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Find the area bounded by the curve y = sinx between x = 0 and x = 2 \pi

Answers (1)

Below is the graph with the shaded region whose area has to be calculated 

From the plot, it is seen that the area from 0 to \pi is positive whereas the area from \pi to 2\pi is negative.

But both the areas are equal in magnitude with opposite signs. 

If we integrate them, the areas will cancel each other and the answer will be 0. 


Therefore, we integrate the two areas separately and split the limits in two ways – 


1) Find area under sinx from 0 to  \pi   and multiple by 2

2)Split the limit 0 to 2  \pi   into 0 to   \pi    and   \pi    to 2  \pi

Here we will solve by splitting the limits 
Y = sinx \\
\begin{aligned} &\text { Integrating from } 0 \text { to } 2 \pi\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{2 \pi} \sin \mathrm{xdx}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi}^{2 \pi} \sin \mathrm{xdx}\\ &\text { Because } \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x} \text { where } c \in(\mathrm{a}, \mathrm{b}) \end{aligned} \\

Now the limits become negative 
\begin{aligned} &\begin{array}{l} \text { Hence for } \pi \text { to } 2 \pi \sin x \text { will become }-\sin x \end{array}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{y} \mathrm{dx}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi}^{2 \pi}-\sin \mathrm{x} \mathrm{d} \mathrm{x}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{\pi} \sin \mathrm{xdx}-\int_{\pi}^{2 \pi} \sin \mathrm{xdx}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=[-\cos \mathrm{x}]_{0}^{\pi}-[-\cos \mathrm{x}]_{\pi}^{2 \pi}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=(-(\cos \pi-\cos 0))-(-(\cos 2 \pi-\cos \pi))\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=(-((-1)-1))-(-(1-(-1))) \end{aligned} \\
\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=4 $$

Hence area bounded by sinx from 0 to 2 \pi \text{ is }4 unit $^{2}$ \\

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