Proving Definite Integral Property 2

To Prove

If $f+g$ is integrable on $[a,b]$, then ${\int }_a^b[f(x)+g(x)]dx-{\int }_a^{b}f(x)dx={\int }_a^{b}g(x)dx$

Proof

Suppose $f$ and $g$ are integrable on $[a,b]$

• Let $P={x_i^{\ast }}_{i=0}^n$ be a riemann partition Then, by definition of $f(x)$ is integrable: ${\int }_a^{b}f(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^{n}f(x_i\ast )△x,\forall x_i^{\ast }\in [x_1\dots x_i]$ Then, by definition of $g(x)$ is integrable: ${\int }_a^{b}g(x)dx={\lim }_{n\to \infty }{\sum }_{i=1}^{n}g(x_i\ast )△x,\forall x_i^{\ast }\in [x_1\dots x_i]$ Then, ${\int }_a^{b}f(x)dx+{\int }_a^{b}g(x)dx$ … ${\lim }_{n\to \infty }{\sum }_{i=1}^n(f+g)(x_i\ast )△x,\forall x_i^{\ast }\in [x_1\dots x_i]$