let $f$ and $g$ be Riemann integrable functions on $[a,b]$ and $c\in R$ be a constant.

Question

I am having trouble with the following problem.

If we let $f$ and $g$ be Riemann integrable functions on $[a,b]$ and $c\in R$ be a constant.

I need to show that $$\displaystyle \int_a^b cf(x)dx=c\int_a^b f(x)dx$$

What happens if we consider the cases $c>0$ and $c<0$ separately?

Answer 1

Let $\phi(x)=cf(x)$. Then the Riemann integral of $\phi$ on $[a,b]$ is by definition (I use the usual $1/n$ length partition for convenience, I also assume you can take $\phi$ to be Riemann integrable), $$ \int_a^bcf(x)dx=\int_a^b \phi(x)dx=\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=0}^n\phi(x^*)(x_{k+1}-x_k) $$ where $x^*\in (x_{k},x_{k+1})$. But this is $$ \int_a^bcf(x)dx=\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=0}^ncf(x^*)(x_{k+1}-x_k)=c\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=0}^nf(x^*)(x_{k+1}-x_k)=c\int_{a}^bf(x)dx $$

Answer 2

It is straightforward to prove from the definition:

Suppose $c>0$. Let $P$ be a partition. Then $L(cf, P) = c L(f,P)$ and $U(cf,P) = c U(f,P)$. It follows from this that $\int (cf) = c (\int f)$.

Now suppose $c<0$. Then $L(cf,P) = cU(f,P) $ and $U(cf,P) = cL(f,P) $. Again, it follows from this that $\int (cf) = c (\int f)$.