Calculate definite integrals with given interval.

[yungman]

I just want to verify is this the way to calculate the result of a definite integral with the given interval. Say the result of the integral over [0,$\frac{\pi}{2}$] is
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}$$

It should be:
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}=\left[\sin\left(\frac{\pi}{2}\right)-\sin 0\right]\left[\cos\left(\frac{\pi}{2}\right)-\cos 0\right]=[1-0][0-1]=1$$

NOT
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}=\left[\sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)\right]-[\sin 0\cos 0]= [0][0]=0$$

 

[Mark44]

I just want to verify is this the way to calculate the result of a definite integral with the given interval. Say the result of the integral over [0,$\frac{\pi}{2}$] is
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}$$

If the above is the antiderivative, there shouldn't be a factor of $d\theta$. What was the original integral?

It should be:
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}=\left[\sin\left(\frac{\pi}{2}\right)-\sin 0\right]\left[\cos\left(\frac{\pi}{2}\right)-\cos 0\right]=[1-0][0-1]=1$$

NOT
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}=\left[\sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)\right]-[\sin 0\cos 0]= [0][0]=0$$

Based on what you show as the result of integration, the NOT example is the correct way of evaluating the antiderivative. The "should be" example is incorrect.

Again, it would help to see where you're starting from.

 

[yungman]

If the above is the antiderivative, there shouldn't be a factor of $d\theta$. What was the original integral?Based on what you show as the result of integration, the NOT example is the correct way of evaluating the antiderivative. The "should be" example is incorrect.

Again, it would help to see where you're starting from.

Thanks, It is part of a longer equation:
$$\int \sin^m\theta d\theta=\frac{1}{m}\sin^{m-1}\theta\cos\theta+\frac{m-1}{m}\int \sin^{m-2}\theta d\theta$$

I just take out the first term for the question.

Can you explain why the second way is the correct way instead? I thought you should take care of the interval term by term!

Thanks

 

[Mark44]

Let's look at your integral with m = 2. Your equation would simplify to this:
$$ \int_0^{\pi/2} \sin^2(\theta)d\theta = \left.(1/2) \sin(\theta)\cos(\theta)\right|_0^{\pi/2} + (1/2)\int_0^{\pi/2} d\theta $$

To evaluate the antiderivative just to the right of the =, above, do just what you did in your NOT example.

 

[yungman]

Let's look at your integral with m = 2. Your equation would simplify to this:
$$ \int_0^{\pi/2} \sin^2(\theta)d\theta = \left.(1/2) \sin(\theta)\cos(\theta)\right|_0^{\pi/2} + (1/2)\int_0^{\pi/2} d\theta $$

To evaluate the antiderivative just to the right of the =, above, do just what you did in your NOT example.

Thanks
This is really part of the question on the Bessel Function where this appears to be the pivotal part in the integral representation of Bessel Function.

 

[HallsofIvy]

I just want to verify is this the way to calculate the result of a definite integral with the given interval. Say the result of the integral over [0,$\frac{\pi}{2}$] is
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}$$

It should be:
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}=\left[\sin\left(\frac{\pi}{2}\right)-\sin 0\right]\left[\cos\left(\frac{\pi}{2}\right)-\cos 0\right]=[1-0][0-1]=1$$

First, [1- 0][0-1]= -1, not 1. More importantly $$\left[f(x)g(x)\right]_a^b= f(b)g(b)- f(a)g(a)$$ which is NOT equal to (f(b)- f(a))(g(b)- g(a))= f(b)g(b)- f(a)g(b)- f(b)g(a)+ f(a)g(a).

NOT
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}=\left[\sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)\right]-[\sin 0\cos 0]= [0][0]=0$$