Question

The value of $c + 2$ for which the area of the figure bounded by the curve $y = 8{x^2} - {x^5}$, the straight lines $x = 1$ and $x = c$ and $x - axis$ is equal to $\dfrac{{16}}{3}$,
A) $1$
B) $3$
C) $ - 1$
D) $4$

Answer 1

Hint: In this question we are given the equation curve and the lines. We have to find the value of $c + 2$. We’ll calculate the value in two cases firstly we’ll take $c < 1$ in which we integrate the equation with respect to $dx$ from $c$ to $1$ and equate it to $\dfrac{{16}}{3}$. In this second case, assume $c \geqslant 1$ and integrate the equation with respect to $dx$ from $1$ to $c$ and equate it to $\dfrac{{16}}{3}$. Solve it further to calculate the value by using the integration formula.

Formula Used: Integration formula –
$\int {cdx = cx} $ where $c$ is the constant
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $

Complete step by step Solution:
Given that,
The area bounded by the curve $y = 8{x^2} - {x^5}$ and the lines $x = 1$ and $x = c$ is $\dfrac{{16}}{3}$.
We’ll take two cases to calculate the value of $c + 1$
Case 1:
Let $c < 1$,
Therefore, the area of the bounded region will be the integration of the equation of the curve from $x = c$ to $x = 1$
So, $A = \int\limits_c^1 {\left( {8{x^2} - {x^5}} \right)dx} $
According to the question,
$\int\limits_c^1 {\left( {8{x^2} - {x^5}} \right)dx} = \dfrac{{16}}{3}$
As we know that, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
It implies that,
$\left[ {\dfrac{{8{x^3}}}{3} - \dfrac{{{x^6}}}{6}} \right]_c^1 = \dfrac{{16}}{3}$
Resolving the limits, we get
$\dfrac{{8{{\left( 1 \right)}^3}}}{3} - \dfrac{{{{\left( 1 \right)}^6}}}{6} - \left( {\dfrac{{8{c^3}}}{3} - \dfrac{{{c^6}}}{6}} \right) = \dfrac{{16}}{3}$
\[\dfrac{8}{3} - \dfrac{1}{6} - \dfrac{{8{c^3}}}{3} + \dfrac{{{c^6}}}{6} = \dfrac{{16}}{3}\]
L.C.M of $3$ and $6$ is $6$
It implies that,
$\dfrac{{16 - 1 - 16{c^3} + {c^6}}}{6} = \dfrac{{16}}{3}$
$\dfrac{{15 - 16{c^3} + {c^6}}}{6} = \dfrac{{16}}{3}$
Cross-multiplying both the sides, we get
$45 - 48{c^3} + 3{c^6} = 96$
$45 - 48{c^3} + 3{c^6} - 96 = 0$
$3{c^6} - 48{c^3} - 51 = 0$
${c^6} - 16{c^3} - 17 = 0$
Using middle term split factorization, we get $\left( {{c^3} + 1} \right)\left( {{c^3} - 17} \right) = 0$
So, $c = - 1,{17^{\dfrac{1}{3}}}$
Adding $2$ on both sides,
$c + 2 = 1,{17^{\dfrac{1}{3}}} + 2$
Here, $c = 1$ satisfying the given condition.
Case 2:
Let, $c \geqslant 1$,
Area will be, $\int\limits_1^c {\left( {8{x^2} - {x^5}} \right)dx} = \dfrac{{16}}{3}$
$\left[ {\dfrac{{8{x^3}}}{3} - \dfrac{{{x^6}}}{6}} \right]_1^c = \dfrac{{16}}{3}$ (Solved above)
$\dfrac{{8{{\left( c \right)}^3}}}{3} - \dfrac{{{{\left( c \right)}^6}}}{6} - \left( {\dfrac{{8{{\left( 1 \right)}^3}}}{3} - \dfrac{{{{\left( 1 \right)}^6}}}{6}} \right) = \dfrac{{16}}{3}$
$\dfrac{{16{c^3} - {c^6} - 16 + 1}}{6} = \dfrac{{16}}{3}$
$ - 45 + 48{c^3} - 3{c^6} = 96$
Here, none of the values satisfies the given condition.

Therefore, the correct option is (A).

Note:Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.