10. Find the area bounded by the curve x = 4y and the line x = 4y -2

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Vishal Baghel | Contributor-Level 10

4 months ago

Given curve is $x^{2} = 4 y$ and the equation of line is $x = 4 y - 2$

The point of intersection of the curve and the line can be determine as follows.

Put, $x = 4 y - 2 \Rightarrow x + 2 = 4 y \Rightarrow y \frac{x + 2}{4}$

In $x^{2} = 4 y$ to determine value of x

i.e, $x^{2} = 4 \times \frac{(x + 2)}{4} = x + 2$

$\begin{array}{l} \Rightarrow x^{2} - x - 2 = 0 \Rightarrow x^{2} + x - 2 x - 2 = 0 \\ \Rightarrow x (x + 1) - 2 (x + 1) = 0 \\ \Rightarrow (x + 1) (x - 2) = 0 \end{array}$

$\Rightarrow x = 2$ and $x = - 1$

$x = 2$ , we have ${(2)}^{2} = 4 y \Rightarrow 4 - 4 y \Rightarrow y = 1$

And at $x = - 1$ we have ${(- 1)}^{2} = 4 y \Rightarrow y = \frac{1}{4}$

So, the coordinates A and B are (2,1) and ( $- 1, \frac{1}{4}$ )

$∴$ The required area before the line & the curve is area $B D^{ι} A B$ = area of trapezium (BNMAB)- area under curve BDA

$\begin{array}{l} = \int_{- 1}^{2} y_{l i n e} d x - \int_{- 1}^{2} y_{c u r v e} d x \\ = \int_{- 1}^{2} \frac{x + 2}{4} d a - \int_{- 1}^{2} \frac{x^{2}}{4} \\ = \frac{1}{4} \int_{- 1}^{2} x d a + \frac{2}{4} \int_{- 1}^{2} d x - \frac{1}{4} \int_{- 1}^{2} x^{2} d x \end{array}$

$\begin{array}{l} = \frac{1}{4} {[\frac{x^{2}}{2}]}_{- 1}^{2} + \frac{2}{4} {[x]}_{- 1}^{2} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{- 1}^{2} \\ = \frac{1}{8} [2^{2} - {(- 1)}^{2}] + \frac{1}{2} [2 - (- 1)] - \frac{1}{12} [2^{3} - {(- 1)}^{3}] \end{array}$

$\begin{array}{l} = \frac{3}{8} + \frac{3}{2} - \frac{9}{12} = \frac{3}{8} + \frac{3}{2} - \frac{3}{4} = \frac{3 + 4 \times 3 - 2 \times 3}{8} \\ = \frac{3 + 12 - 6}{8} = \frac{9}{8} u n i t^{2} \end{array}$

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