# Homework Help: Calculate definite integrals with given interval.

1. Jul 16, 2013

### 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
$$\sin(\theta)\cos(\theta)d\theta|_0^{\frac{\pi}{2}}=\left[\sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)\right]-[\sin 0\cos 0]= [0][0]=0$$

2. Jul 16, 2013

### Staff: Mentor

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.

3. Jul 16, 2013

### yungman

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

4. Jul 16, 2013

### Staff: Mentor

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.

5. Jul 16, 2013

### yungman

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.

6. Jul 16, 2013

### HallsofIvy

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).