A) 8
B) 9
C) 10
D) 11
Correct Answer: B
Solution :
Area of shaded portion \[=\left| \int_{2}^{4}{\left( \frac{y+4}{2} \right)dy} \right|-\left| \int_{-2}^{4}{\frac{{{y}^{2}}}{4}dy} \right|\] \[=\left| \frac{1}{2}\left[ \frac{{{y}^{2}}}{2}+4y \right]_{-2}^{4} \right|-\left| \frac{1}{4}\left[ \frac{{{y}^{3}}}{3} \right]_{-2}^{4} \right|\] \[=\left| \frac{1}{2}\left[ \{8+16\}-\{2-8\} \right] \right|-\left| \frac{1}{4}\left\{ \frac{64}{3}+\frac{8}{3} \right\} \right|\] = 15 ? 6 = 9 sq unitsYou need to login to perform this action.
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