The integral 0 (x+3) x1+3 2 x dx is equal to
- 3( +1)
- 3 3( +6)
- 3( +2)
- 2 3( +4)
Answer
(2) 3 3( +6)
Solution
Let I = I1 + 3I2, where I1 = 0 x x1+3 2 x dx and I2 = 0 x1+3 2 x dx.
For I2: put x = t, - x dx = dt; x = 0 t = 1, x = t = -1.
I2 = -11 dt1+3t2 = 1 3 [ -1( 3 t) ]-11
= 1 3 ( 3+ 3 ) = 2 3 3
For I1: using 0a f(x) dx = 0a f(a-x) dx with x - x ( x and 2 x do not change):
I1 = 0 ( - x) x1+3 2 x dx = I2 - I1
I1 = 2 I2 = 2 2 3 3 = 23 3
I = 23 3 + 3 2 3 3
= 2 + 6 3 3 = 3 3( +6)
Answer: (2)