Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 328: 46

$$\sqrt{3}$$

Given $$y=\sec^2 (x), \ \ 0\leqslant x \leqslant \pi/3$$ From the graph of the region, we can observe that area of the bounded region approximately equal to ( area of the rectangle with width 1 and length $\pi/3$) +$\frac{1}{10}$ ( area of the rectangle with width 3 and length $\pi/3$) $$\text{Area} \approx \frac{1}{10} (3)(\pi/3)+ (\pi/3)( 1)\approx1.36$$ Now we use integration \begin{aligned} \text{Area}&= \int_0^{\pi/3}\sec^2{x}dx\\ &= \tan(x)\bigg|_{0}^{\pi/3} \\ &=\sqrt{3}\end{aligned}