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Wize AP Calculus (AB) Textbook > Integration & Accumulation of Change

Integrating Even and Odd Functions

The Fundamental Theorem of Calculus - Part 2Previous Section
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Integrating Even and Odd Functions


Wize Concept
A function f(x)f\left(x\right)is odd if f(−x)=−f(x)f(-x)=-f(x). Odd functions are symmetric about the origin.

A function f(x)f\left(x\right)is even if f(−x)=f(x)f(-x)=f(x). Even functions are symmetric about the y-axis.

Integrating Odd Functions on Symmetric Intervals

Since on one side of the origin the function is negative but has the same y-values as on the other side of the origin, areas under the curve cancel out over symmetric intervals:



If f(x)f(x)is an odd function then
∫−aaf(x) dx=0\boxed{ \displaystyle \int_{-a}^a f(x) \ dx=0}


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Integrating Even Functions on Symmetric Intervals

Since for corresponding (±\pm) x-values the graph has the same y-values on both sides of the y-axis, areas under the curve on each side are the same over symmetric intervals.



If f(x)f(x)is an even function then
∫−aaf(x) dx=2∫0af(x) dx\boxed{\int_{-a}^a f(x)\ dx=2\int_0^a f(x)\ dx}




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Example: Integrating Even/Odd Functions

Evaluate the following definite integral

∫−11(5x5+2x4) dx\displaystyle \int_{-1}^1 (5x^5+2x^4)\ dx

Notice that multiplying an even or an add function does not change whether it is even or odd, so we can use the fact that x5x^5 is odd and x4x^4 is even to make this easier:
∫−11(5x5+2x4) dx=∫−115x5 dx+∫−112x4 dx=0+2∫012x4 dx=2[2x5/5]01=2(2/5)=45\begin{array}{rl}\displaystyle \int_{-1}^1 (5x^5+2x^4)\ dx &= \displaystyle \int_{-1}^1 5x^5 \ dx + \int_{-1}^1 2x^4\ dx \\[+1em] & = 0 + \displaystyle 2\int_{0}^1 2x^4 \ dx \\[+1em] & = 2[2x^5/5]_0^1 \\[+1em] & = 2(2/5) \\[+1em] &=\displaystyle \frac{4}{5} \end{array}

Evaluate the following definite integral

∫−ππ(2sin⁡3x−4cos⁡x) dx\displaystyle \int_{-\pi}^{\pi} (2\sin^3x - 4\cos x)\ dx