Question

The area between x=y^2 and x=4 is divided into two equal parts by the line x=a such that the ratio of the respective parts is 2:3. find the value of a​?

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is y² = x, which is right handed parabola whose vertex is at the origin i.e. (0, 0).

Now, the area between x = y² and x = 4 is divided into two equal parts by the line x = a such that the ratio of the respective parts is 2 : 3.

It means, the area between the curve x = y² and the line x = a is 2/5 times the area between the curve x = y² and the line x = 4 with x - axis.

So,

$\rm \: \displaystyle\int_{0}^{a}\rm y \: dx \: = \: \frac{2}{5}\displaystyle\int_{0}^{4}\rm y \: dx$

$\rm \: \displaystyle\int_{0}^{a}\rm \sqrt{x} \: dx \: = \: \frac{2}{5}\displaystyle\int_{0}^{4}\rm \sqrt{x} \: dx$

We know,

$\boxed{ \rm{ \:\displaystyle\int\rm {x}^{n}dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

So, using this result, we get

$\rm \: \dfrac{2}{3} \bigg | {\bigg(x \bigg) }^{\dfrac{3}{2} }\bigg |_{0}^{a} = \dfrac{2}{5} \times \dfrac{2}{3} \bigg | {\bigg(x \bigg) }^{\dfrac{3}{2} }\bigg |_{0}^{4}$

$\rm \: {\bigg(a\bigg) }^{\dfrac{3}{2} } = \dfrac{2}{5} \times{\bigg(4 \bigg) }^{\dfrac{3}{2} }$

$\rm \: {\bigg(a\bigg) }^{\dfrac{3}{2} } = \dfrac{2}{5} \times{\bigg( {2}^{2} \bigg) }^{\dfrac{3}{2} }$

$\rm \: {\bigg(a\bigg) }^{\dfrac{3}{2} } = \dfrac{2}{5} \times8$

$\rm \: {\bigg(a\bigg) }^{\dfrac{3}{2} } = \dfrac{16}{5}$

$\rm\implies \:a \: = \:{\bigg(\dfrac{16}{5} \bigg) }^{\dfrac{2}{3}} \:\:square\:units$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$