#### To determine

**To evaluate:**

∫014t2+1dt

#### Answer

∫014t2+1dt= π

#### Explanation

**1) Concept:**

The Fundamental Theorem of Calculus: Suppose f is continuous on [a, b], then

∫abfxdx=Fb-F(a), where F is antiderivative of f, that is F'=f.

**3) Formula:**

ddxtan-1x=11+x2

**4) Calculations:**

Consider,

∫014t2+1dt

Using the Fundamental theorem of Calculus and anti derivative of 11+x2 we have

*∫014t2+1dt=4∫011t2+1dt=4tan-11-4tan-10*

Substituting the values for tan-11 and tan-10,

4π4-40=π

**Conclusion:**

Therefore,

**∫014t2+1dt= π**