Answer
See proof
Work Step by Step
Suppose that the antiderivative of $kf(x)$ is $F(x)$ so:
$$F'(x)=kf(x)$$
Using the $FTC$ it follows:
$$\int_{a}^{b}kf(x)dx=F(b)-F(a)$$
$$k\int_{a}^{b}f(x)dx=k \int_{a}^{b}\frac{F'(x)}{k}dx=k\left(\frac{F(b)}{k}-\frac{F(a)}{k}\right)=k\frac{F(b)-F(a)}{k}=F(b)-F(a)$$
so:
$$\int_{a}^{b}kf(x)dx=k\int_{a}^{b}f(x)dx$$
