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 $f''(c) < 0$.