A) \[\frac{4}{3}\]
B) \[{{e}^{2/3}}\]
C) \[\frac{3}{2}\]
D) \[{{e}^{3/2}}\]
Correct Answer: B
Solution :
Are bounded by the curves is the region ABCD. Therefore, area \[=\int_{0}^{1}{{{x}^{2}}dx+\int_{1}^{t}{\frac{1}{x}dx}}\] \[=\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{1}+[\ln (x)]_{1}^{t}\] \[=\frac{1}{3}+\ln (t)\] It is given that area enclosed is 1 \[\Rightarrow \]\[\frac{1}{3}+\ln (t)=1\] \[\Rightarrow \]\[\ln (t)=\frac{2}{3}\] \[\Rightarrow \]\[t={{e}^{\frac{2}{3}}}\]You need to login to perform this action.
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