Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 54

$\dfrac{5}{3}$

The area by using two integrals is given by: $A=\int_{-(\pi/4)}^{0} \sec^2 t dt+\int_{0}^{1} (1-t^2)dt$ $\implies \int_{-(\pi/4)}^{0} \sec^2 t dt+\int_{0}^{1} (1-t^2)dt=[\tan t]_{-(\pi/4)}^{0}+[t-t^3]_{0}^{1}$ $\implies A=0-(-1)+(1-\dfrac{1}{3})$ $\implies A=1+\dfrac{2}{3}=\dfrac{5}{3}$