Wemazi

Topic 6.8

Find an antiderivative of a power or sum

Builds on interpreting the definite result. 8 minutes.

Raise the power, divide by the new power

The top graph shows \(y = x^2\), and orange lines mark its height at four inputs. Below, the teal curve is \(y = kx^3\), and the slider changes the coefficient \(k\).

At three of the inputs, a short orange segment on the teal curve has the height there as its slope. At \(x = 1.5\) the longer orange line is the teal curve's own slope, which the readouts compare with the height 2.25.

An antiderivative of \(x^2\) has slope \(x^2\) at every input, so its graph runs along all the segments. At \(k = 1\) the curve is too steep, because the slope of \(x^3\) is \(3x^2\), three times the height of \(x^2\).

At \(k = \tfrac13\) the segments lie along the curve and the readouts match, so \(\tfrac{x^3}{3}\) is an antiderivative of \(x^2\). Its exponent is one more than that of \(x^2\), and the divisor is that new exponent.

The same rule for every power

The top curve is \(f(x) = x^n\) and the teal curve is \(\tfrac{x^{n+1}}{n+1}\). Slide \(n\) anywhere from 0 to 3, including fractions, and the orange segments stay along the teal curve.

The derivative of \(x^{n+1}\) is \((n+1)x^n\), and \(n+1\) is a constant, so dividing by it leaves \(x^n\).

Press "Multiply f by 3". The teal curve is multiplied by 3 too, because tripling every slope triples the curve.

Press "Add a second term". The top curve gains the term \(2x\), and the teal curve gains its antiderivative \(x^2\), because slopes add.

Notation

  1. To find an antiderivative of \(x^n\), raise the exponent by one and divide by the new exponent. The rule needs \(n \ne -1\), because then the divisor \(n + 1\) would be 0.

    \[f(x) = x^n \;\Rightarrow\; \cb{F(x) = \frac{x^{n+1}}{n+1}}\]

    Every antiderivative of \(f\) differs from this \(F\) by a constant.

  2. A constant factor stays in front, because multiplying a function by a constant multiplies every slope by that constant.

    \[f(x) = k\,x^n \;\Rightarrow\; F(x) = k\,\frac{x^{n+1}}{n+1}\]

  3. Handle a sum one term at a time. If \(G\) and \(H\) are antiderivatives of \(g\) and \(h\), then \(G + H\) is an antiderivative of \(g + h\), because slopes add.

    \[f = g + h \;\Rightarrow\; F = G + H\]

  4. Write roots and quotients as powers first. A constant is a power too, because \(7 = 7x^0\).

    \[\begin{aligned} \sqrt{x} &= x^{1/2} \\ \frac{1}{x^3} &= x^{-3} \\ 7 &= 7x^0 \end{aligned}\]

When it fails

The rule works on a sum of terms, each a constant times a power of \(x\). It does not work on a product of two factors that both contain \(x\). For \(f(x) = x^2(x + 3)\), the teal curve is the product of the antiderivatives of the two factors, and the dashes turn red where it cuts across them.

At \(x = 1\) the slope of this curve is \(\tfrac{29}{6}\), but \(f(1) = 4\). Multiply out first: \(x^2(x+3) = x^3 + 3x^2\). Then apply the rule to each term, and the dashes lie along the curve \(\tfrac{x^4}{4} + x^3\).

Example 1: A polynomial

Find an antiderivative of \(f(x) = 6x^5 - 4x + 7\).

  1. Treat each term on its own. The exponents are 5, 1 and 0, because \(4x\) is \(4x^1\) and 7 is \(7x^0\).

  2. For \(6x^5\), raise the exponent to 6 and divide by 6, so the factor 6 cancels.

    \[6\cdot\frac{x^6}{6} = \cb{x^6}\]

  3. For \(-4x\), raise the exponent to 2 and divide by 2.

    \[-4\cdot\frac{x^2}{2} = \cb{-2x^2}\]

  4. For 7, raise the exponent to 1 and divide by 1. With all three terms, every segment lies along the curve.

    \[7\cdot\frac{x^1}{1} = \cb{7x}\]

  5. Add the three results, and differentiate to check that the derivative of \(F\) is \(f\).

    \[\begin{aligned} F(x) &= x^6 - 2x^2 + 7x \\ F'(x) &= 6x^5 - 4x + 7 \end{aligned}\]

Example 2: A root and a quotient

Find an antiderivative of \(f(x) = 6\sqrt{x} + \dfrac{4}{x^3}\) for \(x > 0\).

  1. Write each term as a power of \(x\).

    \[f(x) = 6x^{1/2} + 4x^{-3}\]

  2. For \(6x^{1/2}\), the new exponent is \(\tfrac32\). Dividing by \(\tfrac32\) means multiplying by \(\tfrac23\), so the term becomes \(4x^{3/2}\).

    \[6\cdot\frac{x^{3/2}}{3/2} = \cb{4x^{3/2}}\]

  3. A tempting wrong move: For \(4x^{-3}\), the new exponent is \(-2\). Divide by 2.

    \[4\cdot\frac{x^{-2}}{2} = 2x^{-2}\]

    The orange segments cut across the red curve. The term \(\tfrac{4}{x^3}\) is positive when \(x > 0\), so its antiderivative must rise, but \(\tfrac{2}{x^2}\) falls. The new exponent is \(-2\), so the divisor is \(-2\), not 2.

  4. Divide by \(-2\). The divisor is negative, so the sign of the term changes.

    \[4\cdot\frac{x^{-2}}{-2} = \cb{-2x^{-2}}\]

  5. Add the two results. Differentiate to check.

    \[\begin{aligned} F(x) &= 4x^{3/2} - \frac{2}{x^2} \\ F'(x) &= 6x^{1/2} + 4x^{-3} \end{aligned}\]

Practice

Find an antiderivative of \(f(x) = 12x^2 + 9\sqrt{x} - \dfrac{5}{x^2}\) for \(x > 0\).

Write the root and the quotient as powers first. Then slide the coefficients until every orange segment lies along the teal curve. A negative new exponent gives a negative divisor.

Solution

  1. Write each term as a power of \(x\).

    \[f(x) = 12x^2 + 9x^{1/2} - 5x^{-2}\]

  2. The new exponents are 3, \(\tfrac32\) and \(-1\). Divide each term by its new exponent.

    \[12\cdot\frac{x^3}{3} + 9\cdot\frac{x^{3/2}}{3/2} - 5\cdot\frac{x^{-1}}{-1}\]

  3. Simplify, and differentiate to check.

    \[\begin{aligned} F(x) &= 4x^3 + 6x^{3/2} + \frac{5}{x} \\ F'(x) &= 12x^2 + 9x^{1/2} - \frac{5}{x^2} \end{aligned}\]

Questions

Questions 1 and 2 use the curve \(f(x) = x^4\) on top. Below, the teal curve is a candidate for an antiderivative, and its dashes show the slope that \(f\) asks for at each \(x\).

  1. The teal candidate is \(G(x) = x^5/4\). Its dashes are red. What is wrong?

    • The divisor should be 5, the new exponent
    • The exponent should stay 4
    • A constant must be added
    Answer

    The divisor should be 5, the new exponent

  2. What is the slope of the correct candidate \(F(x) = x^5/5\) at \(x = 2\)?

    • 16
    • 6.4
    • 80
    • 32
    Answer

    16

  3. Which function is an antiderivative of \(f(x) = \dfrac{6}{x^4} - 2\sqrt{x}\) for \(x > 0\)?

    • \(-\dfrac{2}{x^3} - \dfrac{4}{3}x^{3/2}\)
    • \(\dfrac{2}{x^3} - \dfrac{4}{3}x^{3/2}\)
    • \(-\dfrac{24}{x^5} - \dfrac{1}{\sqrt{x}}\)
    Answer

    \(-\dfrac{2}{x^3} - \dfrac{4}{3}x^{3/2}\)

Summary

You raise the exponent of each term by one, divide by the new exponent, and keep each constant factor in front.

This works because the derivative of \(x^{n+1}\) is \((n+1)x^n\), so dividing by \(n+1\) leaves \(x^n\), and the slopes of a sum add.

You use it for sums of powers, after you write roots and quotients as powers.

This is one of five skills in topic 6.8 of Which & Why: AP Calculus.