Question

Find the area bounded by the curve $y = \sin x$ and $y = \cos x$ between any two consecutive points of intersection.

Answer 1

Hint : As we can see that the above question contains trigonometry as sine and cosine are trigonometric ratios. We have to find the area , so will integrate the values . We know that to calculate the area under the curves is by integration method which is $A = \int\limits_{x = {x_1}}^{x = {x_2}} {({y_1} - {y_2})dx}$ , where ${y_1},{y_2}$ are the upper and lower curves.

Complete step by step solution:
We know that $y = \sin x = \cos x$ when $x = \dfrac{\pi }{4},\dfrac{{5\pi }}{4}$ .
So we can say that $x = {x_2} = \dfrac{{5\pi }}{4}$ and $x = {x_1} = \dfrac{\pi }{4}$ . We have ${y_1} = \sin x$ and ${y_2} = \cos x$ .
Now by putting the values in the formula we have
$A = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{5\pi }}{4}} {(\sin x - \cos x)dx}$
We will break them into parts:
$A = \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{5\pi }}{4}} {\sin xdx - } \int\limits_{\dfrac{\pi }{4}}^{\dfrac{{5\pi }}{4}} {\cos xdx}$ .
We know that integration of $\sin x$ is $- \cos x$ and integration of $\cos x = \sin x$ .
So we can write the above as ${[ - \cos x]_{\dfrac{\pi }{4}}}^{\dfrac{{5\pi }}{4}} - {[\sin x]_{\dfrac{\pi }{4}}}^{\dfrac{{5\pi }}{4}}$ . From the above we know that
$x = {x_2} = \dfrac{{5\pi }}{4}$ and $x = {x_1} = \dfrac{\pi }{4}$ ,
so we will put the values of $x$ in the expression. It can be written as
$\left( {\cos \dfrac{\pi }{4} - \cos \dfrac{{5\pi }}{4}} \right) - \left( {\sin \dfrac{{5\pi }}{4} - \sin \dfrac{\pi }{4}} \right)$ .
We know the trigonometric values of
$\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{{5\pi }}{4} = \cos \dfrac{{5\pi }}{4} = - \dfrac{1}{{\sqrt 2 }}$ .
Now we substitute the values of these in the above expression:
$\dfrac{1}{{\sqrt 2 }} - \left( { - \dfrac{1}{{\sqrt 2 }}} \right) - \left( { - \dfrac{1}{{\sqrt 2 }}} \right) + \dfrac{1}{{\sqrt 2 }}$ .
We will simplify them now:
$\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} = \dfrac{4}{{\sqrt 2 }}$ . We can write it as $\dfrac{{2 \times \sqrt 2 \times \sqrt 2 }}{{\sqrt 2 }} = 2\sqrt 2$ .
Hence the required area is $2\sqrt 2$ .
So, the correct answer is “$2\sqrt 2$ SQ units”.

Note : Before solving this kind of question we should have the full knowledge of trigonometric values and their functions. In the formula used above ${x_1}$ and ${x_2}$ are the upper and lower limits. In this type of question we should always remember the basic integration formula such as $\int {\sin x dx = - \cos x + C}$ and $\int {\cos xdx = \sin x + C}$