c
\end{align*}\]

Given \(\dfrac{dy}{dx}=1\), find the antiderivative.

Here, \(n=0\) because \(x^{0}=1\). We add \(1\) to the power of \(0\), divide by the new power, and add a constant \(c\).
\[\begin{align*} y & = \frac{x^{0+1}}{0+1}+c,\,c\in\mathbb{R}\\
& = x+c
\end{align*}\]

Given \(\dfrac{dy}{dx}=x^{-3}\), find the antiderivative.

Here, \(n=-3\). The new power will be \(-2\).
\[\begin{align*} y & = \frac{x^{-3+1}}{-3+1}+c,\,c\in\mathbb{R}\\
& = \frac{x^{-2}}{-2}+c\\
& = -\frac{1}{2x^{2}}+c
\end{align*}\]

Given \(\dfrac{dy}{dx}=\sqrt{x}\), find the antiderivative.

Here, \(n=\dfrac{1}{2}\) because \(\sqrt{x}=x^{\frac{1}{2}}\). The new power will be \(\dfrac{1}{2}+1=\dfrac{3}{2}\).
\[\begin{align*} y & = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c,\,c\in\mathbb{R}\\
& = \frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c\\
& = \frac{2}{3}x^{\frac{3}{2}}+c
\end{align*}\]

Exercise – finding antiderivatives

  1. Find antiderivatives for the following.
    1. \(x^{3}\)
    2. \(s^{8}\)
    3. \(\sqrt[3]{x}\)
    4. \(x^{-5}\)
    5. \(6\)
    6. \(m^{-2}\)
    7. \(p^{-\frac{1}{2}}\)
  2. Integrate the following.
    1. \(\int 3dx\)
    2. \(\int 5xdx\)

    1. \(\dfrac{1}{4}x^{4}+c\)
    2. \(\dfrac{1}{9}s^{9}+c\)
    3. \(\dfrac{3}{4}x^{\frac{4}{3}}+c\)
    4. \(-\dfrac{1}{4x^{4}}+c\)
    5. \(6x+c\)
    6. \(-\dfrac{1}{m}+c\)
    7. \(2\sqrt{p}+c\)
    1. \(3x+c\)
    2. \(\dfrac{5}{2}x^{2}+c\)

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