Chapter 6, Problem 48RE

To determine

To find: the function of $y\left(x\right)$ as a definite integral.

Answer to Problem 48RE

The function of $y\left(x\right)$ as a definite integral is: ${{\displaystyle \int }}_5^x\frac{\sin t}{t}dt+3$

Explanation of Solution

Given information:

Given expression is: $\frac{dy}{dx}=\frac{\sin x}{x}$ and $y(5)=3$

Calculation:

Given that,

  $\frac{dy}{dx}=\frac{\sin x}{x}$

  $y(5)=3$

By fundamental theorem of calculus, part $1,$ it is given that-

If $f$ is continuous on $[a,b],$ then

  $\begin{array}{cc}\frac{dF}{dx}& =\frac{d}{dx}{{\displaystyle \int }}_a^{x}f(t)dt\\ & =f(x)\end{array}$

Proof-

Apply the definite of the derivative directly to the function $F$ . That is,

  $\begin{array}{cc}\frac{dF}{dx}& =\underset{h\to 0}{\lim }\frac{F(x+h)-F(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{{{\displaystyle \int }}_{x+h}^{++}f(t)dt-{{\displaystyle \int }}_a^{x}f(t)dt}{h}\\ & =\underset{h\to 0}{\lim }\frac{{{\displaystyle \int }}_{x+h}^{x}f(t)dt}{h}\\ & =\underset{h\to 0}{\lim }\left[\frac{1}{h}{{\displaystyle \int }}_x^{x+h}f(t)dt\right]\end{array}$

The expression in brackets in the last line is the average value of $f$ from $x$ to $x+h,$ we know

from the mean value theorem for definite integrals. That $f$ being continuous, takes on its average

value at least once in the interval; that is,

  $\frac{1}{h}{{\displaystyle \int }}_x^{x+h}f(t)dt=f(c)$ for some $c$ between $x$ and $x+h,$

So,

  $\frac{dF}{dx}=\underset{h\to 0}{\lim }\left[\frac{1}{h}{{\displaystyle \int }}_x^{x+h}f(t)dt\right]$

  $=\underset{h\to 0}{\lim }f(c),$ where $c$ lies between $x$ and $x+h,$

What happens to $c$ as $h$ goes to zero? As $x+h$ gets closer to $x$ it carries $c$ along with it like a bead on a wire, forcing $c$ to approach $x$ ,since $f$ is continuous, this means that $c$ approaches $=f(x)$ :

  $\underset{h\to 0}{\lim }f(c)=f(x)$

Putting it all together,

  $\begin{array}{cc}\frac{dF}{dx}& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{{{\displaystyle \int }}_{x+h}^{x+h}f(t)dt}{h}\end{array}$

  $=\underset{h\to 0}{\lim }f(c)$ for some $c$ between $x$ and $x+h$

  $=f(x)$

So,

  $\frac{d}{dx}{{\displaystyle \int }}_a^{x}f(t)dt=f(x)$

Use above proof to express $\frac{dy}{dx}$ as a definite integral,

Here,

  $\frac{dy}{dx}=\frac{\sin x}{x}$

  $y(5)=3$

Hence,

  $f(t)=\frac{\sin t}{t}$

  $a=3$

So,

  $y(x)={{\displaystyle \int }}_5^x\frac{\sin t}{t}dt+3$

Hence, the function of $y\left(x\right)$ as a definite integral is: ${{\displaystyle \int }}_5^x\frac{\sin t}{t}dt+3$