A) \[\frac{16}{5}(3-\sqrt{2})\]
B) \[\frac{32}{5}(3-\sqrt{2})\]
C) \[\frac{8}{5}(3-\sqrt{2})\]
D) \[\frac{64}{5}(3-\sqrt{2})\]
Correct Answer: B
Solution :
Let \[I=\int_{3}^{4}{x\sqrt{6-x}dx}\] By using the property \[\therefore \] \[I=\int_{2}^{4}{(6-x)}\sqrt{x}dx\] \[=\int_{2}^{4}{(6{{x}^{1/2}}-{{x}^{3/2}})d}x\] \[=\left[ \frac{6{{x}^{3/2}}}{3/2}-\frac{{{x}^{5/2}}}{5/2} \right]_{2}^{4}\] \[=\left[ 4\times 8-\frac{2}{5}\times 32-\left( 4\times 2\sqrt{2}-\frac{2}{5}\times 4\sqrt{2} \right) \right]\] \[=\left[ 32-\frac{64}{5}-\frac{32\sqrt{2}}{5} \right]\] \[=\frac{32}{5}(3-\sqrt{2})\]
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