Question

Let \[f:\left[ { - 3,1} \right] \to \mathbb{R}\] be given as
\[f\left( x \right) = \left\{ \begin{array}{l}\min \left\{ {x + 6,{x^2}} \right\}, - 3 \le x \le 0\\\max \left\{ {\sqrt x ,{x^2}} \right\},0 \le x \le 1\end{array} \right.\]
If the area bounded by \[y = f\left( x \right)\] and \[x\]-axis is \[A\], then find the value of \[6A\].

Answer 1

Hint: Find the point(s) of intersection of the curves \[y = x + 6\] and \[y = {x^2}\] over the interval \[\left[ { - 3,0} \right]\]. It will be at \[\left( { - 2,4} \right)\]. Then divide the interval \[\left[ { - 3,0} \right]\] into two subintervals \[\left[ { - 3, - 2} \right)\] and \[\left( { - 2,0} \right]\] and check which of the functions \[y = x + 6\] and \[y = {x^2}\] takes minimum value on the subintervals \[\left[ { - 3, - 2} \right)\] and \[\left( { - 2,0} \right]\]. Thus find \[\min \left\{ {x + 6,{x^2}} \right\}\] over the interval \[\left[ { - 3,0} \right]\]. After that find the point(s) of intersection of the curves \[y = \sqrt x \] and \[y = {x^2}\] over the interval \[\left[ {0,1} \right]\]. It will be at \[\left( {0,0} \right)\] and \[\left( {1,1} \right)\] and hence find \[\max \left\{ {\sqrt x ,{x^2}} \right\}\] over the interval \[\left[ {0,1} \right]\]. Finally, you’ll obtain \[\min \left\{ {x + 6,{x^2}} \right\} = x + 6\] over the interval \[\left[ { - 3, - 2} \right)\],\[\min \left\{ {x + 6,{x^2}} \right\} = {x^2}\] over the interval \[\left( { - 2,0} \right]\] and \[\max \left\{ {\sqrt x ,{x^2}} \right\} = \sqrt x \] over the interval \[\left[ {0,1} \right]\]. Then find the area \[A\] using the formula of finding area using definite integral.

Formula Used:
Area of the region bounded by \[y = f\left( x \right)\] and \[x\]-axis within the interval \[\left[ {a,b} \right]\] is \[A = \int_a^b {f\left( x \right)dx} \]
\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\]
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \], where \[C\] is integrating constant.

Complete step by step solution:
The given function is
\[f\left( x \right) = \left\{ \begin{array}{l}\min \left\{ {x + 6,{x^2}} \right\}, - 3 \le x \le 0\\\max \left\{ {\sqrt x ,{x^2}} \right\},0 \le x \le 1\end{array} \right.\]
Let us find \[\min \left\{ {x + 6,{x^2}} \right\}\] over the interval \[\left[ { - 3,0} \right]\].
First of all, let us find the point of intersection of the curves \[y = x + 6\] and \[y = {x^2}\]
At the point of intersection, \[x + 6 = {x^2}\]
\[ \Rightarrow {x^2} = x + 6\]
\[ \Rightarrow {x^2} - x - 6 = 0\]
\[ \Rightarrow {x^2} - 3x + 2x - 6 = 0\]
\[ \Rightarrow x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0\]
\[ \Rightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\]
\[ \Rightarrow \left( {x - 3} \right) = 0\] or \[\left( {x + 2} \right) = 0\]
\[ \Rightarrow x = 3\] or \[x = - 2\]
\[3 \notin \left[ { - 3,0} \right]\] but \[ - 2 \in \left[ { - 3,0} \right]\]
At \[x = - 2\], \[y = 4\]
So, the two curves meet at the point \[\left( { - 2,4} \right)\] on the interval \[\left[ { - 3,0} \right]\].
Let us divide the interval \[\left[ { - 3,0} \right]\] into two subintervals \[\left[ { - 3, - 2} \right)\] and \[\left( { - 2,0} \right]\].
Let us check which of the functions \[y = x + 6\] and \[y = {x^2}\] takes minimum value on the subinterval \[\left[ { - 3, - 2} \right)\].
Let \[x = - 3\].
Then the functional value of the function \[y = x + 6\] at \[x = - 3\] is \[y = - 3 + 6 = 3\]
And the functional value of the function \[y = {x^2}\] at \[x = - 3\] is \[y = {\left( { - 3} \right)^2} = 9\]
\[\therefore \min \left\{ {x + 6,{x^2}} \right\} = x + 6\] over the interval \[\left[ { - 3, - 2} \right)\].
Now, let us check which of the functions \[y = x + 6\] and \[y = {x^2}\] takes minimum value on the subinterval \[\left( { - 2,0} \right]\].
Let \[x = - 1\].
Then the functional value of the function \[y = x + 6\] at \[x = - 1\] is \[y = - 1 + 6 = 5\]
And the functional value of the function \[y = {x^2}\] at \[x = - 1\] is \[y = {\left( { - 1} \right)^2} = 1\]
\[\therefore \min \left\{ {x + 6,{x^2}} \right\} = {x^2}\] over the interval \[\left( { - 2,0} \right]\].
Let us find \[\max \left\{ {\sqrt x ,{x^2}} \right\}\] over the interval \[\left[ {0,1} \right]\].
First of all, let us find the point of intersection of the curves \[y = \sqrt x \] and \[y = {x^2}\]
At the point of intersection, \[\sqrt x = {x^2}\]
Squaring both sides, we get \[x = {x^4}\]
\[ \Rightarrow {x^4} - x = 0\]
Take \[x\] common from left-hand-side
\[ \Rightarrow x\left( {{x^3} - 1} \right) = 0\]
\[ \Rightarrow x = 0\] or \[\left( {{x^3} - 1} \right) = 0\]
Use the identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\]
\[\therefore {x^3} - 1 = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\]
\[\left( {{x^3} - 1} \right) = 0\]
\[ \Rightarrow \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) = 0\]
\[ \Rightarrow \left( {x - 1} \right) = 0\] or \[\left( {{x^2} + x + 1} \right) = 0\]
\[ \Rightarrow x = 1\]
Discriminant of the quadratic equation \[{x^2} + x + 1 = 0\] is \[D = {\left( 1 \right)^2} - 4\left( 1 \right)\left( 1 \right) = 1 - 4 = - 3 < 0\]
So, it has no real roots.
\[\therefore \]The roots of the equation \[\sqrt x = {x^2}\] are \[0,1\].
\[0 \in \left[ {0,1} \right]\] and \[1 \in \left[ {0,1} \right]\]
At \[x = 0\], \[y = 0\]
At \[x = 1\], \[y = 1\]
So, the two curves meet at the points \[\left( {0,0} \right)\] and \[\left( {1,1} \right)\] on the interval \[\left[ {0,1} \right]\].
Let us check which of the functions \[y = \sqrt x \] and \[y = {x^2}\] takes maximum value on the interval \[\left[ {0,1} \right]\].
Let us take \[x = \dfrac{1}{2} \in \left[ {0,1} \right]\]
Then the functional value of the function \[y = \sqrt x \] at \[x = \dfrac{1}{2}\] is \[y = \dfrac{1}{{\sqrt 2 }}\]
And the functional value of the function \[y = {x^2}\] at \[x = \dfrac{1}{2}\] is \[y = {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4}\]
\[\therefore \max \left\{ {\sqrt x ,{x^2}} \right\} = \sqrt x \] over the interval \[\left[ {0,1} \right]\].
Finally, we get
\[\min \left\{ {x + 6,{x^2}} \right\} = x + 6\] over the interval \[\left[ { - 3, - 2} \right)\]
\[\min \left\{ {x + 6,{x^2}} \right\} = {x^2}\] over the interval \[\left( { - 2,0} \right]\]
\[\max \left\{ {\sqrt x ,{x^2}} \right\} = \sqrt x \] over the interval \[\left[ {0,1} \right]\]
Now, the required area is \[\int\limits_{ - 3}^1 {f\left( x \right)dx} \]
\[ = \int\limits_{ - 3}^{ - 2} {\left( {x + 6} \right)dx + } \int\limits_{ - 2}^0 {\left( {{x^2}} \right)dx + \int\limits_0^1 {\left( {\sqrt x } \right)dx} } \]
Use the formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \], where \[C\] is integrating constant.
\[\int\limits_{ - 3}^{ - 2} {\left( {x + 6} \right)dx} = \int\limits_{ - 3}^{ - 2} {xdx} + 6\int\limits_{ - 3}^{ - 2} {dx} \]
\[ = \left[ {\dfrac{{{x^2}}}{2}} \right]_{ - 3}^{ - 2} + 6\left[ x \right]_{ - 3}^{ - 2}\]
\[ = \left\{ {\dfrac{{{{\left( { - 2} \right)}^2}}}{2} - \dfrac{{{{\left( { - 3} \right)}^2}}}{2}} \right\} + 6\left\{ {\left( { - 2} \right) - \left( { - 3} \right)} \right\}\]
\[ = \left( {2 - \dfrac{9}{2}} \right) + 6\left( { - 2 + 3} \right)\]
\[ = - \dfrac{5}{2} + 6\]
\[ = \dfrac{7}{2}\]
\[\int\limits_{ - 2}^0 {\left( {{x^2}} \right)dx} = \left[ {\dfrac{{{x^3}}}{3}} \right]_{ - 2}^0 = \left\{ {\dfrac{{{{\left( 0 \right)}^3}}}{3} - \dfrac{{{{\left( { - 2} \right)}^3}}}{3}} \right\} = 0 - \dfrac{{ - 8}}{3} = \dfrac{8}{3}\]
\[\int_{ - 2}^0 {\left( {{x^2}} \right)dx} = \left[ {\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]_0^1 = \dfrac{2}{3}\left[ {{x^{\dfrac{3}{2}}}} \right]_0^1 = \dfrac{2}{3}\left[ {{1^{\dfrac{3}{2}}} - {0^{\dfrac{3}{2}}}} \right] = \dfrac{2}{3}\]
\[\therefore A = \dfrac{7}{2} + \dfrac{8}{3} + \dfrac{2}{3} = \dfrac{{21 + 16 + 4}}{6} = \dfrac{{41}}{6}\]
\[ \Rightarrow 6A = 41\]

Hence the required value is \[41\].

Note:The given function is a multi-valued function defined on the interval \[\left[ { - 3,1} \right]\]. The function takes different forms over the intervals \[\left[ { - 3,0} \right]\] and \[\left[ {0,1} \right]\]. You have to find \[\min \left\{ {x + 6,{x^2}} \right\}\] over the interval \[\left[ { - 3,0} \right]\] and \[\max \left\{ {\sqrt x ,{x^2}} \right\}\] over the interval \[\left[ {0,1} \right]\] very carefully.