The area enclosed by the curves y=8x-x^2 and | vedclass

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(A) Given equations of curves are: $y = 8x - x^2$ $(i)$ $8x - 4y + 11 = 0$ (ii) From (ii),$4y = 8x + 11 \Rightarrow y = \frac{8x+11}{4} = 2x + \frac{1

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$\frac{125}{6}$

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(A) Given equations of curves are: $y = 8x - x^2$ $(i)$ $8x - 4y + 11 = 0$ (ii) From (ii),$4y = 8x + 11 \Rightarrow y = \frac{8x+11}{4} = 2x + \frac{11}{4}$. To find the intersection points,substitute $y$ from (ii) into $(i)$: $2x + \frac{11}{4} = 8x - x^2$ $x^2 - 6x + \frac{11}{4} = 0$ $4x^2 - 24x + 11 = 0$ $4x^2 - 22x - 2x + 11 = 0$ $2x(2x - 11) - 1(2x - 11) = 0$ $(2x - 1)(2x - 11) = 0$ So,$x = \frac{1}{2}$ and $x = \frac{11}{2}$. The area is given by: $A = \int_{1/2}^{11/2} [y_{	ext{parabola}} - y_{	ext{line}}] dx$ $A = \int_{1/2}^{11/2} [(8x - x^2) - (2x + \frac{11}{4})] dx$ $A = \int_{1/2}^{11/2} (-x^2 + 6x - \frac{11}{4}) dx$ $A = [-\frac{x^3}{3} + 3x^2 - \frac{11}{4}x]_{1/2}^{11/2}$ Evaluating the definite integral: $A = [-\frac{1}{3}(\frac{1331}{8} - \frac{1}{8}) + 3(\frac{121}{4} - \frac{1}{4}) - \frac{11}{4}(\frac{11}{2} - \frac{1}{2})]$ $A = [-\frac{1}{3}(\frac{1330}{8}) + 3(\frac{120}{4}) - \frac{11}{4}(5)]$ $A = [-\frac{665}{12} + 90 - \frac{55}{4}] = [-\frac{665}{12} + \frac{1080}{12} - \frac{165}{12}] = \frac{250}{12} = \frac{125}{6} 	ext{ sq units}$.

The area enclosed by the curves $y=8x-x^2$ and $8x-4y+11=0$ is

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