Area Under Curves Question 11
Question 11 - 31 January - Shift 2
Let the area of the region $\lbrace (x, y):|2 x-1| \leq y \leq \mid x^{2}-x \mid, 0 \leq x \leq 1 \rbrace $ be A. Then $(6 A+11)^{2}$ is equal to
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Answer: 125
Solution:
Formula: Area between two curves - Area enclosed between two curves intersecting at two different points
$ y \geq|2 x-1|, y \leq|x^{2}-x| $
Both curve are symmetric about $x=\frac{1}{2}$ Hence
$ \begin{aligned} & A=2 \int _{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}((x-x^{2})-(1-2 x)) d x \\ & A=2 \int _{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}(-x^{2}+3 x-1) d x=2(\frac{-x^{3}}{3}+\frac{3}{2} x^{2}-x) _{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}} \end{aligned} $
On solving $6 A+11=5 \sqrt{5}$
$(6 A+11)^2=(5 \sqrt{5})^2= 125$
