Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 401: 79

a.Because $x\geq 0$,$x^3\geq 0$ $1+x^3\geq 1$ Because $y=\sqrt x$ is an increasing function. then $\sqrt{1+x^3}\geq1 $ Let $1+x^3=t$ Let $y=t-\sqrt t$,$t\geq 1$ $y(1)=0$ $y'=1-\frac{1}{2\sqrt t}=\frac{2\sqrt t-1}{2\sqrt t}$ Because $t\geq 1,y'>0$ $t\in(1,+\infty)$,y is an increasing function. so $y>y(1)=0$ for $t\geq 0$ So $t-\sqrt t \geq 0$ $t\geq\sqrt t$ Thus,$1+x^3\geq \sqrt{1+x^3}$ In conclusion ,$1\leq \sqrt{1+x^3}\leq 1+x^3$ b.Because $1\leq \sqrt{1+x^3}\leq 1+x^3$ then $\int_0^11dx\leq \int_0^1\sqrt{1+x^3}dx\leq \int_0^11+x^3dx$ $\int_0^11dx=1$ $\int_0^11+x^3dx=x+\frac14x^4\vert_0^1=1+\frac14=\frac54=1.25$ so $1\leq \int_0^1\sqrt{1+x^3}dx\leq 1.25$

a.Because $x\geq 0$,$x^3\geq 0$ $1+x^3\geq 1$ Because $y=\sqrt x$ is an increasing function. then $\sqrt{1+x^3}\geq1 $ Let $1+x^3=t$ Let $y=t-\sqrt t$,$t\geq 1$ $y(1)=0$ $y'=1-\frac{1}{2\sqrt t}=\frac{2\sqrt t-1}{2\sqrt t}$ Because $t\geq 1,y'>0$ $t\in(1,+\infty)$,y is an increasing function. so $y>y(1)=0$ for $t\geq 0$ So $t-\sqrt t \geq 0$ $t\geq\sqrt t$ Thus,$1+x^3\geq \sqrt{1+x^3}$ In conclusion ,$1\leq \sqrt{1+x^3}\leq 1+x^3$ b.Because $1\leq \sqrt{1+x^3}\leq 1+x^3$ then $\int_0^11dx\leq \int_0^1\sqrt{1+x^3}dx\leq \int_0^11+x^3dx$ $\int_0^11dx=1$ $\int_0^11+x^3dx=x+\frac14x^4\vert_0^1=1+\frac14=\frac54=1.25$ so $1\leq \int_0^1\sqrt{1+x^3}dx\leq 1.25$