Some Properties of Definite Integrals: Overview, Questions, Preparation

Property 6:  ${\int }_0^{\mathrm{2a}\mathrm{<br>}}$f(x) dx = 2 ${\int }_0^a$f(x) dx; if f(2a - x) = f(x)

  ${\int }_0^{\mathrm{2a}\mathrm{<br>}}$f(x) dx =  0; if f(2a - x) = -f(x) 

Proof: 

Let's assume a definite integral function ${\int }_0^{2a}\text{<br>}$f(x) dx, we can use property 5 to write it as:

${\int }_0^{\mathrm{2a}\mathrm{<br>}}$f(x) dx = ${\int }_0^a$f(x) dx +  ${\int }_0^a$(2a - x) dx  .............(1)

Suppose f(2a - x) = f(x) 

Then, by substituting the value we get:

${\int }_0^{2a}$f(x) dx = ${\int }_0^a$f(x) dx +  ${\int }_0^a$f(x) dx ⇒ ${\int }_0^{2a}$f(x) dx = 2  ${\int }_0^a$f(x) dx

If f(2a - x) = -f(x),

The above function will result in,

${\int }_0^{2a}$f(x) dx= ${\int }_0^a$f(x) dx -  ${\int }_0^a$f(x) dx ⇒ ${\int }_0^{2a}$f(x) dx = 0

Property 7:  ${\int }_{-a}^a$f(x) dx = 2 ${\int }_0^a$f(x) dx; if a function is even i.e. f(-x) = f(x)         

                    ${\int }_{-a}^a$f(x) dx = 0; if a function is odd i.e. f(-x) = -f(x)

Proof: 

We know, as per property 3:

${\int }_{-a}^a$f(x) dx =  ${\int }_{-a}^0$f(x) dx +  ${\int }_0^a$f(x) dx  .........(1)

Suppose 

${\int }_{-a}^0$f(x) dx = I₁ ......(2)

Now, assume y is a variable such that x + y = 0

⇒ x = -y

⇒ dx = -dy

If x = -a, then y= -(-a) = a  .... ...  .... (i)

And if x = 0, then y = 0  ........  ... .....    .... (ii)

If we substitute the values in equation (2), we get, 

I₁ =  $-{\int }_a^0$f(-y)dy ⇒ I₁ =  ${\int }_0^a$f(-y)dy ( Proprety 2)

⇒ I₁ =  ${\int }_0^a$f(-x)dx

Using I_{1 in equation (1),}

${\int }_{-a}^a$f(x) dx =  ${\int }_0^a$f(-x) dx + ${\int }_0^a$f(x) dx   ...   ....   ...  (3)

If f(-x) = f(x):

Then, 

${\int }_{-a}^a$f(x) dx =  ${\int }_0^a$f(x) dx +  ${\int }_0^a$f(x) dx = 2 ${\int }_0^a$f(x) dx

If  f(-x) = -f(x)

Then,

${\int }_{-a}^a$f(x) dx = - ${\int }_0^a$f(x) dx + ${\int }_0^a$f(x) dx = 0

Solved Example 1: $I=\displaystyle\int_{0}^{2} x(1-x)^{100}\,dx$

Solution: Suppose $t=1-x$ then, $x=1-t$⇒ $dt=-dx$

Lower limits: $x=0\Rightarrow t=1$

$x=2\Rightarrow t=-1$

Then, 

$$I=\int_{1}^{-1} (1-t)t^{100}(-dt)=\int_{-1}^{1} (1-t)t^{100}\,dt =\int_{-1}^{1} \big(t^{100}-t^{101}\big)\,dt.$$

We know that, $t^{100}$is even and $t^{101}$is odd,

$${\int }_{-1}^1{t}^{100}dt=2{\int }_0^1{t}^{100}dt=2\cdot \frac{1}{101}=\frac{2}{101},$$

$$\int_{-1}^{1} t^{100}\,dt=2\int_{0}^{1} t^{100}\,dt=2\cdot\frac{1}{101}=\frac{2}{101}, \qqtad \int_{-1}^{1} t^{101}\,dt=0.$$Adding both the values we get,

$$I=\frac{2}{101}\mathrm{.}$$

Solved Example 2: I =  ${\int }_0^{50}\text{<br>}$[x] dx

Solution: If we break this function using property 3, we get...

$I={\displaystyle {\int }_0^{50}\mathrm{[x}]\text{<br>}d\mathrm{x\; =}}$ ${\int }_0^1\text{<br>}$0 dx +  ${\int }_1^2\text{<br>}$1 dx +  ${\int }_2^3\text{<br>}$2 dx + .............  + ${\int }_{49}^{50}\text{<br>}$50dx

= 0 + (2 - 1) + 2(3 - 2) + ....... + 49(50- 49)

= 0 + 1 + 2 +...   ....   ...+ 50

= 1275

Here is the complete Class 12 study material based on the NCERT textbook.