Question

If \[f(x)\] is monotonic in \[(a,b)\], then prove that the area bounded by the ordinates at \[x = a:x = b:y = f(x)\] and \[y = f(c)\], \[c \in (a,b)\] is minimum when \[c = \dfrac{{a + b}}{2}\]. Hence, if the area bounded by the graph the graph of \[f(x) = \dfrac{{{x^3}}}{3} - {x^2} + a\], the straight lines \[x = 0,x = 2\] and the x-axis is minimum then find the value of \['a'.\]
A) \[\dfrac{2}{3}\]
B) \[\dfrac{2}{5}\]
C) \[\dfrac{7}{3}\]
D) \[\dfrac{4}{3}\]

Answer 1

Hint: At first, we will try to find out the area of the bounded region. Then by condition of minimum of an area we will find the value of \[a.\]

Complete step by step answer:
Let us consider the function \[f(x)\] which is monotonically increasing in \[(a,b)\].
Let us consider a point \[c\]between \[(a,b)\].
Since, \[a < c < b\]
Then, \[f(a) < f(c) < f(b)\]
Here, \[f(c)\] is constant.
The area bounded by the ordinates at \[x = a:x = b\]
Let us consider the area of the shaded region is \[A\].
So, the area of the shaded region
\[A = \int\limits_a^c {f(c)dx - } \int\limits_a^c {f(x)dx + } \int\limits_c^b {f(x)dx - } \int\limits_c^b {f(c)dx} \]
Simplifying we get,
\[A = f(c)\int\limits_a^c {dx - } f(c)\int\limits_c^b {dx + } \int\limits_c^a {f(x)dx + } \int\limits_c^b {f(x)dx} \]
Simplifying again we get,
\[A = f(c)[(c - a) - (b - c)] + \int\limits_c^a {f(x)dx + } \int\limits_c^b {f(x)dx} \]
Simplifying again we get,
\[A = f(c)[2c - (a + b)]f(c) + \int\limits_c^a {f(x)dx + } \int\limits_c^b {f(x)dx} \]
We know that, \[\int\limits_c^a {f(x)dx + } \int\limits_c^b {f(x)dx} \]is positive.
Now, the shaded area \[A\] will be minimum when, \[[2c - (a + b)]f(c) = 0\]
But, \[f(c) \ne 0\]
Then, \[2c - (a + b) = 0\]
Simplifying we get, \[c = \dfrac{{(a + b)}}{2}\]
Now, we have,
\[A = \int\limits_0^2 {f(x)dx} \]
Substitute the value of \[f(x)\] we get,
\[ = \int\limits_0^2 {\left[ {\dfrac{{{x^3}}}{3} - {x^2} + a} \right]dx} \]
We know that, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\]
Applying the formula, we get,
\[ = {\left[ {\dfrac{{{x^4}}}{{12}} - \dfrac{{{x^3}}}{3} + ax} \right]_0}^2\]
Simplifying we get,
\[ = \dfrac{{16}}{{12}} - \dfrac{8}{3} + 2a\]
Solving we get,
\[ = 2a - \dfrac{4}{3}\]
Since, the area is minimum so, \[2a - \dfrac{4}{3} = 0\]
Hence, \[a = \dfrac{2}{3}\]
Therefore, the correct option is (A) \[\dfrac{2}{3}\].
Note – A function is monotonic in a certain interval if the function is increasing or decreasing.
A function \[f(x)\] is minimum \[x = c\]if the second derivative of the function at the point is positive that is \[f''(c) > 0\] and the function \[f(x)\] is maximum \[x = c\]if the second derivative of the function at the point is positive that is \[f''(c) < 0\].