Rewrite radicals and powers

A root in the bottom

The chalk curve is \(\dfrac{x+2}{\sqrt{x}}\) for \(x > 0\), and the teal region is its area from \(x = 1\) to \(x = 4\). The power rule integrates terms such as \(x^n\), not this fraction as it stands. Drag the orange line to read the height of the curve.

A root is a fractional power, for example \(\sqrt{x} = x^{1/2}\). A power in the bottom of a fraction is a negative power, for example \(\dfrac{1}{\sqrt{x}} = x^{-1/2}\).

The bottom divides each term of the top, so the fraction splits into two terms. To divide powers of the same base, subtract the exponents: \(\dfrac{x}{x^{1/2}} = x^{1/2}\) and \(\dfrac{2}{x^{1/2}} = 2x^{-1/2}\).

Press the button. The orange dashed curve is \(x^{1/2} + 2x^{-1/2}\). It lies on the chalk curve, because at every \(x\) the two terms add up to the curve's height.

Then the region splits in two. The teal region under \(\sqrt{x}\) has area \(\tfrac{14}{3}\). The green region above it has area 4, so the total is \(\tfrac{26}{3}\).

Any root or fraction

The integrand is any fraction or root built from powers of x. Write it as a sum of terms, where each term is a number times a power of x.

Press the button. The orange dashed curve is that sum, and it lies on the chalk curve at every x, so the rewrite does not change the integrand.

Then the region splits. The teal region lies under the first term. The green region on top has the height of the second term, so together they fill the area under the chalk curve.

The power rule integrates each term: raise the exponent by 1, then divide by the new exponent.

Notation

1. A root is a fractional power. The \(n\)th root of \(x^m\) is \(x\) to the power \(m/n\).

   \[\sqrt[n]{x^{m}} = x^{m/n}, \qquad \sqrt{x} = x^{1/2}\]

2. A power in the bottom of a fraction moves to the top, and its exponent changes sign.

   \[\dfrac{1}{x^{k}} = x^{-k}\]

3. Divide a sum term by term. When you divide powers of the same base, subtract the exponents. When you multiply them, add the exponents. The orange dashed curve is the result, \(x^{1/2} + 2x^{-1/2}\).

   \[\begin{aligned} \dfrac{a+b}{c} &= \dfrac{a}{c} + \dfrac{b}{c} \\ \dfrac{x^m}{x^n} &= x^{m-n} \\ x^m x^n &= x^{m+n} \end{aligned}\]

4. Integrate each term with the power rule. The rule needs \(k \neq -1\), because it divides by \(k + 1\), and you cannot divide by 0.

   \[\int \cy{x^{k}}\,dx = \dfrac{x^{k+1}}{k+1} + C, \quad k \neq -1\]

When it fails

If the integrand is \(\dfrac{x+2}{x}\), it splits into \(1 + 2x^{-1}\), so the second term has \(k = -1\). The power rule would divide by \(k + 1 = 0\), so it fails for this term.

The green region's area comes from a logarithm instead, because an antiderivative of \(x^{-1}\) is \(\ln|x|\).

Example 1: a product and a quotient

Evaluate \(\displaystyle\int_1^4 \left(x\sqrt{x} + \dfrac{4}{x^2}\right)dx\).

1. Write the root as a power. When you multiply powers of \(x\), add the exponents.

   \[x\sqrt{x} = x^{1}\cdot x^{1/2} = x^{3/2}\]

2. Write the fraction as a negative power. The orange dashed curve is the sum of the two powers, and it lies on the chalk curve.

   \[\dfrac{4}{x^2} = 4x^{-2}\]

3. Integrate each term with the power rule. For \(4x^{-2}\) the new exponent is \(-1\), so the term becomes \(4x^{-1} \div (-1) = -4x^{-1}\).

   \[\begin{aligned} &\int \left(x^{3/2} + 4x^{-2}\right)dx \\ &\quad = \dfrac{2}{5}x^{5/2} - 4x^{-1} + C \end{aligned}\]

4. Evaluate from 1 to 4, where \(4^{5/2} = 32\). The teal and green areas, \(\tfrac{62}{5}\) and 3, add up to the same \(\tfrac{77}{5}\).

   \[\begin{aligned} \left[\dfrac{2}{5}x^{5/2} - \dfrac{4}{x}\right]_1^4 &= \dfrac{59}{5} + \dfrac{18}{5} \\ &= \cb{\dfrac{77}{5}} \end{aligned}\]

Example 2: a square

Evaluate \(\displaystyle\int_0^4 \left(\sqrt{x} + 1\right)^2 dx\).

1. The power rule integrates single powers of \(x\), not the square of a sum, so expand the square first.

2. A tempting wrong move: Square each term of the sum.

   \[\left(\sqrt{x}+1\right)^2 = \cr{x + 1}\]

   At \(x = 4\) the chalk curve has height \((2+1)^2 = 9\), and the red line \(x + 1\) has height 5. The red line is lower than the chalk curve for every \(x > 0\), so the area under it is too small.

3. A square of a sum has three terms: the two squares and twice the product. The missing middle term is \(2 \cdot \sqrt{x} \cdot 1 = 2x^{1/2}\), and with it the orange dashed curve lies on the chalk curve.

   \[\left(\sqrt{x}+1\right)^2 = x + 2x^{1/2} + 1\]

4. Integrate each term and evaluate from 0 to 4, where \(4^{3/2} = 8\) and every term is 0 at \(x = 0\). The three regions have areas 8, \(\tfrac{32}{3}\) and 4.

   \[\begin{aligned} &\left[\dfrac{x^2}{2} + \dfrac{4}{3}x^{3/2} + x\right]_0^4 \\ &\quad = 8 + \dfrac{32}{3} + 4 = \cb{\dfrac{68}{3}} \end{aligned}\]

Practice

Evaluate \(\displaystyle\int_1^8 \dfrac{2x+5}{\sqrt[3]{x}}\,dx\).

Write the cube root as a power of \(x\). Then divide each term of the top by it.

Questions

The graph shows a curve \(y = 2x^{k}\) for \(x > 0\). The orange dots mark its heights at \(x = 1\) and \(x = 4\).

1. Which formula gives the height of the curve at \(x = 1\) and at \(x = 4\)?

   • \(2x^{-1/2}\)
   • \(2x^{1/2}\)
   • \(2x^{-2}\)
   Answer

   \(2x^{-1/2}\)

2. The teal region from \(x = 1\) to \(x = 4\) fits between the two faint rectangles, which are 3 wide and 1 and 2 high. Which value is its area \(\int_1^4 2x^{-1/2}\,dx\)?

   • 4
   • 2
   • \(-4\)
   Answer

   4

3. Which function is an antiderivative of \(\dfrac{x^3+6}{x^2}\)?

   • \(\dfrac{x^2}{2} - \dfrac{6}{x}\)
   • \(\dfrac{x^2}{2} + \dfrac{6}{x}\)
   • \(\dfrac{x^4/4 + 6x}{x^3/3}\)
   Answer

   \(\dfrac{x^2}{2} - \dfrac{6}{x}\)

Summary

You write every root as a fractional power, every power in the bottom as a negative power, and you expand products and squares.

The sum of powers has the same height as the chalk curve at every x, so the power rule can integrate it term by term.

The power rule handles every exponent except −1, and the exponent −1 needs a logarithm.

This is one of four skills in topic 6.10 of Which & Why: AP Calculus.