Question

Find the area bounded by the curve y = cosx, the x-axis and the ordinates x= 0 and $x=2\pi$.

Answer 1

Hint: Use the fact that the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x= b is given by $\int_{a}^{b}{\left| f\left( x \right) \right|dx}$. Hence the required area is given by $\int_{0}^{2\pi }{\left| \cos x \right|dx}$. Draw the graph of $y=\cos x$ and observe that in the intervals $\left( 0,\dfrac{\pi }{2} \right)$ and $\left( \dfrac{3\pi }{2},2\pi \right)$ cosx is positive and in the interval $\left( \dfrac{\pi }{2},\dfrac{3\pi }{2} \right)$ cosx is negative. Use the fact that $\forall c\in \left( a,b \right)\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{c}{f\left( x \right)dx}+\int_{c}^{b}{f\left( x \right)dx}$. Hence prove that $\int_{0}^{2\pi }{\left| \cos x \right|dx}=\int_{0}^{\dfrac{\pi }{2}}{\cos xdx}-\int_{\dfrac{\pi }{2}}^{\dfrac{3\pi }{2}}{\cos xdx}+\int_{\dfrac{3\pi }{2}}^{2\pi }{\cos xdx}$. Hence find the required area.

Complete step-by-step answer:

We know that the area bounded by the curve y = f(x), the x-axis and the ordinates x = a and x= b is given by $\int_{a}^{b}{\left| f\left( x \right) \right|dx}$.

Hence, we have

Required area $=\int_{0}^{2\pi }{\left| \cos x \right|dx}$

Now from the graph, it is evident that in the intervals [0, A] and [B, C] cosx is positive, and in the interval [A, B] cosx is negative.Here $A=\dfrac{\pi }{2},B=\dfrac{3\pi }{2},C=2\pi$

Hence, we have in the interval [0,A] and [B,C] ,|cosx| = cosx and in the interval [A,B] ,|cosx|=-cosx

Now know that $\forall c\in \left( a,b \right)\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{c}{f\left( x \right)dx}+\int_{c}^{b}{f\left( x \right)dx}$. Hence, we have

$\int_{0}^{2\pi }{\left| \cos x \right|dx}=\int_{0}^{\dfrac{\pi }{2}}{\left| \cos x \right|dx}+\int_{\dfrac{\pi }{2}}^{\dfrac{3\pi }{2}}{\left| \cos x \right|dx}+\int_{\dfrac{3\pi }{2}}^{2\pi }{\left| \cos x \right|dx}$

Now, we know that in the intervals $\left( 0,\dfrac{\pi }{2} \right)$ and $\left( \dfrac{3\pi }{2},2\pi \right),\left| \cos x \right|=\cos x$ and in the interval $\left( \dfrac{\pi }{2},\dfrac{3\pi }{2} \right),\left| \cos x \right|=-\cos x$.

Hence, we have $\int_{0}^{2\pi }{\left| \cos x \right|dx}=\int_{0}^{\dfrac{\pi }{2}}{\cos xdx}-\int_{\dfrac{\pi }{2}}^{\dfrac{3\pi }{2}}{\cos xdx}+\int_{\dfrac{3\pi }{2}}^{2\pi }{\cos xdx}$.

Hence, we have

Total area $=\int_{0}^{\dfrac{\pi }{2}}{\left| \cos x \right|dx}+\int_{\dfrac{\pi }{2}}^{\dfrac{3\pi }{2}}{\left| \cos x \right|dx}+\int_{\dfrac{3\pi }{2}}^{2\pi }{\left| \cos x \right|dx}$

Now, we know that $\int{\cos xdx}=\sin x$

Hence, we have

Total area $$\begin{aligned}

& =\left( \left. \sin x \right|{0}^{\dfrac{\pi }{2}} \right)-\left( \left. \sin x \right|{\dfrac{\pi }{2}}^{\dfrac{3\pi }{2}} \right)+\left( \left. \sin x \right|_{\dfrac{3\pi }{2}}^{2\pi } \right)=\left( \sin \dfrac{\pi }{2}-0 \right)-\left( \sin \left( \dfrac{3\pi }{2} \right)-\sin \left( \dfrac{\pi }{2} \right) \right)+\left( \sin \left( 2\pi \right)-\sin \left( \dfrac{3\pi }{2} \right) \right) \\

& =\left( 1-0 \right)-\left( -1-1 \right)-\left( 0-1 \right)=1+2+1=4 \\

\end{aligned}$$

Hence the total area = 4 square units

Note: [1] Alternative Solution: Best method

As is evident from the graph the total area is four times the area in the interval [0, A]

Hence, we have

Total area $=4\int_{.0}^{\dfrac{\pi }{2}}{\cos xdx}=4\left( \sin \left( \dfrac{\pi }{2} \right)-\sin \left( 0 \right) \right)=4$, which is the same as obtained above.