Wemazi

Topic 6.6

Use symmetry

Builds on applying linearity to integrals. 8 minutes.

Two parts that cancel

The figure shows the curve \(f(x) = x^3\) and the region between it and the axis from \(-b\) to \(b\).

Move \(b\) to change the width. The right part, from 0 to \(b\), lies above the axis, so it is teal. At \(b = 2\), its area is 4.

The left part, from \(-b\) to 0, lies below the axis, so it is red and counts as negative. At \(b = 2\), its signed area is \(-4\).

The two parts have the same size and opposite signs, so they cancel: \(\int_{-2}^{2} x^3\,dx = 4 + (-4) = 0\).

Even and odd functions

An odd function looks the same after a half-turn about the origin, so \(f(-x) = -f(x)\). Move \(b\). The left part keeps the same size as the right part and the opposite sign. The two parts cancel, so the integral from \(-b\) to \(b\) is 0.

Press Make the curve even. An even function is its own mirror image across the vertical axis, so \(f(-x) = f(x)\). The left part now copies the right part and its sign, so the integral from \(-b\) to \(b\) is twice the right part's integral.

An interval from −b to b is symmetric about 0. Both rules need one, because then every point on the left has a matching point on the right.

Notation

  1. A function is even if replacing \(x\) by \(-x\) does not change it, and odd if replacing \(x\) by \(-x\) changes its sign.

    \[\begin{aligned} \text{even: } f(-x) &= f(x) \\ \text{odd: } f(-x) &= -f(x) \end{aligned}\]

  2. To test a formula, replace \(x\) by \(-x\). For \(\tfrac{x^3}{2}\), the result is \(\tfrac{(-x)^3}{2} = -\tfrac{x^3}{2}\), so it is odd. For \(x^2\), the result is \((-x)^2 = x^2\), so it is even.

  3. If \(f\) is even, its integral over a symmetric interval is twice its integral over the right half, from 0 to \(a\).

    \[\int_{-a}^{a} f(x)\,dx = 2\,\cb{\int_0^a f(x)\,dx}\]

  4. If f is odd, its integral over a symmetric interval is 0, because its two parts cancel.

    \[\int_{-a}^{a} f(x)\,dx = 0\]

When it fails

The rules need a function that is even or odd. Take \(f(x) = \tfrac{x^3}{2} + 1\). Then \(f(-x) = -\tfrac{x^3}{2} + 1\), which is neither \(f(x)\) nor \(-f(x)\).

The odd part \(\tfrac{x^3}{2}\) cancels. The constant 1 adds the green band, 1 tall and 4 wide, so the integral from \(-2\) to 2 is 4. The odd rule gives 0, and the even rule doubles the right half to \(2(4) = 8\). Neither is correct.

Example 1: an even part and an odd part

Evaluate \(\displaystyle\int_{-2}^{2} (6x^2 + x^5)\,dx\).

  1. Split the integral into the two terms. The interval is the same for both, and the green band shows the term \(6x^2\).

    \[\int_{-2}^{2} 6x^2\,dx + \int_{-2}^{2} x^5\,dx\]

  2. The term \(x^5\) is odd, because \((-x)^5 = -x^5\). Its red part on the left cancels its teal part on the right, so its integral from \(-2\) to 2 is 0.

    \[\int_{-2}^{2} x^5\,dx = 0\]

  3. The term \(6x^2\) is even, so its integral is twice the integral over the right half. That half is 16, because the antiderivative \(2x^3\) is 16 at \(x = 2\) and 0 at \(x = 0\).

    \[\begin{aligned} \int_{-2}^{2} 6x^2\,dx &= 2\int_0^2 6x^2\,dx \\ &= 2(16) = \cg{32} \end{aligned}\]

  4. Add the two results.

    \[0 + 32 = 32\]

Example 2: an interval that is not symmetric

\(f\) is odd. \(\displaystyle\int_0^3 f(x)\,dx = 5\) and \(\displaystyle\int_0^2 f(x)\,dx = 2\). Find \(\displaystyle\int_{-2}^{3} f(x)\,dx\).

  1. The interval runs from \(-2\) to 3. It is symmetric only from \(-2\) to 2, so the orange piece from 2 to 3 has no matching piece on the left.

  2. A tempting wrong move: The function is odd, so the integral is 0.

    \[\int_{-2}^{3} f(x)\,dx = 0\]

    The odd rule needs an interval from \(-a\) to \(a\). Nothing cancels the orange piece from 2 to 3, which lies above the axis, so the integral is positive.

  3. Split at 2. The interval from −2 to 2 is symmetric, so its integral is 0. The orange piece from 2 to 3 is the integral from 0 to 3 minus the integral from 0 to 2.

    \[\begin{aligned} \int_{-2}^{3} f\,dx &= \int_{-2}^{2} f\,dx + \int_2^3 f\,dx \\ &= 0 + (5 - 2) \end{aligned}\]

  4. Simplify.

    \[\int_{-2}^{3} f(x)\,dx = \cy{3}\]

Practice

\(g\) is even. \(\displaystyle\int_0^5 g(x)\,dx = 7\) and \(\displaystyle\int_0^2 g(x)\,dx = 3\). Find \(\displaystyle\int_{-5}^{2} g(x)\,dx\).

Split at 0. The piece from −5 to 0 is the mirror image of the piece from 0 to 5, so it has the same integral.

Solution

  1. Split at 0. The teal piece runs from −5 to 0, and the orange piece runs from 0 to 2.

    \[\int_{-5}^{2} g\,dx = \int_{-5}^{0} g\,dx + \int_0^2 g\,dx\]

  2. The function is even, so the teal piece mirrors the piece from 0 to 5 and has the same integral, 7.

    \[\int_{-5}^{0} g\,dx = \int_0^5 g\,dx = 7\]

  3. The orange piece is given as 3. Add the two pieces.

    \[7 + 3 = 10\]

Questions

The figure shows the odd function \(f(x) = x^3 - 4x\) from \(-3\) to 3. The signed areas of its four parts are \(-6.25\) on \([-3, -2]\), 4 on \([-2, 0]\), \(-4\) on \([0, 2]\) and 6.25 on \([2, 3]\).

  1. What is \(\displaystyle\int_{-3}^{3} f(x)\,dx\)?

    • 0
    • 4.5
    • 6.25
    Answer

    0

  2. What is \(\displaystyle\int_{-3}^{2} f(x)\,dx\)?

    • \(-6.25\)
    • 0
    • 6.25
    Answer

    \(-6.25\)

  3. What is \(\displaystyle\int_{-2}^{2} (x^4 - 3x^3 + 5)\,dx\)?

    • \(\dfrac{164}{5}\)
    • 0
    • \(\dfrac{82}{5}\)
    Answer

    \(\dfrac{164}{5}\)

Summary

You test whether f is even or odd, then double the right half for an even function and use 0 for an odd function.

This works because the left half is the mirror image of the right half for an even function, and its opposite for an odd function.

You use it when the interval is symmetric about zero, and you split any other interval so that one part is symmetric.

This is one of four skills in topic 6.6 of Which & Why: AP Calculus.