Question

Find the area of the region bounded by the curve y2 – 6y + x + 5 = 0 and y-axis.​

Answer 1

SOLUTION

TO DETERMINE

The area of the region bounded by the curve

$\sf{ {y}^{2} - 6y + x + 5 = 0 \: \: and \: \: y \: axis }$

CONCEPT TO BE IMPLEMENTED

The area of the region bounded by the curve y = f (x), y axis and the lines y = a and y = b is

$= \displaystyle \int\limits_{a}^{b} x \, dy$

EVALUATION

Here the given given equation of the curve is

$\sf{ {y}^{2} - 6y + x + 5 = 0 \: }$

Which can be rewritten as below

$\sf{ {y}^{2} - 6y + 5 = - x \: }$

$\implies \: \sf{ (y - 1)(y - 5) = - x \: }$

This is an equation of an parabola

The above curve cuts y axis at ( 0,1) & ( 0,5)

So the region bounded by the curve and y axis is ABCA

Hence required area of the region bounded by the curve and y axis

$= - \displaystyle \int\limits_{1}^{5} ( {y}^{2} - 6y + 5) \, dy$

$= \displaystyle - \bigg[\frac{ {y}^{3} }{3} - 3 {y}^{2} + 5y \bigg]_{1}^{5}$

$\displaystyle \sf{ = - \bigg[\frac{ {5}^{3} }{3} - 75 + 25 \bigg] + \bigg[\frac{ {1}^{3} }{3} - 3 + 5 \bigg]}$

$\displaystyle \sf{ = - \bigg[\frac{ 125 }{3} - 50 \bigg] + \bigg[\frac{ 1 }{3} + 2 \bigg]}$

$\displaystyle \sf{ = \frac{32}{3} \: \: sq \: unit }$

Graph : For figure refer to the attachment

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