Area under the curve y -sin 2 x -cos 2 x between x -0 and x -4- isA- 2 sq- unitB- 1 sq- unitC- 3 sq- unitD- 4 sq- unit

The correct option is B 1 sq. unitRequired area = ∫_0^π/4(sin2x+cos2x)dx = [ cos2x/2 +sin2x/2 ] _0^π/4 =1/2 [ cosπ/2+sinπ/2+cos0 sin0 ]=1 sq. unit ...

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Standard XII

Mathematics

Property 7

Area under th...

Question

Area under the curve $y = s i n 2 x + c o s 2 x$ between $x = 0$ and $x = \frac{π}{4}$ , is

2 sq. unit

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1 sq. unit

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3 sq. unit

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4 sq. unit

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Solution

The correct option is B 1 sq. unit
Required area = $\int_{0}^{\frac{π}{4}} (s i n 2 x + c o s 2 x) d x$
$= {[- \frac{c o s 2 x}{2} + \frac{s i n 2 x}{2}]}_{0}^{\frac{π}{4}}$
$= \frac{1}{2} [- c o s \frac{π}{2} + s i n \frac{π}{2} + c o s 0 - s i n 0] = 1$ sq. unit

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Area under the curve y=sin2x+cos2x between x=0 and x=π4, is

The correct option is B 1 sq. unitRequired area = ∫π40(sin2x+cos2x)dx =[−cos2x2+sin2x2]π40 =12[−cosπ2+sinπ2+cos0−sin0]=1 sq. unit

Area under the curve $y = s i n 2 x + c o s 2 x$ between $x = 0$ and $x = \frac{π}{4}$ , is

The correct option is B 1 sq. unit
Required area = $\int_{0}^{\frac{π}{4}} (s i n 2 x + c o s 2 x) d x$
$= {[- \frac{c o s 2 x}{2} + \frac{s i n 2 x}{2}]}_{0}^{\frac{π}{4}}$
$= \frac{1}{2} [- c o s \frac{π}{2} + s i n \frac{π}{2} + c o s 0 - s i n 0] = 1$ sq. unit