NDA 2015
Exam NDA
Year 2015
Subject Mathematics
Chapter Area Under Curves NDA 2015 · Mathematics · Area Under Curves
The area bounded by the coordinate axes and the curve x + y =1 , is
The area bounded by the coordinate axes and the curve \sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=1$, is
Options A 1 square unit B 1/2 square unit C 1/3 square unit D 1/6 square unit Correct answer D. 1/6 square unit
Step-by-step solution
Area
= ∫ y d x = ∫ 0 1 ( 1 − x ) 2 d x =\int \mathrm{ydx}=\int_0^1(1-\sqrt{x})^2 d x = ∫ ydx = ∫ 0 1 ( 1 − x ) 2 d x [\becausecurve makes the
i n t e r c e p t o f 1 o n b o t h a x e s ] intercept of 1 on both axes] in t erce pt o f 1 o nb o t ha x es ] = ∫ 0 1 ( 1 + x − 2 x ) d x = [ x ] 0 1 + 1 2 [ x 2 ] 0 1 − 4 3 [ x 3 2 ] 0 1 = 1 + 1 2 − 4 3 = 3 2 − 4 3 = 1 6 sq unit \begin{aligned}
& =\int_0^1(1+x-2 \sqrt{x}) d x \\
& =[x]_0^1+\frac{1}{2}\left[x^2\right]_0^1-\frac{4}{3}\left[x^{\frac{3}{2}}\right]_0^1 \\
& =1+\frac{1}{2}-\frac{4}{3} \\
& =\frac{3}{2}-\frac{4}{3}=\frac{1}{6} \text{ sq unit }
\end{aligned} = ∫ 0 1 ( 1 + x − 2 x ) d x = [ x ] 0 1 + 2 1 [ x 2 ] 0 1 − 3 4 [ x 2 3 ] 0 1 = 1 + 2 1 − 3 4 = 2 3 − 3 4 = 6 1 sq unit $$
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