Given that k0(x+2)dx=6 , where k>0, find the value of k.?

1 Answer
Apr 16, 2018

The value of k=2

Explanation:

The indefinite integral is

(x+2)dx=12x2+2x+C

The definite integal is

k0(x+2)dx=[12x2+2x]k0

=12k2+2k

But

k0(x+2)dx=6

Therefore,

12k2+2k=6

k2+4k12=0

Solving this quadratic equation in k

k=4±424(1)(12)2

=4±642

k1=6

and

k2=2

As k>0, k=2