Question

Use geometry (i.e., areas of triangles, rectangles, and circles) to find the exact values of each of the definite integrals in Exercises $21-28$.

${\int }_{-1}^1\sqrt{1-x^2}dx$.

Short answer

The exact value of${\int }_{-1}^1\sqrt{1-x^2}dx$is,$\frac{\mathrm{\pi }}{2}$.

Step-by-step solution

Step 1. Given information

${\int }_{-1}^1\sqrt{1-x^2}dx$.

Step 2. The following figure shows the graph of 1-x2.

Step 3. Find the area.

The shaded region can be considered to be a half circle or radius$1$ unit.

The area of the shaded half circle is,

$$\begin{gathered} \frac{{\mathrm{\pi r}}^2}{2}=\frac{\mathrm{\pi }{\left(1\right)}^2}{2} \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Therefore, the exact value is,$\frac{\mathrm{\pi }}{2}$.