A) \[\frac{30}{7}\] sq. unit
B) \[\frac{31}{7}\] sq. unit
C) \[\frac{32}{3}\] sq. unit
D) \[\frac{34}{3}\] sq. unit
Correct Answer: C
Solution :
We have \[y=4x-{{x}^{2}}\] and \[y=0\];\[\therefore \] \[x=0\], \[4\] Required area \[=\int_{0}^{4}{(4x-{{x}^{2}})dx=\left[ \frac{4{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]_{0}^{4}}\] \[=32-\frac{64}{3}=\frac{32}{3}\]sq. unit.
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