Question 199 of 230MediumMCQ
If the area of the region bounded by the curves, \[y = {x^2},y = {1 \over x}\] and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :
Solutions · 1
Official solution
Point of intersection of y = x2 and y = \[{1 \over x}\].
put y = \[{1 \over x}\] in y = x2, then we get,
\[{1 \over x} = {x^2}\]
\[ \Rightarrow \] \[\,\,\,\] x3 - 1 = 0
\[ \Rightarrow \] \[\,\,\,\] x = 1
\[\therefore\,\,\,\] y = 1
\[\therefore\,\,\,\] point B = (1, 1)
Area of region ABCDA
= \[\int\limits_0^1 {{x^2}} \] dx + \[\int\limits_1^t {{1 \over x}} \] dx
= \[\left[ {{{{x^3}} \over 3}} \right]_0^1\] + \[\left[ {\ell n\,x} \right]_1^t\]
= \[{1 \over 3}\] + \[\ell n\,t\] - \[\ell n\] 1
= \[{1 \over 3}\] + \[\ell n\,t\] [ as \[\ell n\] 1 = 0]
given this Area = 1 sq unit.
\[\therefore\,\,\,\] \[{1 \over 3}\] + \[\ell n\,t\] = 1
\[ \Rightarrow \] \[\ell n\,t\] = \[{2 \over 3}\]
\[ \Rightarrow \] t = e\[{^{{2 \over 3}}}\] Answer the question to unlock