Integration of even function

• An even function $ f(x) $ satisfies the condition $ f(x) = f(-x) $ for all $ x $ in its domain .This symmetry means that the function’s graph is the same on both sides of the y-axis.
• Therefore The integration of even function over a symmetric interval can be simplified due to the symmetry of the function. .

Integration of even function over a symmetric intervalcan be simplified as follows:

$
\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx
$

This simplification is possible because the area under the curve from $-a$ to $0$ is the same as the area from $0$ to $a$ due to the symmetry of the function.

Example 1
$
\int_{-a}^{a} x^2 \, dx
$

Solution
Since $ x^2 $ is an even function, the integral can be simplified to:

$
2 \int_{0}^{a} x^2 \, dx
$

Calculating this integral gives:

$
= 2 \left[ \frac{x^3}{3} \right]_{0}^{a}
= 2 \left( \frac{a^3}{3} – 0 \right)
= \frac{2a^3}{3}
$

Example 2
$
\int_{-\pi/2}^{\pi/2} \cos^2(x) \, dx
$

Solution
The function $ \cos^2(x) $ is an even function because $ \cos^2(x) = \cos^2(-x) $.

\ Since $ f(x) $ is even, we can simplify the integral:

$
\int_{-\pi/2}^{\pi/2} \cos^2(x) \, dx = 2 \int_{0}^{\pi/2} \cos^2(x) \, dx
$

To integrate $ \cos^2(x) $, use the identity:

$
\cos^2(x) = \frac{1}{2}(1 + \cos(2x))
$

Thus, the integral becomes:

$
2 \int_{0}^{\pi/2} \frac{1}{2}(1 + \cos(2x)) \, dx
$

$
= \int_{0}^{\pi/2} (1 + \cos(2x)) \, dx
= \left[ x + \frac{1}{2} \sin(2x) \right]_{0}^{\pi/2}
$

$
= \left( \frac{\pi}{2} + 0 \right) – \left( 0 + 0 \right)
= \frac{\pi}{2}
$

Solved examples on Integration of even function

Question 1
$
\int_{-1}^{1} x^4 – 6x^2 + 8 \, dx
$

Solution
The function $ f(x) = x^4 – 6x^2 + 8 $ is even since all terms are even powers of $ x $.

$
\int_{-1}^{1} (x^4 – 6x^2 + 8) \, dx = 2 \int_{0}^{1} (x^4 – 6x^2 + 8) \, dx
$

$
= 2 \left[ \frac{x^5}{5} – 2x^3 + 8x \right]_{0}^{1}
$

$
= 2 \left( \frac{1^5}{5} – 2 \cdot 1^3 + 8 \cdot 1 \right) – 0 =62/5
$

Question 2

$
\int_{-\pi}^{\pi} \sin^2(x) \, dx
$
Solution

$ \sin^2(x) $ is even because $ \sin^2(x) = \sin^2(-x) $.

$
\int_{-\pi}^{\pi} \sin^2(x) \, dx = 2 \int_{0}^{\pi} \sin^2(x) \, dx
$

$
\sin^2(x) = \frac{1}{2}(1 – \cos(2x))
$

$
= 2 \int_{0}^{\pi} \frac{1}{2}(1 – \cos(2x)) \, dx
= \int_{0}^{\pi} (1 – \cos(2x)) \, dx
$

$
= \left[ x – \frac{1}{2} \sin(2x) \right]_{0}^{\pi} = \pi
$

Question 3

$
\int_{-1}^{1} \frac {x^3 + |x| + 1}{x^2 + 2|x| +1} \, dx
$

Solution
$
\int_{-1}^{1} \frac {x^3 + |x| + 1}{x^2 + 2|x| +1} \, dx = \int_{-1}^{1} \frac {x^3}{x^2 + 2|x| +1} \, dx + \int_{-1}^{1} \frac {|x| + 1}{x^2 + 2|x| +1} \, dx
$

Now
$\int_{-1}^{1} \frac {x^3}{x^2 + 2|x| +1} \, dx = 0$ as this is odd function
$\int_{-1}^{1} \frac {|x| + 1}{x^2 + 2|x| +1} \, dx= 2\int_{-1}^{1} \frac {|x| + 1}{x^2 + 2|x| +1} \, dx$ as this is even function
Hence

$
=\int_{-1}^{1} \frac { |x| + 1}{x^2 + 2|x| +1} \, dx \\
= 2 \int_{0}^{1} \frac { |x| + 1}{x^2 + 2|x| +1} \, dx \\
=2 \int_{0}^{1} \frac {|x|+1}{(|x|+1)^2} \, dx \\
=2 \int_{0}^{1} \frac {1}{|x| + 1} \, dx \\
= 2 \int_{0}^{1} \frac {1}{x + 1} \, dx \\
=2 \log 2
$

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