Integrals

For 0<a<1, the value of the integral 0 dx1-2a x+a2 is :

1. 2 +a2
2. 2 -a2
3. 1-a2
4. 1+a2

Answer

(3) 1-a2

Solution

I = 0 dx1 - 2a x + a2; by 0 f(x) dx = 0 f( - x) dx, I = 0 dx1 + 2a x + a2

Adding the two forms gives 2I = 0 2(1 + a2) dx(1 + a2)2 - 4a2 2x.

Since this integrand depends only on 2x, it is symmetric about 2, so I = 0 /2 2(1 + a2) dx(1 + a2)2 - 4a2 2x.

Dividing by 2x and using 2x = 1 + 2x, (1 + a2)2 - 4a2 = (1 - a2)2:

I = 0 /2 2(1 + a2) 2x dx(1 + a2)2 2x + (1 - a2)2

Now put t = x (t: 0 ), which gives I = 21 + a2 0 dtt2 + ( 1 - a21 + a2 )2 = 21 + a2 1 + a21 - a2 2 = 1 - a2

(1), (2) and (4) do not equal 1 - a2 (at a 0 the integral is , which (1) and (2) do not give), so they are wrong.

Answer: (3)