Proof of Various Integral Properties — Question 1

Question 1

Assume that $f$ is integrable on the interval $[a,b]$ and that $c$ is a constant. Prove that $$\int_a^b c\,f(x)\,dx = c\int_a^b f(x)\,dx.$$

Question 1 - Solution

We prove this property using the definition of the definite integral in terms of Riemann sums.

Let $P$ be a partition of $[a,b]$: $$a=x_0<x_1<\cdots<x_n=b,$$ and let $\Delta x_i = x_i-x_{i-1}$. Choose a sample point $x_i^*$ in each subinterval $[x_{i-1},x_i]$.

A Riemann sum for $f$ on $[a,b]$ is $$\sum_{i=1}^n f(x_i^*)\,\Delta x_i.$$

A Riemann sum for $c f$ on $[a,b]$ is $$\sum_{i=1}^n c f(x_i^*)\,\Delta x_i.$$

Factor out the constant $c$: $$\sum_{i=1}^n c f(x_i^*)\,\Delta x_i = c\sum_{i=1}^n f(x_i^*)\,\Delta x_i.$$

Now take the limit as the norm of the partition $\|P\|\to 0$. Since $f$ is integrable on $[a,b]$, the limit of its Riemann sums exists: $$\lim_{\|P\|\to 0}\sum_{i=1}^n f(x_i^*)\,\Delta x_i = \int_a^b f(x)\,dx.$$

Therefore, $$\lim_{\|P\|\to 0}\sum_{i=1}^n c f(x_i^*)\,\Delta x_i = c\int_a^b f(x)\,dx.$$

By definition of the definite integral, this limit equals $$\int_a^b c\,f(x)\,dx.$$

Hence, $$\int_a^b c\,f(x)\,dx = c\int_a^b f(x)\,dx.$$

$$\boxed{\int_a^b c\,f(x)\,dx = c\int_a^b f(x)\,dx}$$