The area enclosed by the curves xy + 4y = 16 | vedclass

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Question

(C) Given curves are $xy + 4y = 16$ and $x + y = 6$.From the first equation,$y(x + 4) = 16$,so $y = \frac{16}{x + 4}$.From the second equation,$y = 6

Vedclass · Accepted Answer

$30 - 32 \log_e 2$

Vedclass · Answer

(C) Given curves are $xy + 4y = 16$ and $x + y = 6$.From the first equation,$y(x + 4) = 16$,so $y = \frac{16}{x + 4}$.From the second equation,$y = 6 - x$.To find the points of intersection,set the two expressions for $y$ equal:$6 - x = \frac{16}{x + 4}$$(6 - x)(x + 4) = 16$$6x + 24 - x^2 - 4x = 16$$-x^2 + 2x + 8 = 0$$x^2 - 2x - 8 = 0$$(x - 4)(x + 2) = 0$So,$x = 4$ and $x = -2$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -2$ to $x = 4$:$A = \int_{-2}^{4} \left( (6 - x) - \frac{16}{x + 4} \right) dx$$A = \left[ 6x - \frac{x^2}{2} - 16 \ln|x + 4| \right]_{-2}^{4}$$A = \left( 6(4) - \frac{16}{2} - 16 \ln(8) \right) - \left( 6(-2) - \frac{4}{2} - 16 \ln(2) \right)$$A = (24 - 8 - 16 \ln(2^3)) - (-12 - 2 - 16 \ln 2)$$A = (16 - 48 \ln 2) - (-14 - 16 \ln 2)$$A = 16 - 48 \ln 2 + 14 + 16 \ln 2$$A = 30 - 32 \ln 2$

The area enclosed by the curves $xy + 4y = 16$ and $x + y = 6$ is equal to:

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