If 0 4 2x1+ x x dx= 1a e ( a3 )+ b 3, where a, b , then a+b is equal to .
Answer
8
Solution
Since 2 2x = 1 - 2x and 2(1 + x x) = 2 + 2x, the integrand is 1 - 2x2 + 2x, so I = I1 - I2 with
I1 = 0 /4 dx2 + 2x and I2 = 0 /4 2x dx2 + 2x
In I2 the numerator is half the derivative of the denominator, hence I2 = 12 [ e(2 + 2x) ]0 /4 = 12 e 32
In I1 write 2x = 2 x1 + 2x and put x = t, so that 2x dx = dt:
I1 = 01 dt2t2 + 2t + 2 = 12 01 dt (t + 12 )2 + 34
= 1 3 [ -1 2t + 1 3 ]01 = 1 3 ( 3 - 6 ) = 6 3
Thus I = 12 e ( 23 ) + 6 3, which gives a = 2, b = 6.
a + b = 8
Answer: 8