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**Hint:**According to the question given in the question we have to determine the region bounded by $y = x$ and $y = {x^2}$. So, first of all we have to determine the values of coordinates of point P as mentioned in the diagram, given below:

Now, we can obtain the coordinates or point of x so, we have to substitute the value of x in the expression to find the required area.

Now, to find the required area we have to find the integration with the limit from 0 to 1 and after finding the integration we can obtain the area required.

**Formula used:**$ \int {xdx = \dfrac{{{x^2}}}{2} + c..................(A)} $

**Complete Step by Step Solution:**

Step 1: First of all to find the coordinates of point P we have to substitute the value of y in the given expression as mentioned in the solution hint and we can also understand it with the help of diagram as mentioned below:

$ \Rightarrow {x^2} = x$

Now, on solving the expression obtained just above,

$ \Rightarrow x = 0,1$

Step 2: As we have obtained the value of $x = 0$ which corresponds to origin as mentioned in the diagram in step 1 O(0,0) and for p we have to put x = 1

Step 3: Now, to obtain the required area we have to find the integration from 0 to 1 as mentioned in the solution hint. Hence,

$A = \int\limits_0^1 {({y_1} - {y_2}} )dx$

Step 4: Now, on substituting the values of the given expressions $y = x$ and $y = {x^2}$. In the expression as obtained in the solution step 3. Hence,

$ = \int\limits_0^1 {x - {x^2}} dx$

Now, to solve the integration as obtained just above we have to apply the formula (A) as mentioned in the solution hint.

\[

= \left[ {\dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3}} \right]_0^1 \\

= \left[ {\dfrac{1}{2} - \dfrac{1}{3} - 0} \right] \\

= \dfrac{{3 - 2}}{6} \\

= \dfrac{1}{6}

\]

Hence, with the help of formula (A) as mentioned in the solution hint we have obtained the required area which is \[ = \dfrac{1}{6}\].

**Therefore correct option is (D)**

**Note:**To find the required area it is necessary to find the coordinates of point P so that we can determine the area covered by the lines $y = x$ and $y = {x^2}$

Integration is the best way to determine the area of undefined shapes by substituting the values of the limits obtained.

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