Question

The area bounded by the curve \[y = {x^4} - 2{x^3} + {x^2} + 3\] the axis of abscissa and two ordinates corresponding to the points of minimum of the function \[y(x)\] is
A.\[\dfrac{{10}}{3}sq.unit\]
B.\[\dfrac{{27}}{{10}}sq.unit\]
C.\[\dfrac{{21}}{{10}}sq.unit\]
D.\[\dfrac{{91}}{{30}}sq.unit\]

Answer 1

Hint: In this problem of area bounded by the curve we will to find the first and second derivative of the given function or curve. To get the points of minima and maxima we will find the first derivative and for finding the points of minima exactly we will find the second derivative.

Complete step-by-step answer:
Given that the curve is
\[y = {x^4} - 2{x^3} + {x^2} + 3\]
To find the first derivative
\[\dfrac{{dy}}{{dx}} = 4{x^3} - 6{x^2} + 2x\]
Now equating this to zero we will get the point of maxima and minima
\[\begin{gathered}
  \dfrac{{dy}}{{dx}} = 0 \\
  4{x^3} - 6{x^2} + 2x = 0 \\
\end{gathered} \]
Taking 2x common from all terms
\[2x\left( {2{x^2} - 3x + 1} \right) = 0\]
So equation in bracket is a quadratic equation .So we have to find the roots of the equation.
\[ \Rightarrow \left( {2{x^2} - 3x + 1} \right) = 0\]
Here we have to find the factors in such a way that -3 is the sum of the two roots and product is 2.
\[ \Rightarrow \left( {2{x^2} - 2x - x + 1} \right) = 0\]
The two numbers are -2 and -1.
Taking 2x common from first two terms and -1 common from rest two terms,
\[ \Rightarrow \left( {2x(x - 1) - 1(x - 1)} \right) = 0\]
The two roots are,
\[ \Rightarrow \left( {2x - 1} \right)\left( {x - 1} \right) = 0\]
So overall roots of the cubic equation above are
\[\left( {2x} \right)\left( {2x - 1} \right)\left( {x - 1} \right) = 0\]
So,
\[2x = 0 \Rightarrow x = 0\]
\[\left( {2x - 1} \right) = 0 \Rightarrow 2x = 1 \Rightarrow x = \dfrac{1}{2}\]
\[x - 1 = 0 \Rightarrow x = 1\]
Thus the roots are \[x = 0\], \[x = \dfrac{1}{2}\] and \[x = 1\].
Now to find the points of minima we will find the second derivative of the function.
\[\dfrac{{{d^2}y}}{{d{x^2}}} = 12{x^2} - 12x + 2\]
Taking 2 common,
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2\left( {6{x^2} - 6x + 1} \right)\]
Now putting the values of roots for second derivative we will find the points of minima
\[ \Rightarrow {\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]_{x = 0}} = 2\left( {6{x^2} - 6x + 1} \right) = 2\left( {0 + 0 + 1} \right) = 2\]
\[ \Rightarrow {\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]_{x = 1}} = 2\left( {6{x^2} - 6x + 1} \right) = 2\left( {6{{(1)}^2} - 6 \times 1 + 1} \right) = 2\left( {6 - 6 + 1} \right) = 2\]
\[ \Rightarrow {\left[ {\dfrac{{{d^2}y}}{{d{x^2}}}} \right]_{x = \dfrac{1}{2}}} = 2\left( {6{{\left( {\dfrac{1}{2}} \right)}^2} - 6\left( {\dfrac{1}{2}} \right) + 1} \right) = 2\left( {6 \times \dfrac{1}{4} - 6 \times \dfrac{1}{2} + 1} \right) = 2\left( {\dfrac{3}{2} - 3 + 1} \right) = 2\left( {\dfrac{{3 - 6 + 2}}{2}} \right) = - 1\]
Now we observe that at x=0 and at x=1 we have found the minima of the curve .So let’s find the area bounded by the curve.
Area bounded \[ = \int\limits_0^1 {{x^4} - 2{x^3} + {x^2} + 3dx} \]
 Applying the integration
\[ \Rightarrow \mathop {\left[ {\dfrac{{{x^5}}}{5} - 2\dfrac{{{x^4}}}{4} + \dfrac{{{x^3}}}{3} + 3x} \right]}\nolimits_0^1 \]
Putting the upper limit and lower limit
\[ \Rightarrow \left[ {\dfrac{1}{5} - \dfrac{1}{2} + \dfrac{1}{3} + 3} \right] - 0\]
Taking LCM as 30
\[ \Rightarrow \left[ {\dfrac{{1 \times 6 - 1 \times 15 + 1 \times 10 + 3 \times 30}}{{30}}} \right]\]
\[ \Rightarrow \dfrac{{6 - 15 + 10 + 90}}{{30}}\]
Performing the sum we get
\[ \Rightarrow \dfrac{{106 - 15}}{{30}}\]
\[ \Rightarrow \dfrac{{91}}{{30}}\]
Thus this is in simplest form. So area bounded by the curve is \[\dfrac{{91}}{{30}}sq.units\]
Thus the correct option is D.

Note: Here differentiation step is the most important step .Students should know when we have to take single differentiation and when to take double differentiation. Also note that since the lower limit is zero no need to repeat the bracket once its value is written to be zero.