Find a suitable antiderivative
Guess a function, then test its slope
The top graph shows the curve \(y = 2x^3\). A function \(\cb{F}\) is an antiderivative of this curve if its slope at every \(x\) equals the height of the curve there.
The teal curve below is a first guess, \(\cb{F(x) = x^4}\). The short dashes along it show the slope that the top curve asks for at each \(x\). A red dash cuts across the guess where its slope is wrong.
At \(x = 1\) the top curve has height \(\cy{2}\), and the orange line below has that slope. The red line has the slope of the guess, \(4x^3\), which is 4 at \(x = 1\), so it is twice as steep as the orange line.
Move the slider to halve the guess. The dashes become orange when \(F(x) = x^4/2\), because its slope, \(2x^3\), equals the height of the top curve at every \(x\).
A function whose slope follows the curve
The top graph shows a curve \(f\). The teal curve \(\cb{F}\) is an antiderivative of \(f\) when every dash along it is orange.
Shift the teal curve up or down. The dashes stay orange, because a shift does not change the slope at any point.
Every shifted curve is also an antiderivative, and any two antiderivatives of the same curve differ by a constant. If you tilt the candidate, the dashes become red where its slope is wrong.
Notation
A function \(\cb{F}\) is an antiderivative of \(f\) when the derivative of \(\cb{F}\) is \(f\).
\[\cb{F}'(x) = \cy{f(x)}\]Here \(F'(x)\) is the slope of the teal curve at \(x\), and \(f(x)\) is the height of the top curve.
The derivative of a power multiplies by the exponent and lowers the exponent by one. An antiderivative reverses both steps, so you raise the exponent by one and divide by the new exponent.
\[\frac{d}{dx}\,\cb{\frac{x^{n+1}}{n+1}} = \cy{x^n} \qquad (n \neq -1)\]The exponent \(-1\) is the exception, because the formula would divide by 0. Instead, an antiderivative of \(1/x\) is \(\ln|x|\), because its derivative is \(1/x\).
\[\frac{d}{dx}\,\cb{\ln|x|} = \cy{\frac{1}{x}}\]A constant factor stays in front, and a sum is handled one term at a time. If \(f\) is a quotient, first divide each term of the top by the bottom to make a sum of powers.
\[\big(k\,\cb{F} + \cb{G}\big)' = k\,\cy{f} + \cy{g}\]Here \(F\) and \(G\) are antiderivatives of \(f\) and \(g\), and \(k\) is a constant.
To check an answer, differentiate it. The result must be the function you started with.
When it fails
Move \(n\) toward \(-1\). The candidate divides by \(n + 1\), which approaches 0, so the teal curve moves up and leaves the picture.
At \(n = -1\) the divisor is 0, so the formula does not exist. Use \(\ln|x|\) for the exponent \(-1\) instead.
Example 1: a power plus a reciprocal
Find an antiderivative of \(f(x) = 6x^2 + \dfrac{2}{x}\) for \(x > 0\).
The function is a sum of two terms, drawn faint and dashed, so you find an antiderivative of each.
Raise the exponent of \(6x^2\) to 3. Then divide by 3. The factor 6 stays in front.
\[6\cdot\frac{x^3}{3} = \cb{2x^3}\]The term \(2/x\) is 2 times \(1/x\), and an antiderivative of \(1/x\) is \(\ln|x|\).
\[2\cdot\ln|x| = \cb{2\ln|x|}\]Add the two results. The teal curve below is the sum.
\[F(x) = \cb{2x^3 + 2\ln|x|}\]Differentiate to check. The result is \(f\), so every dash along \(F\) is orange.
\[F'(x) = 6x^2 + \frac{2}{x}\]
Example 2: a quotient
Find an antiderivative of \(f(x) = \dfrac{x^2 + 4}{x^2}\) for \(x > 0\).
There is no formula for a quotient. Divide each term of the top by \(x^2\), which writes \(f\) as a sum of powers.
\[\frac{x^2 + 4}{x^2} = 1 + 4x^{-2}\]A tempting wrong move: Find an antiderivative of the top and of the bottom. Then divide.
\[\frac{x^3/3 + 4x}{x^3/3} = 1 + \frac{12}{x^2}\]At \(x = 2\), the function \(f\) has height 2, so an antiderivative must rise there. This candidate has slope \(-3\) at \(x = 2\), so it falls, and the red dashes cut across it.
Use the sum instead. An antiderivative of 1 is \(x\). For \(4x^{-2}\), raise the exponent to \(-1\) and divide by \(-1\).
\[4\cdot\frac{x^{-1}}{-1} = -\frac{4}{x}\]Add the terms. The derivative of the result is \(1 + 4/x^2\), which is \(f\), so the dashes are orange.
\[F(x) = \cb{x - \frac{4}{x}}\]
Practice
Find an antiderivative of \(f(x) = \dfrac{x^3 + 6}{x^2}\) for \(x > 0\).
Divide each term of the top by \(x^2\) first, so that every term is a power. When your candidate is correct, all the dashes along it are orange.
Solution
Write \(f\) as a sum of powers.
\[f(x) = x + 6x^{-2}\]An antiderivative of \(x\) is \(x^2/2\). For \(6x^{-2}\), raise the exponent to \(-1\) and divide by \(-1\).
\[F(x) = \frac{x^2}{2} - \frac{6}{x}\]Differentiate to check. The result is \(x + 6/x^2\), which is \(f\).
\[F'(x) = x + \frac{6}{x^2}\]
Questions
The top graph shows the curve \(y = 2x^3\). The teal curve below is a candidate \(F\), with short dashes that show the slope the top curve asks for at each \(x\).
The teal candidate is \(F(x) = x^4\). Red dashes cut across it. Which change makes it an antiderivative of \(2x^3\)?
- Divide it by 2
- Multiply it by 2
- Add 2 to it
- Divide it by 4
Answer
Divide it by 2
The teal candidate is \(F(x) = x^4/2 - 3\). Is it an antiderivative of \(2x^3\)?
- Yes, because the shift does not change the slope
- No, because \(F(0)\) is not 0
- No, because the slope is now \(2x^3 - 3\)
Answer
Yes, because the shift does not change the slope
Which function is an antiderivative of \(f(x) = 10x^4 - \dfrac{5}{x}\) for \(x > 0\)?
- \(2x^5 - 5\ln|x|\)
- \(40x^3 + \dfrac{5}{x^2}\)
- \(\dfrac{10x^5}{4} - 5\ln|x|\)
Answer
\(2x^5 - 5\ln|x|\)
Summary
You find an antiderivative term by term, with each constant factor kept in front, the power formula for a power and \(\ln|x|\) for \(1/x\).
It works because the slope of the answer equals the height of the given function at every x, and differentiating the answer tests this.
You use it as the first step in evaluating a definite integral, after you rewrite any quotient as a sum of powers.
This is one of four skills in topic 6.7 of Which & Why: AP Calculus.