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2015-04-25T01:08:54+05:30

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Setup an integral for the area between curves :

\text{Area}=\int\limits_0^1\sqrt{x}-x^3\,dx\\=\frac{x^{1/2+1}}{1/2+1}-\frac{x^{3+1}}{3+1}\Bigg|_0^1\\=\frac{2}{3}x^{3/2}-\frac{x^4}{4}\Bigg|_0^1\\=\frac{2}{3}-\frac{1}{4}-0\\=\boxed{\frac{5}{12}}
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2015-04-25T08:29:52+05:30

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The curve for y = √x  and the currve for y = x³ meet at  a point P (x, y) where,
 
  y = √x  = x³   
  =>  x  = x⁶
  =>  x⁵ =  1    Or,  x = 0
  =>  x = 1 and 0  are  real solutions.    perhaps the others are imaginary solutions.

The curve y = √x  is above the curve for  y= x³  in the interval  [ 0, 1 ].

The integral for  x^n=\frac{x^{1+n}}{1+n}

The area enclosed by the curves in the first quadrant :

=\int\limits^1_0 {(\sqrt{x}-x^3)} \, dx\\\\=\int\limits^1_0 {x^{\frac{1}{2}} \, dx -  \int\limits^1_0 {x^3} \, dx

=[\frac{x^{1+\frac{1}{2}}}{1+\frac{1}{2}}-\frac{1}{1+3}x^{1+3}]_0^1\\\\=\frac{2}{3}-\frac{1}{4}\\\\=\frac{5}{12}

.Not very difficult.. right.
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