To find the values of
a that satisfy the inequality
(√x+1−)dx<4, we need to first evaluate the integral and then solve the inequality.
Consider the integral
(√x+1−)dx We will solve this integral term by term:
For the term
√x :
For the term 1 :
∫1dx=xFor the term
− :
∫−dx=−∫x−dx=−2x1∕2Now, combining these results, we have:
Evaluating this at the bounds
x=1 and
x=a (x+x−2x)|1a=(a+a−2a)−(1+1−2.1) =a3∕2+a−2a1∕2−(1+1−2)=a3∕2+a−2a1∕2−0=a3∕2+a−2a1∕2 Now, plugging this back into the inequality:
(a3∕2+a−2a1∕2)<4 Simplify the expression inside the inequality:
Now, let
√a=t. Hence,
a=t2. Substituting
t into the inequality:
t2+t−2<4Rearranging the inequality:
t2+t−6<0We solve the quadratic equation
t2+t−6=0 by factoring it:
(t+3)(t−2)=0The roots are
t=−3 and
t=2. The quadratic inequality
t2+t−6<0 is satisfied between the roots:
−3<t<2Since
√a=t, this translates to:
−3<√a<2However, since
√a represents a real, non-negative quantity, the inequality reduces to:
0≤√a<2Squaring both sides to return to
a-terms:
0≤a<4The appropriate interval for
a is:
Therefore, the values of
a that satisfy the original inequality lie in the interval
[0,4). Since
a cannot be exactly 0 (as it would make the original integral undefined), the interval we need is:
Option C:
(0,4)