How do you evaluate the definite integral $\int{\sin xdx}$ from $\left[ 0,\dfrac{\pi }{2} \right]$ ?
Answer
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Hint: We have been given to calculate the definite integral of trigonometric function which is a sine function. It has to be integrated with respect to $dx$. Since we have to perform definite integration, thus after writing the integral sine function equal to cosine function, we shall also apply the upper limit and lower limits of integration. Further we will calculate the final solution after putting the values of $\cos 0$ and $\cos \dfrac{\pi }{2}$.
Complete step by step solution:
Given that we have to find a definite integral of$\int{\sin xdx}$ within the interval $\left[ 0,\dfrac{\pi }{2} \right]$.
From this statement, we understand that the upper limit of the definite integral is $\dfrac{\pi }{2}$ and the lower limit of definite integral is 0.
Thus, the definite integral is given as
$\int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}$
We know that the integral of sine function is negative of cosine function, that is, $\int{\sin xdx}=-\cos x+C$.
Substituting this value, we get
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=\left. -\cos x \right|_{0}^{\dfrac{\pi }{2}}$
Applying the upper and lower limits of integration, we get
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=-\left( \cos \dfrac{\pi }{2}-\cos 0 \right)$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=\cos 0-\cos \dfrac{\pi }{2}$
We know that $\cos 0=1$ and $\cos \dfrac{\pi }{2}=0$. Putting these values, we get
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=1-0$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=1$
Therefore, the definite integral $\int{\sin xdx}$ from $\left[ 0,\dfrac{\pi }{2} \right]$ is equal to 1.
Note:
Definite integral of a function $f\left( x \right)$ is the area bound under the graph of function, $y=f\left( x \right)$and above the x-axis which is bound between two bounds as $x=a$ and $x=b$. Here, $a=$ 0 and $b=\dfrac{\pi }{2}$. One of the powers of integral calculus is that the one of the two boundaries of the area to be found is a curve and the other boundary is the x-axis when the function is being integrated with respect to $dx$.
Complete step by step solution:
Given that we have to find a definite integral of$\int{\sin xdx}$ within the interval $\left[ 0,\dfrac{\pi }{2} \right]$.
From this statement, we understand that the upper limit of the definite integral is $\dfrac{\pi }{2}$ and the lower limit of definite integral is 0.
Thus, the definite integral is given as
$\int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}$
We know that the integral of sine function is negative of cosine function, that is, $\int{\sin xdx}=-\cos x+C$.
Substituting this value, we get
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=\left. -\cos x \right|_{0}^{\dfrac{\pi }{2}}$
Applying the upper and lower limits of integration, we get
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=-\left( \cos \dfrac{\pi }{2}-\cos 0 \right)$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=\cos 0-\cos \dfrac{\pi }{2}$
We know that $\cos 0=1$ and $\cos \dfrac{\pi }{2}=0$. Putting these values, we get
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=1-0$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sin xdx}=1$
Therefore, the definite integral $\int{\sin xdx}$ from $\left[ 0,\dfrac{\pi }{2} \right]$ is equal to 1.
Note:
Definite integral of a function $f\left( x \right)$ is the area bound under the graph of function, $y=f\left( x \right)$and above the x-axis which is bound between two bounds as $x=a$ and $x=b$. Here, $a=$ 0 and $b=\dfrac{\pi }{2}$. One of the powers of integral calculus is that the one of the two boundaries of the area to be found is a curve and the other boundary is the x-axis when the function is being integrated with respect to $dx$.
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