Consider the two curves y = 1/ x 2 \& y = 1/ [ 4 ( x − 1 )] . 1.At what value of 'a' (a 2) is the reciprocal of the area of the fig. bounded by the curves, the lines x = 2& x = a equal to

Question

Consider the two curves y = 1/ x 2 \& y = 1/ [ 4 ( x − 1 )] . 1.At what value of 'a' (a > 2) is the reciprocal of the area of the fig. bounded by the curves, the lines x = 2& x = a equal to 'a' itself ? 2. At what value of 'b' ( 1 < b < 2 ) the area of the figure bounded by these curves, the lines x = b & x = 2 equal to 1-1/b.

TestHub · Accepted Answer

For x>1 the second curve is above the first. Integrate their difference: 4 ( x − 1 ) 1 ​ − x 2 1 ​ = 4 x 2 ( x − 1 ) ( x − 2 ) 2 ​ ≥ 0. F ( x ) = 4 1 ​ ln ( x − 1 ) + 1/ x . F ( a ) − F ( 2 ) = 1/ a ⇒ ln ( a − 1 ) = 2 ⇒ a = 1 + e 2 . F ( 2 ) − F ( b ) = 1 − 1/ b ⇒ ln ( b − 1 ) = − 2 ⇒ b = 1 + e − 2 . ​

Mathematics - Area Under Curves Question with Solution | TestHub

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