Question

The value of $c$ for which the area of the figure boundary by the curve $y=8{{x}^{2}}-{{x}^{5}}$ , the straight line is $x=1$ and $x=c$ and the $x-axis$ is equal to $\dfrac{16}{3}$ is
A) 2
B) $\sqrt{8-\sqrt{17}}$
C) 3
D) -1

Answer 1

Hint: The question will be solved by the concept of definite integration. The integral formula used for ${{x}^{n}}$which is represented by $\int{{{x}^{n}}dx}$ is $\dfrac{{{x}^{n+1}}}{n+1}$ . To solve this function given in the question the same formula will be used as given above. We will solve the question in two parts, one when $c$ will act as the lower limit while in other $c$will be the upper limit.

Complete step by step solution:
In the question, area under the curve is given when the curve given is $y=8{{x}^{2}}-{{x}^{5}}$which is bounded by the two straight which are $x=1$ and $x=c$ . To solve this kind of problem we will have to integrate the function given where the limits will be $1$ and $c$.
The formula of the area in integral form is denoted by$\int\limits_{a}^{b}{ydx}$ , here $a$ and $b$ are the limits of the integration, which means $a$ and $b$ are the two extreme values of $x$ under which the area need to be found while $y$ is the function of which the area need to be found.
For $c<1$ :
Applying the same formula in the question, we get:
$\int\limits_{c}^{1}{\left( 8{{x}^{2}}-{{x}^{5}} \right)}dx$
On further calculating it we get:
$\Rightarrow \int\limits_{c}^{1}{8{{x}^{2}}dx-\int\limits_{c}^{1}{{{x}^{5}}dx}}$
For integrating the formula which we will apply is, if the function given is ${{x}^{n}}$ then integration of the function,$\int{{{x}^{n}}dx}$ becomes $\dfrac{{{x}^{n+1}}}{n+1}$ . Applying the same in the given formula we get:
$\Rightarrow \left[ \dfrac{8{{x}^{3}}}{3}-\dfrac{{{x}^{6}}}{6} \right]_{c}^{1}$
On putting the limits in the integrated part, we get:
$\Rightarrow \left[ \dfrac{8{{\left( 1 \right)}^{3}}}{3}-\dfrac{{{\left( 1 \right)}^{6}}}{6}-\left( \dfrac{8{{c}^{3}}}{3}-\dfrac{{{c}^{6}}}{6} \right) \right]$
$\Rightarrow \left[ \dfrac{8}{3}-\dfrac{1}{6}-\dfrac{8{{c}^{3}}}{3}+\dfrac{{{c}^{6}}}{6} \right]$
In the question given the area is equal to $\dfrac{16}{3}$ , so we will equate the above integral with the value $\dfrac{16}{3}$ to find the value of $c$ which a lower limit.
$\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{8{{c}^{3}}}{3}+\dfrac{{{c}^{6}}}{6}=\dfrac{16}{3}$
$\Rightarrow -\dfrac{8{{c}^{3}}}{3}+\dfrac{{{c}^{6}}}{6}=\dfrac{16}{3}-\dfrac{8}{3}+\dfrac{1}{6}$
Taking ${{c}^{3}}$ to be common from L.H.S we get:
$\Rightarrow {{c}^{3}}\left[ -\dfrac{8}{3}+\dfrac{{{c}^{3}}}{6} \right]=\dfrac{16}{3}-\dfrac{8}{3}+\dfrac{1}{6}$
Calculating the above equation, by finding the L.C.M we get:
$\Rightarrow {{c}^{3}}\left[ -\dfrac{16}{6}+\dfrac{{{c}^{3}}}{6} \right]=\dfrac{17}{6}$
Removing $6$ from both L.H.S and R.H.S, we get:
$\Rightarrow {{c}^{6}}-16{{c}^{3}}-17=0$
On factorization ${{c}^{6}}-16{{c}^{3}}-17=0$ we get:S
$\Rightarrow \left( {{c}^{3}}+1 \right)\left( {{c}^{3}}-17 \right)=0$
So the value of ${{c}^{3}}$ will be
$\Rightarrow {{c}^{3}}=-1,17$
$\Rightarrow c=-1,{{17}^{\dfrac{1}{3}}}$
$\Rightarrow c=-1$satisfy the above equation.
For $c\ge 1$ none of the values of $c$ satisfies the required condition that
$\int\limits_{1}^{c}{\left( 8{{x}^{3}}-{{x}^{6}} \right)dx}=\dfrac{16}{3}$
$\therefore $ The value of $c=-1$ for the given problem.

Note: In the given problem, we have solved the question in two parts, that is by taking $c<1$ and $c\ge 1$. This is done because in both the cases the value of the limits differ. For $c<1$, $c$ acts as a lower limit while for $c\ge 1$ ,$c$ is the upper limit. Area under the curve is one of the most important applications for integration.