Chapter 6.3, Problem 43E

To determine

To find: The lower bound for the given expression value.

Answer to Problem 43E

The lower bound for the given is $\frac{7}{6}$ .

Explanation of Solution

Giveninformation :

  $\sec x\ge 1+\left(\frac{x^2}{2}\right)$

Calculation :

Since it has been given that

  $\sec x\ge 1+\left(\frac{x^2}{2}\right)$

Take integral both sides with interval $\left[0,1\right]$

So,

  ${{\displaystyle \int }}_0^1\sec xdx\ge {{\displaystyle \int }}_0^1\left[1+\left(\frac{x^2}{2}\right)\right]dx$

Now, calculate right-side integral

Since $1+\frac{x^2}{2}$ is the rate of change of the quantity

  $F(x)=x+\frac{x^3}{6}$

That is,

  $F^{\prime }(x)=1+\frac{x^2}{2}$

Since,

  ${{\displaystyle \int }}_a^{x}f(t)dt=F(x)-F(a)$

Where F is any anti-derivative of f

So,

  $\begin{array}{cc}{{\displaystyle \int }}_0^1\left[1+\frac{x^2}{2}\right]dx& =\left[1+\frac{1^3}{6}\right]-\left[0+\frac{0^3}{6}\right]\\ & =1+\frac{1}{6}\\ & =\frac{7}{6}\end{array}$

Hence,

  $\begin{array}{l}{{\displaystyle \int }}_0^1\sec xdx\ge {{\displaystyle \int }}_0^1\left[1+\left(\frac{x^2}{2}\right)\right]dx\hfill \\ {{\displaystyle \int }}_0^1\sec xdx\ge \frac{7}{6}\hfill \end{array}$