Definite Integral - Common Tricks


In this section we will cover a few tricks that you can use for solving definite integrals. These tricks can provide easy solutions to some otherwise very complex integrations. We will understand the proof each method with some examples.

King's Rule

The kings rule involves a variable substitution base on the given limits of the integral.
The rule is simple and states that:
$\displaystyle \int \limits_a^b f(x) dx = \int \limits_a^b f(a + b - x) dx$

We will first look at the proof of this followed by a few examples.
Let
$I = \displaystyle \int \limits_a^b f(x) dx$

Substituting $u = a + b - x$
we get $du = -dx$ and the limits as $x = a, u = b$ and $x = b, u = a$ 
which gives us:
$I = \displaystyle \int \limits_b^a f(a + b - u) (-du)$
$\Rightarrow I = \displaystyle -\int \limits_b^a f(a + b - u) (du)$

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Switching the limits and removing the negative sign, we get:
$\displaystyle \int \limits_a^b f(a + b - u) (du)$
Changing the name of the variable from $u$ to $x$, we get:
$I = \displaystyle \int \limits_a^b f(a + b - x) (dx)$

Therefore,
$\displaystyle \int \limits_a^b f(x) dx = I = \displaystyle \int \limits_a^b f(a + b - x) (dx)$

$\therefore \displaystyle \int \limits_a^b f(x) dx = \displaystyle \int \limits_a^b f(a + b - x) (dx)$

Now, try the following example:
--------- Reference to question: b791971d-d93e-4bb8-bb4b-ec7e22023bbd ---------



King's Rule (Extended)

This rule is a direct consequence of the King's rule explained in the previous section.
This rule says that the integration:

$\displaystyle \int \limits_a^b \dfrac{f(x)}{f(x) + f(a+b -x)} dx = \int \limits_a^b \dfrac{f(a + b - x)}{f(x) + f(a+b -x)} dx = \dfrac{a - b}{2}$

Let:
$I = \displaystyle \int \limits_a^b \dfrac{f(x)}{f(x) + f(a+b -x)} dx$      $...(i)$

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Using the King's rule from the previous section:
$I = \displaystyle \int \limits_a^b \dfrac{f(a + b - x)}{f(a + b - x) + f(a+b -(a + b -x))} dx$

$\Rightarrow I = \displaystyle \int \limits_a^b \dfrac{f(a + b - x)}{f(a + b - x) + f(x)} dx$

Adding eqns $(i)$ and $(ii)$, we get:
$2I = \displaystyle \int \limits_a^b \dfrac{f(x)}{f(x) + f(a+b -x)} dx + \int \limits_a^b \dfrac{f(a + b - x)}{f(a + b - x) + f(x)} dx$

$\Rightarrow 2I = \displaystyle \int \limits_a^b \dfrac{\cancel{f(x) + f(a + b - x)}}{\cancel{f(x) + f(a+b -x)}} dx$

$\Rightarrow 2I = \displaystyle \int \limits_a^b 1 dx = x \Biggr|_a^b$

$I = \dfrac{b-a}{2}$



Let us try the next example:

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 Rules and Methods of Integration