The area of the plane bounded by the curves y=\sqrt{x}, x \in[0,1], y=x^{2}, x \in[1,2] and y=-x^{2}+2 x+4, x \in[0,2] is

Correct Answer is : 193 sq units

NIMCET 2008 Question Paper with solutions

Question : 38 of 120

Marks: +1, -0

The area of the plane bounded by the curves

y=√{x}

x ∈[0,1], y=x^{2}, x ∈[1,2]

and

y=-x^{2}+2 x+4, x ∈[0,2]

Solution:

y=-x^{2}+2 x+4=5-(x-1)^{2}

The required area will be

∫↙{0}↖{1}[(-x^{2}+2 x+4)-√{x}] d x+∫↙{1}↖{2}[(-x^{2}+2 x+4)-x^{2}] d x

=[\;{-x^{3}}/{3}+x^{2}+4 x-\;{x^{3 \/ 2}}/{3 \/ 2}]_{0}^{1}+[\;{-2 x^{3}}/{3}+x^{2}+4 x]_{1}^{2}

=-\;{1}/{3}+1+4-\;{2}/{3}+[(-\;{16}/{3}+4+8)-(\;{-2}/{3}+1+4)]

=4+[\;{20}/{3}-\;{13}/{3}]=4+\;{7}/{3}=\;{19}/{3} \sq

units

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