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Question: The area enclosed by the curves \(y = \sin x + \cos x\) and \(y = \left| {\cos x - \sin x} \right|\)...

The area enclosed by the curves y=sin⁡x+cos⁡xy = \sin x + \cos x and y=∣cos⁡x−sin⁡x∣y = \left| {\cos x - \sin x} \right| over the interval [0,π2]\left[ {0,\dfrac{\pi }{2}} \right]is
(A). 4(2−1)4\left( {\sqrt 2 - 1} \right)
(B). 22(2−1)2\sqrt 2 \left( {\sqrt 2 - 1} \right)
(C). 2(2+1)2\left( {\sqrt 2 + 1} \right)
(D). 22(2+1)2\sqrt 2 \left( {\sqrt 2 + 1} \right)

Explanation

Solution

Before attempting this question, one should have prior knowledge about the concept of area between two curves and also remember to use the method of integration to find the area enclosed by the equation of given curves.

Complete step-by-step answer :
According to the given information we have two curves with equations y=sin⁡x+cos⁡xy = \sin x + \cos x and y=∣cos⁡x−sin⁡x∣y = \left| {\cos x - \sin x} \right|with the intervals [0,π2]\left[ {0,\dfrac{\pi }{2}} \right]
Let take y=sin⁡x+cos⁡xy = \sin x + \cos x as equation 1 and y=∣cos⁡x−sin⁡x∣y = \left| {\cos x - \sin x} \right|as equation 2
Multiplying and dividing the R.H.S of equation 1 by 2\sqrt 2 we get
y=2(sin⁡x12+cos⁡x12)y = \sqrt 2 \left( {\sin x\dfrac{1}{{\sqrt 2 }} + \cos x\dfrac{1}{{\sqrt 2 }}} \right)
As we know that sin⁡π4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}and cos⁡π4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}also we know that sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y
Therefore, y=2(sin⁡(x+π4))y = \sqrt 2 \left( {\sin \left( {x + \dfrac{\pi }{4}} \right)} \right) (equation 3)
Now multiplying and dividing R.H.S equation 2 by 2\sqrt 2 we get
y=2∣cos⁡x12−sin⁡x12∣y = \sqrt 2 \left| {\cos x\dfrac{1}{{\sqrt 2 }} - \sin x\dfrac{1}{{\sqrt 2 }}} \right|
As we know that sin⁡π4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}and cos⁡π4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}also we know that cos⁡(x+y)=sin⁡xcos⁡y−cos⁡xsin⁡y\cos \left( {x + y} \right) = \sin x\cos y - \cos x\sin y
Therefore, y=2∣cos⁡(x+π4)∣y = \sqrt 2 \left| {\cos \left( {x + \dfrac{\pi }{4}} \right)} \right| (equation 4)
For x = 0
Value of equation 3 i.e. 2(sin⁡(x+π4))\sqrt 2 \left( {\sin \left( {x + \dfrac{\pi }{4}} \right)} \right)will be 1
For x = π4\dfrac{\pi }{4}
Value of equation 3 i.e. 2(sin⁡(x+π4))\sqrt 2 \left( {\sin \left( {x + \dfrac{\pi }{4}} \right)} \right)will be 2\sqrt 2
For x = π2\dfrac{\pi }{2}
Value of equation 3 i.e. 2(sin⁡(x+π4))\sqrt 2 \left( {\sin \left( {x + \dfrac{\pi }{4}} \right)} \right)will be 1
The curve will be

For x = 0
Value of equation 4 i.e. 2∣cos⁡(x+π4)∣\sqrt 2 \left| {\cos \left( {x + \dfrac{\pi }{4}} \right)} \right|will be 1
For x = π4\dfrac{\pi }{4}
Value of equation 4 i.e. 2∣cos⁡(x+π4)∣\sqrt 2 \left| {\cos \left( {x + \dfrac{\pi }{4}} \right)} \right|will be 0
For x = π2\dfrac{\pi }{2}
Value of equation 4 i.e. 2∣cos⁡(x+π4)∣\sqrt 2 \left| {\cos \left( {x + \dfrac{\pi }{4}} \right)} \right|will be 1
Hence, the curve will be

So, the resulting curve of the both the equation will be

So, the resulting area enclosed by the curves will be ABCD
Let's find the area enclosed by the given curves over the interval [0,π2]\left[ {0,\dfrac{\pi }{2}} \right]i.e. ABCD
From the graph, we want the area of ABCD and to find that we can find the area of ABD and BCD separately
So, the area of ABD over the interval [0,π4]\left[ {0,\dfrac{\pi }{4}} \right]= upper curve (AB) – lower curve (AD)
Therefore, area of ABD = ∫0π4((sin⁡x+cos⁡x)−(cos⁡x−sin⁡x))dx\int_0^{\dfrac{\pi }{4}} {\left( {\left( {\sin x + \cos x} \right) - \left( {\cos x - \sin x} \right)} \right)dx}
⇒\RightarrowArea of ABD = ∫0π4(sin⁡x+cos⁡x−cos⁡x+sin⁡x)dx\int_0^{\dfrac{\pi }{4}} {\left( {\sin x + \cos x - \cos x + \sin x} \right)dx}
⇒\RightarrowArea of ABD = ∫0π42sin⁡xdx\int_0^{\dfrac{\pi }{4}} {2\sin xdx}
⇒\RightarrowArea of ABD = 2∫0π4sin⁡xdx2\int_0^{\dfrac{\pi }{4}} {\sin xdx}
As we know that ∫sin⁡x=−cos⁡x\int {\sin x = - \cos x}
Therefore, Area of ABD = 2[−cos⁡x]0π42\mathop {\left[ { - \cos x} \right]}\nolimits_0^{\dfrac{\pi }{4}}
Also, we know that ∫baf(x)dx=F(a)−F(b)\int_b^a {f\left( x \right)dx} = F\left( a \right) - F\left( b \right)
Therefore, area of ABD = 2(−cos⁡(π4)−(−cos⁡(0)))2\left( { - \cos \left( {\dfrac{\pi }{4}} \right) - \left( { - \cos \left( 0 \right)} \right)} \right)
Since, cos⁡(π4)=12\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} and cos⁡(0)=1\cos \left( 0 \right) = 1
So, Area of ABD = 2(−12+1)2\left( { - \dfrac{1}{{\sqrt 2 }} + 1} \right)
Now the area of BCD over the interval [π4,π2]\left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right]= upper curve (BC) – lower curve (CD)
Since, we know that for interval [π4,π2]\left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right] value of sin x dominates the value of cos x
Therefore, area of BCD = ∫π4π2((sin⁡x+cos⁡x)−(sin⁡x−cos⁡x))dx\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {\left( {\left( {\sin x + \cos x} \right) - \left( {\sin x - \cos x} \right)} \right)dx}
⇒\RightarrowArea of BCD = ∫π4π2(sin⁡x+cos⁡x−sin⁡x+cos⁡x)dx\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {\left( {\sin x + \cos x - \sin x + \cos x} \right)dx}
⇒\RightarrowArea of BCD = ∫π4π22cos⁡xdx\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {2\cos xdx}
⇒\RightarrowArea of BCD = 2∫π4π2cos⁡xdx2\int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {\cos xdx}
As we know that ∫cos⁡x=sin⁡x\int {\cos x = \sin x}
Therefore, Area of BCD = 2[sin⁡x]π4π22\mathop {\left[ {\sin x} \right]}\nolimits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}
Also, we know that ∫baf(x)dx=F(a)−F(b)\int_b^a {f\left( x \right)dx} = F\left( a \right) - F\left( b \right)
Therefore, area of BCD = 2(sin⁡(π2)−sin⁡(π4))2\left( {\sin \left( {\dfrac{\pi }{2}} \right) - \sin \left( {\dfrac{\pi }{4}} \right)} \right)
Since, sin⁡(π4)=12\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} and sin⁡(π2)=1\sin \left( {\dfrac{\pi }{2}} \right) = 1
So, Area of BCD = 2(1−12)2\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right)
We know that the area of ABCD = area of ABD + area of BCD
Substituting the value in the above equation we get
Area of ABCD = 2(−12+1)+2(1−12)2\left( { - \dfrac{1}{{\sqrt 2 }} + 1} \right) + 2\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right)
⇒\Rightarrow Area of ABCD = 2(2−22)2\left( {2 - \dfrac{2}{{\sqrt 2 }}} \right)
⇒\Rightarrow Area of ABCD = 4(2−12)4\left( {\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}} \right)
⇒\Rightarrow Area of ABCD = 22(2−1)2\sqrt 2 \left( {\sqrt 2 - 1} \right)
Therefore, area enclosed by the curves y=sin⁡x+cos⁡xy = \sin x + \cos x and y=∣cos⁡x−sin⁡x∣y = \left| {\cos x - \sin x} \right|is equal to 22(2−1)2\sqrt 2 \left( {\sqrt 2 - 1} \right)
Hence, option B is the correct option.

Note :In the above solution we used the term “curves” which can be explained as the path produced by the continuity of moving point. The path produced by the moving point is created by the equation there are different types of curves such as simple curve, closed curve, algebraic curves, etc.