A) 2/7
B) 5/2
C) 3/2
D) \[-3/2\]
Correct Answer: B
Solution :
\[I=\int_{0}^{3}{|2-x|dx}\]\[=\int_{0}^{2}{(2-x)}\,dx+\int_{2}^{3}{-(2-x)\,dx}\] \[=\int_{0}^{2}{(2-x)}\,dx-\int_{2}^{3}{\,(2-x)\,dx}=\left[ 2x-\frac{{{x}^{2}}}{2} \right]_{0}^{2}-\left[ 2x-\frac{{{x}^{2}}}{2} \right]_{2}^{3}\] \[\int_{0}^{\pi }{\text{ }\left| \text{ }{{\sin }^{4}}x\text{ } \right|\text{ }dx=2\int_{0}^{\pi /2}{{{\sin }^{4}}x\,dx}}\]\[=2-\left[ 4-\frac{9}{2} \right]\]\[=\frac{5}{2}\].
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