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Vector Algebra

Question
CBSEENMA12032715

Find the area bounded by the curve  y = x2 and the line y = x.
OR
Find the area of the region {(x. y): x2 ≤ y ≤ x}.

Solution
The given region is {(x, y): x2 ≤ y ≤ x}
This region is the intersection of the following regions:
straight R subscript 1 space equals space open curly brackets open parentheses straight x comma space straight y right parenthesis close parentheses space colon space straight x squared space less or equal than space straight y close curly brackets
straight R subscript 2 space equals space open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight y space less or equal than straight x close curly brackets
Consider the equations
                   straight y space equals straight x squared                  ...(1)
  and            y = x                     ...(2)
From (1) and (2), we get
                          straight x space equals space straight x squared space space or space space straight x squared minus straight x space equals space 0
rightwards double arrow space space space space straight x left parenthesis straight x minus 1 right parenthesis space equals space 0 space space space space rightwards double arrow space space space space straight x space equals space 0 comma space space 1
∴    from (2), y = 0, 1
∴    curves (1) and (2) intersect in the points O (0, 0) and A (1, 1)
Required area = area of the shaded region
equals space integral subscript 0 superscript 1 straight x space dx space minus space integral subscript 0 superscript 1 straight x squared space dx space equals space open square brackets straight x squared over 2 close square brackets subscript 0 superscript 1 space minus space open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 1
equals space open parentheses 1 half minus 0 close parentheses space minus space open parentheses 1 third minus 0 close parentheses space equals space 1 half minus 1 third space equals space fraction numerator 3 minus 2 over denominator 6 end fraction space equals space 1 over 6 space sq. space units.

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