#### To determine

**To evaluate:**

∫1/21/241-x2 dx

#### Answer

∫1/21/241-x2 dx=π3

#### 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.

**2) Formula:**

d/dx (sin-1x)=11-x2

**3) Calculations:**

Consider

∫1/21/241-x2 dx

Applying the formula in the above equation ant derivative of 11-x2 is (sin-1x).

Thus using the Fundamental theorem of Calculus

∫1/21/241-x2 dx=4sin-112-4sin-112

Now, substituting the values for sin-112 and sin-112,

=4·π4-4·π6

Simplifying the equation gives

∫1/21/241-x2 dx=π-2π3=π3

**Conclusion:**

Therefore,

∫1/21/241-x2 dx=π3