Reimann integral proof by induction

Question:

Suppose that f(x)=x for all x∈[0,b]. Show that f is integrable and that ∫₀^{b}f(x)dx=((bē)/2)

Hint for the proof:

For each n∈ℕ consider the partition P_{n}={0,(b/n),((2b)/n),...,(((n-1)b)/n),b}.

Induction after this but it can't be at n=0 because we end up dividing by 0?

Need some help on this one. Using Reimann.

Thanks.

Re: Reimann integral proof by induction

It should be , and I think it's only a typo. We can compute the integral thanks to this partition, but before that we have to show that the function is Riemann-integrable.

Re: Reimann integral proof by induction

Quote:

Originally Posted by

**CountingPenguins** Question:

Suppose that f(x)=x for all x∈[0,b]. Show that f is integrable and that ∫₀^{b}f(x)dx=((bē)/2)

Hint for the proof:

For each n∈ℕ consider the partition P_{n}={0,(b/n),((2b)/n),...,(((n-1)b)/n),b}.

Induction after this but it can't be at n=0 because we end up dividing by 0?

Need some help on this one. Using Reimann.

Thanks.

Hi CountingPenguins,

There is no strict rule as to include 0 in the natural number set or not (Refer: Natural number - Wikipedia, the free encyclopedia). In this context they must have meant the natural numbers as the positive integers, because including 0 would be meaningless.