Question

The area of the region enclosed by the curves \[y = xlogx\;\] and \[y = 2x - 2{x^2}\] is
A) $\dfrac{5}{{12}}$
B) $\dfrac{7}{{12}}$
C) $1$
D) $\dfrac{4}{7}$

Answer 1

Hint: In this question, we are given the equation of the curves and we have to find the area bounded by them. Firsly, plot the graph of the curves by taking $x = 0$ then $y = 0$. Then, calculate the point of minima of the first curve i.e., \[y = xlogx\;\] by differentiating the equation of curve with respect to $x$ and equating it to $0$. Now, to find the area of the bounded region integrate the difference of the second curve and the first curve with respect $dx$ from $x = 0$ to $x = 1$. In last, apply the ILATE rule of integration to solve integrate and then resolve the limits.



Formula Used:Chain rule of differentiation –
$\dfrac{d}{{dx}}f\left( x \right)g\left( x \right) = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)$
ILATE Rule of integration
Here, the full form of ILATE is Inverse, Logarithms, Algebraic, Trigonometric, Exponential. We use this series to select which of the first will be $\left( {u\left( x \right)} \right)$ or $\left( {v\left( x \right)} \right)$. Here, the first function will be the function which comes first in this series.
$\int {u(x)v\left( x \right)dx = u\left( x \right)\int {v\left( x \right)dx} - \int {\left( {\dfrac{{du\left( x \right)}}{{dx}}\int {v\left( x \right)dx} } \right)} dx} $
Lorartithm formula –
$\log 1 = 0$
Integration formula –
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $



Complete step by step solution:Given that,
Equation of the two curves are \[y = xlogx\;\]and \[y = 2x - 2{x^2}\]
Graph of the given equations is attached below (figure 1);


Figure 1: A Graph contains the equation of the given curves
Now, as you can see that in graph curve \[y = xlogx\;\] indicates that $x > 0$. But for $0 < x < 1$, $x\log x < 0$.
Similarly, for $x > 1$, $x\log x > 0$
Now, at $x\log x = 0$ we get $x = 1$
Further for the point of minima on the curve \[y = xlogx\;\] (also written as \[y = xlo{g_e}x\]), Differentiate the curve with respect to $x$ and equate it to $0$
It will be, $\dfrac{d}{{dx}}\left( {xlo{g_e}x} \right) = 0$
Applying the chain rule of differentiation $\dfrac{d}{{dx}}f\left( x \right)g\left( x \right) = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)$
We get, \[1 + {\log _e}x = 0\]
It Implies that, $x = \dfrac{1}{e}$
Therefore, the area of the bounded region will be the integration of the difference of \[y = 2x - 2{x^2}\] and \[y = xlogx\;\] with respect to $dx$ from $x = 0$ to $x = 1$
Thus, the required area will be
$A = \int\limits_0^1 {\left( {2x - 2{x^2}} \right) - \left( {x\log x} \right)dx} $
Also, written as $A = \int\limits_0^1 {\left( {2x - 2{x^2}} \right)dx} + \int\limits_0^1 {x\log xdx} $
As we know that, $\int {cdx = cx} $ where $c$ is the constant and $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
So, $A = \left[ {\dfrac{{2{x^2}}}{2} - \dfrac{{2{x^3}}}{3}} \right]_0^1 + \int\limits_0^1 {x\log xdx} $
Applying ILATE rule, i.e., $\int {u(x)v\left( x \right)dx = u\left( x \right)\int {v\left( x \right)dx} - \int {\left( {\dfrac{{du\left( x \right)}}{{dx}}\int {v\left( x \right)dx} } \right)} dx} $
We get, $A = \left[ {\dfrac{{2{x^2}}}{2} - \dfrac{{2{x^3}}}{3}} \right]_0^1 + \left[ {\dfrac{{{x^2}}}{2}\log x - \dfrac{{{x^2}}}{4}} \right]_0^1$
On resolving the limits, we get
$A = \dfrac{{2{{\left( 1 \right)}^2}}}{2} - \dfrac{{2{{\left( 1 \right)}^3}}}{3} - \left( {\dfrac{{2{{\left( 0 \right)}^2}}}{2} - \dfrac{{2{{\left( 0 \right)}^3}}}{3}} \right) + \left[ {\dfrac{{{{\left( 1 \right)}^2}}}{2}\log x - \dfrac{{{{\left( 1 \right)}^2}}}{4} - \left( {\dfrac{{{{\left( 0 \right)}^2}}}{2}\log \left( 0 \right) - \dfrac{{{{\left( 0 \right)}^2}}}{4}} \right)} \right]$
On resolving, we get $A = \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{7}{{12}}$



Option ‘B’ is correct

Note: Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces