ক্ষেত্রফল নির্ণয়

If the area (in sq. units) of the region {(x,y):y2≤4x,x+y≤1,x≥0,y≥0}\{(x, y): y^2\le 4x, x+y\le 1, x\geq 0, y\ge 0 \} is a2+ba\sqrt{2}+b, then a−ba-b is equal to?

হানি নাটস

{(x,y):y2≤4x,x+y≤1,x≥0,y≥0}\{(x, y): y^2\leq 4x, x+y\leq 1, x\geq 0, y\geq 0\}

A∫03−222xdx+12(1−(3−22))(1−(3−22))A\displaystyle\int^{3-2\sqrt{2}}_02\sqrt{x}dx+\dfrac{1}{2}(1-(3-2\sqrt{2}))(1-(3-2\sqrt{2}))

=2[x3/2]03−223/2+12(22−2)(22−2)=\dfrac{2[x^{3/2}]_0^{3-2\sqrt{2}}}{3/2}+\dfrac{1}{2}(2\sqrt{2}-2)(2\sqrt{2}-2)

=823+(−103)=\dfrac{8\sqrt{2}}{3}+\left(-\dfrac{10}{3}\right)

a=83a=\dfrac{8}{3}, b=−103b=-\dfrac{10}{3}

a−b=6a-b=6.

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