Question & Answer
QUESTION

Find the value of the expression given as: $\int\limits_{0}^{\pi }{\sin \theta d\theta }$.

ANSWER Verified Verified
Hint: Use the formulae $\int{\sin xdx}=-\cos x+\,C$ and $\int\limits_{a}^{b}{f(x)dx}=\left[ F(x) \right]_{a}^{b}$ too get the integration and then substitute the values \[\cos 0=1\,and\,\cos \pi =-1\] in the equation to get the final answer.

Complete step-by-step answer:
To solve the above equation we will first write it down and assume it as ‘L’, therefore,
$\Rightarrow L=\int\limits_{0}^{\pi }{\sin \theta d\theta }$ ……………………………………………………. (1)
To solve the above equation we should know the formulae of integration given below,
Formula:
$\int{\sin xdx}=-\cos x+\,C$
$\int\limits_{a}^{b}{f(x)dx}=\left[ F(x) \right]_{a}^{b}$
If we use above two formulae in equation (1) we will get,
$\Rightarrow L=\left[ -\cos \theta \right]_{0}^{\pi }$
Now to proceed further in the solution we should know how to substitute the limits in the given equation and for that we should refer the formula given below,
Formula:
$\left[ f(x) \right]_{a}^{b}=f(b)-f(a)$
If we use the above formula in ‘L’ we will get,
$\Rightarrow L=\left[ \left( -\cos \pi \right)-\left( -\cos 0 \right) \right]$
Further simplification in the above equation will give,
\[\Rightarrow L=\left[ -\cos \pi +\cos 0 \right]\]
As we know, the value of \[\cos 0\] is equal to 1 and the value of \[\cos \pi \] is equal to -1 and if we substitute these values in the above equation we will get,
\[\Rightarrow L=\left[ -\left( -1 \right)+1 \right]\]
Further simplification in the above equation will give,
Therefore, L = 1 + 1
If we simplify the above equation we will get,
Therefore, L = 2.
If we compare the above equation with equation (1) we can write,
\[\Rightarrow \int\limits_{0}^{\pi }{\sin \theta d\theta }=2\]
Therefore the value of \[\int\limits_{0}^{\pi }{\sin \theta d\theta }\] is equal to 2.

Note: Many students commit the mistake of writing the \[\int{\sin \theta d\theta }\] as \[\cos \theta \] in over confidence. But do remember that it is the derivative of \[\sin \theta \] and not the integration.