Chapter 5.4, Problem 22E

To determine

To find the value of ${\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)$ .

Answer to Problem 22E

The value of ${\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)$ is $\frac{\sqrt{5}}{2}$ .

Explanation of Solution

Given information:

The function are $\mathrm{f}\left(\mathrm{x}\right)={\displaystyle \underset{0}{\overset{\sin \mathrm{x}}{\int }}\sqrt{1+{\mathrm{t}}^2}\mathrm{d}\mathrm{t}}$ and $\mathrm{g}\left(\mathrm{y}\right)={\displaystyle \underset{3}{\overset{\mathrm{y}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}}$

Concept used:

The fundamental theorem of calculus, part 1 is defined by,

If $\mathrm{f}$ is continuous on $\left[\mathrm{a},\mathrm{b}\right]$ , then the function $\mathrm{g}$ defined by,

  $\mathrm{g}\left(\mathrm{x}\right)={\displaystyle \underset{\mathrm{a}}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)}\mathrm{d}\mathrm{t}$

  $\mathrm{a}\le \mathrm{x}\le \mathrm{b}$ is an antiderivative of $\mathrm{f}$ , that is ${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ for $\mathrm{a}<\mathrm{x}<\mathrm{b}$ .

Calculation:

Using Part 1 of the Fundamental Theorem of Calculus, the derivative of the function is obtained as:

  $\mathrm{f}\left(\mathrm{x}\right)={\displaystyle \underset{0}{\overset{\sin \mathrm{x}}{\int }}\sqrt{1+{\mathrm{t}}^2}\mathrm{d}\mathrm{t}}$

By, evaluation theorem,

  $\begin{array}{l}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\displaystyle \underset{0}{\overset{\sin \mathrm{x}}{\int }}\sqrt{1+{\mathrm{t}}^2}\mathrm{d}\mathrm{t}}=\mathrm{F}\left(\sin \mathrm{x}\right)\\ \mathrm{F}\left(\mathrm{x}\right)={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\sqrt{1+{\sin }^2\mathrm{x}}\end{array}$

Now, for $\mathrm{g}\left(\mathrm{y}\right)={\displaystyle \underset{3}{\overset{\mathrm{y}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}}$ ,

  $\mathrm{g}\left(\mathrm{y}\right)={\displaystyle \underset{3}{\overset{\mathrm{y}}{\int }}{\displaystyle \underset{0}{\overset{\sin \mathrm{x}}{\int }}\sqrt{1+{\mathrm{t}}^2}\mathrm{d}\mathrm{t}}\mathrm{d}\mathrm{x}}$

Using Part 1 of the Fundamental Theorem of Calculus,

  ${\mathrm{g}}^{\prime }\left(\mathrm{y}\right)={\displaystyle \underset{3}{\overset{\mathrm{y}}{\int }}\mathrm{F}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}}$ [since, $\sqrt{1+{\mathrm{t}}^2}$ is continuous and $0<\mathrm{x}<\mathrm{b}$ ]

Again, Using Part 1 of the Fundamental Theorem of Calculus,

  $\begin{array}{l}{\mathrm{g}}^{″}\left(\mathrm{y}\right)=\mathrm{F}\left(\mathrm{y}\right)\\ {\mathrm{g}}^{″}\left(\mathrm{y}\right)=\sqrt{1+{\sin }^2\mathrm{y}}\\ {\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)=\sqrt{1+{\sin }^2\left(\frac{\mathrm{\pi }}{6}\right)}\\ {\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)=\sqrt{1+{\left(\frac{1}{2}\right)}^2}\\ {\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)=\sqrt{1+\frac{1}{4}}\\ {\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)=\sqrt{\frac{5}{4}}\\ {\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)=\frac{\sqrt{5}}{2}\end{array}$

Therefore,

The value of ${\mathrm{g}}^{″}\left(\frac{\mathrm{\pi }}{6}\right)$ is $\frac{\sqrt{5}}{2}$ .