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} $