Chapter 6, Problem 50RE

To determine

To calculate: The value of the function $\frac{dy}{dx}=2x$

Answer to Problem 50RE

The value of x equal to 0, y equal to 3

Explanation of Solution

Given information:

The given figures are shown below,

  

Formula used:

  $\frac{dF}{dx}=\underset{h\to 0}{\lim }\frac{F\left(x+h\right)-F\left(x\right)}{h}$

Calculation:

Correct option (b)

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

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

  $\begin{array}{l}\frac{dF}{dx}=\frac{d}{dx}{\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)}dt\\ \frac{dF}{dx}=f\left(x\right)\end{array}$

Proof-

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

  $\begin{array}{l}\frac{dF}{dx}=\underset{h\to 0}{\lim }\frac{F\left(x+h\right)-F\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \underset{a}{\overset{x+h}{\int }}f\left(t\right)dt-{\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}}}{h}\\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \underset{x}{\overset{x+h}{\int }}f\left(t\right)dt}}{h}\\ \frac{dF}{dx}=\underset{h\to 0}{\lim }\left[\frac{1}{h}{\displaystyle \underset{x}{\overset{x+h}{\int }}f\left(t\right)dt}\right]\end{array}$

The expression in brackets in the last line is the average value of f from x to x + h : The mean value theorem for definite integrals. That f , being continuous, takes on its average value at least once in the interval; that is,

  $\frac{1}{h}{\displaystyle \underset{x}{\overset{x+h}{\int }}f\left(t\right)}dt=f\left(c\right)\text{for some c between }x\text{ and }x+h$

So,

  $\begin{array}{l}\frac{dF}{dx}=\underset{h\to 0}{\lim }\left[\frac{1}{h}{\displaystyle \underset{x}{\overset{x+h}{\int }}f\left(t\right)dt}\right]\\ \frac{dF}{dx}=\underset{h\to 0}{\lim }f\left(c\right),\text{where c lies between }x\text{ and }x+h\end{array}$

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 f (c) approaches f (x):

  $\underset{h\to 0}{\lim }f\left(c\right)=f\left(x\right)$

Putting it all together,

  $\begin{array}{l}\frac{dF}{dx}=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \underset{x}{\overset{x+h}{\int }}f\left(t\right)dt}}{h}\\ =\underset{h\to 0}{\lim }f\left(c\right)\text{for some }c\text{ between }x\text{and}x+h\\ \frac{dF}{dx}=f\left(x\right)\end{array}$

So,

  $\frac{d}{dx}{\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)}dt=f\left(x\right)$

Here,

  $\begin{array}{l}\frac{dy}{dx}=2x\\ y\left(1\right)=4\end{array}$

Hence,

  $\begin{array}{l}y={\displaystyle \underset{1}{\overset{x}{\int }}2tdt}+4\\ ={\left[t^2\right]}_1^x+4\\ =\left(x^2\right)-\left(1^2\right)+4\\ y=x^2+3\end{array}$

Observe the graphs drawn in the question, in the graph (b), at x equal to 0, y equal to 3.

Conclusion:

The value of x equal to 0, y equal to 3