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Wize University Calculus 1 Textbook > Integration Techniques

Integrating Even and Odd Functions

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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
Extra Practice
Practice: Integration Technique (~F2017 Final Q27)
Evaluate ∫−π2π2cos⁡2x sin⁡x dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\ \sin x\ dx.
Practice: Integration Technique (~F2017 Final Q27)
Practice: Integration Technique
Evaluate ∫−π2π2cos⁡2x sin⁡x dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\ \sin x\ dx.
Property of Definite Integral
Suppose f(−x)=−f(x)f(-x)=-f(x) and g(−x)=g(x)g\left(-x\right)=g\left(x\right) for all xx values.
If ∫01f(x)dx=5  and  ∫−10g(x)dx=7\displaystyle\int_0^1f\left(x\right)dx=5\ \ \text{and}\ \ \int_{-1}^0g\left(x\right)dx=7, evaluate ∫−11[2f(x)+3g(x)]dx\displaystyle \int_{-1}^1\left[2f\left(x\right)+3g\left(x\right)\right]dx
Integrating Even and Odd Functions
Evaluate ∫−π2π2cos⁡2xsin⁡x dx\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\sin x \ dx
Evaluate ∫−7π37π3sin⁡x dx\displaystyle\int_{-\frac{7\pi}{3}}^{\frac{7\pi}{3}}\sin x\ dx
Evaluate∫−7070x3+x5 dx\text{Evaluate} \displaystyle \int_{-70}^{70} x^3+x^5\space dx
Integrating Even or Odd Functions
Evaluate
∫−ππ(2sin⁡3x−4cos⁡x) dx.\int_{-\pi}^{\pi} (2\sin^3x - 4\cos x)\ dx.
Integrating Even and Odd Functions
Evaluate ∫−π2π2cos⁡2xsin⁡x dx\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\sin x \ dx
Suppose that f(x)f(x) is even, ∫06f(x)dx=9\int_{0}^{6}f(x)dx=9 and ∫−33f(x)dx=6\int_{-3}^{3}f(x)dx=6 and g(x)g(x) is odd with ∫−36g(x)dx=3\int_{-3}^{6}g(x)dx=3. What is the value of the integral ∫362f(x)−3g(x)dx\int_{3}^{6}2f(x)-3g(x)dx?
Evaluate ∫−7π37π3sin⁡x dx\displaystyle\int_{-\frac{7\pi}{3}}^{\frac{7\pi}{3}}\sin x\ dx
Integrating Even and Odd Functions: Definite Integrals
Evaluate ∫−114−(x+1)2+sin⁡x⋅e2x4 dx\int_{-1}^1\sqrt{4-\left(x+1\right)^2}+\sin x\cdot e^{2x^4}\ dx.
Integrating Even and Odd Functions
Suppose that f(−x)=f(x)f\left(-x\right)=f\left(x\right) and ∫055⋅f(x)dx=9\int_0^55\cdot f\left(x\right)dx=9. Find ∫−553⋅f(x) dx\int_{-5}^53\cdot f\left(x\right)\ dx.
Property of Definite Integral
Suppose f(−x)=−f(x)f(-x)=-f(x) and g(−x)=g(x)g\left(-x\right)=g\left(x\right) for all xx values.
If ∫01f(x)dx=5  and  ∫−10g(x)dx=7\displaystyle\int_0^1f\left(x\right)dx=5\ \ \text{and}\ \ \int_{-1}^0g\left(x\right)dx=7, evaluate ∫−11[2f(x)+3g(x)]dx\displaystyle \int_{-1}^1\left[2f\left(x\right)+3g\left(x\right)\right]dx
Practice: Integration Technique (~F2017 Final Q27)
Practice: Integration Technique
Evaluate ∫−π2π2cos⁡2x sin⁡x dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2x\ \sin x\ dx.