Somas, Diferenças e Multiplicação por Constantes em Integrandos

We can break up the definite integral (over an interval $[a,b]$) of a complicated combination of functions by applying the following properties of integrals:

$$\int_a^b[f(x)+g(x)]\,\text{d}x=\int_a^bf(x)\,\text{d}x+\int_a^bg(x)\,\text{d}x\qquad(1)$$ $$\int_a^b[f(x)-g(x)]\,\text{d}x=\int_a^bf(x)\,\text{d}x-\int_a^bg(x)\,\text{d}x\qquad(2)$$ $$\int_a^bcf(x)\,\text{d}x=c\int_a^bf(x)\,\text{d}x\qquad(3)$$

For example,

$$\begin{align*} \int_0^1(x^2+4x)\,\text{d}x&=\int_0^1x^2\,\text{d}x+\int_0^14x\,\text{d}x\qquad(1)\\ &=\int_0^1x^2\,\text{d}x+4\int_0^1x\,\text{d}x\qquad(3)\\ &=\left[\frac{x^3}3\right]_0^1+4\left[\frac{x^2}2\right]_0^1\\ &=\left[\frac{1^3}3-\frac{0^3}3\right]+4\left[\frac{1^2}2-\frac{0^2}2\right]\\ &=\frac13+4\cdot\frac12=\frac13+2=\frac73 \end{align*}$$

For two single variable functions $f$ and $g$ that are continuous except at a finite number of points on the interval $[a,b]$, the following linearity principle holds: $$\int_{a}^{b}[f(x)+g(x)]dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$$ To prove this, suppose that $F(x)$ is an antiderivative of $f(x)$ and that $G(x)$ is an antiderivative of $g(x)$. Now, we may use the linearity property of derivatives, which follows from the corresponding property for limits: $$\frac{d}{dx}[F(x)+G(x)]=\frac{d}{dx}F(x)+\frac{d}{dx}G(x)$$ $$\frac{d}{dx}[F(x)+G(x)]=f(x)+g(x)$$ Thus, we see that $F(x)+G(x)$ is an antiderivative of $f(x)+g(x)$.

Using the Fundamental Theorem of Calculus, $$\int_{a}^{b}[f(x)+g(x)]dx=\left[F(x)+G(x)\right]_{a}^{b}$$ $$\int_{a}^{b}[f(x)+g(x)]dx=F(b)+G(b)-F(a)-G(a)$$ $$\int_{a}^{b}[f(x)+g(x)]dx=[F(b)-F(a)]+[G(b)-G(a)]$$ Using the Fundamental Theorem of Calculus one more time, we obtain our desired result: $$\int_{a}^{b}[f(x)+g(x)]dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$$ Another way to prove this is to work directly with the definition of an integral, and to add up the areas of the approximating rectangles. This leads us to exactly the same conclusion.