Question

The area between curve y=x^2 and the line y=2 is:

Answer 1

Given : y=x² and the line y=2

To Find : Area between

Solution:

y=x² and the line y=2

x² = 2

=> x = ±√2

about y axis

y=x²

=> x = √y

y from 0 to 2

area is Symmetric about y axis

Hence Area between

    $=2\int\limits^2 _0 {\sqrt{y} } \, dy$

$=2 ( {\dfrac{y^{ \frac{3}{2} }}{ \frac{3}{2} } }\left\right|_0^2)$

= 4 * 2√2/3

= 8√2/3

The area between curve y=x^2 and the line y=2 is  8√2/3 sq units

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Answer 2

SOLUTION

TO DETERMINE

The area between curve y = x² and the line y = 2

EVALUATION

Here the given equation of the curve is

y = x²

We have to find the area between curve y = x² and the line y = 2

We Draw the figure with the given data

OACO is required region

The region is symmetric about y axis

Hence the required area

= The area of the region OACO

= 2 × The area of the region OABO

$\displaystyle \sf = 2 \times \int\limits_{0}^{2} x \, dy$

$\displaystyle \sf = 2 \times \int\limits_{0}^{2} \sqrt{y} \, dy$

$\displaystyle \sf = 2 \left. \frac{ {y}^{ \frac{3}{2} } }{ \frac{3}{2} } \right|_0^2$

$\displaystyle \sf = \frac{4}{3} \left. {y}^{ \frac{3}{2} } \right|_0^2$

$\displaystyle \sf = \frac{4}{3} \times {2}^{ \frac{3}{2} }$

$\displaystyle \sf = \frac{4 \times 2 \sqrt{2} }{3}$

$\displaystyle \sf = \frac{8 \sqrt{2} }{3} \: \: \: sq.unit$

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