The line x-4 divide the area of the region bounded by y-sin x- y - cos x and X-axis 0 - x -2 into two regions of areas A1 and A2- Then- A1-A2 equals

The correct option is D 1:1 Area A_1=∫_0^π/4sin x dx = [cos x]_0^π/4 =1 1√(2)=√(2) 1√(2) and Area A_2=∫_π/4^π/2cos x dx = [sin x] ...

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Byju's Answer

Standard XII

Mathematics

Area between Two Curves

The line x=...

Question

The line $x = \frac{π}{4}$ divide the area of the region bounded by $y = sin x, y = cos x$ and X-axis $(0 \leq x \leq \frac{π}{2})$ into two regions of areas $A_{1}$ and $A_{2}$ . Then, $A_{1} : A_{2}$ equals

4 : 1

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3 : 1

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2 : 1

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1 : 1

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Solution

The correct option is D $1 : 1$
Area $A_{1} = \int_{0}^{\frac{π}{4}} sin x d x$
$= - [cos x]_{0}^{\frac{π}{4}}$
$= 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

and Area $A_{2} = \int_{\frac{π}{4}}^{\frac{π}{2}} cos x d x$
$= [sin x]_{\frac{π}{4}}^{\frac{π}{2}} = [1 - \frac{1}{\sqrt{2}}] = \frac{\sqrt{2} - 1}{\sqrt{2}}$
$∴ A_{1} : A_{2} = \frac{\sqrt{2} - 1}{\sqrt{2}} : \frac{\sqrt{2} - 1}{\sqrt{2}} = 1 : 1$

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Similar questions

Q. If

A_{1}

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y = cos x, y = sin x

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and

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(0, \frac{π}{2})

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\frac{A_{1}}{A_{2}}

equals to :

Q. The area of the region between the curves

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Q. Let

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axis in the first quadrant. Also, let

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axis and

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Q. The parabola

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The line x=π4 divide the area of the region bounded by y=sinx,y=cosx and X-axis (0≤x≤π2) into two regions of areas A1 and A2. Then, A1:A2 equals

The correct option is D 1:1Area A1=∫π40sinxdx=−[cosx]π40=1−1√2=√2−1√2and Area A2=∫π2π4cosxdx=[sinx]π2π4=[1−1√2]=√2−1√2 ∴A1:A2=√2−1√2:√2−1√2=1:1

The line $x = \frac{π}{4}$ divide the area of the region bounded by $y = sin x, y = cos x$ and X-axis $(0 \leq x \leq \frac{π}{2})$ into two regions of areas $A_{1}$ and $A_{2}$ . Then, $A_{1} : A_{2}$ equals