#### To determine

**To find:**

limh→01h∫22+h1+t3dt

#### Answer

limh→01h∫22+h1+t3dt=3

#### Explanation

**1) Concept:**

i) The second fundamental theorem for calculus: If f is continuous on [a, b], then the function is defined by ∫abftdt=Fb-Fa𝑤here F is an antiderivative of f

ii) ∫aafxdx=0

**2) Given:**

limh→01h∫22+h1+t3dt

**3) Calculation:**

The given expression is limh→01h∫22+h1+t3dt

To evaluate the limit, substitute limit

limh→01h∫22+h1+t3dt

=limh→0∫22+h1+t3dth

=∫22+01+t3dt0

=∫221+t3dt0

But ∫aafxdx=0, therefore the value of ∫221+t3dt=0

=00

This is an indeterminate form of limit; therefore apply L’Hospital rule and second fundamental theorem of the derivative of integral. The derivative of denominator is is ddhh=1. And derivative of numerator is ddh∫22+h1+t3dt =1+(2+h)3 So by L’Hospital rule

limh→0∫22+h1+t3dth=limh→01+(2+h)31

=limh→01+(2+h)31

Substitute the limit

=1+(2+0)31

=1+(2)3

=9

=3

**Conclusion:**

Therefore the value of limit is limh→01h∫22+h1+t3dt=3