A) \[\frac{128}{3}\]
B) \[\frac{64}{3}\]
C) \[64\]
D) \[128\]
Correct Answer: B
Solution :
[b] \[x\,\,dy+2y\,\,dx=x\,(x-3)\,dx\] \[\Rightarrow \,\,\,\,\,{{x}^{2}}dy+2xy\,\,dx=({{x}^{3}}-3{{x}^{2}})dx\] \[\Rightarrow \,\,\,\,\,d({{x}^{2}}y)=({{x}^{3}}-3{{x}^{2}})dx\] \[\Rightarrow \,\,\,\,\,\,\,\,y{{x}^{2}}=\frac{{{x}^{4}}}{4}-{{x}^{3}}+c\] Curve passes through the point \[(4,0)\]. \[\therefore \,\,\,c=0\] Therefore, curve is\[y=\frac{{{x}^{2}}}{4}-x.\] Required area \[=\int\limits_{0}^{8}{\left( x-\left( \frac{{{x}^{2}}}{4}-x \right) \right)}dx\] \[=\left( {{x}^{2}}-\frac{{{x}^{3}}}{12} \right)_{0}^{8}=64-\frac{2}{3}\times 64=\frac{64}{3}\]You need to login to perform this action.
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