My teacher shown me some analysis problems which can be intuitively proved geometrically (and are simple to remeber).
I am trying to prove geometrically that:
$$ \int {f(x)-g(x)\ dx} = \int {f(x) \ dx} - \int {g(x)\ dx} $$
but I think I can't.
If I am trying to draw a random function graph for $f(x)$ and a random one for $g(x)$ and then trying to draw $f(x)-g(x)$ graph I don't reach any conlusion or what to do next.
Can it be "proved" (to understand why $ \int {f(x)-g(x)\ dx} = \int {f(x) \ dx} - \int {g(x)\ dx} $) by drawing? And how?
(For definite or indefinite integrals)
Thank you!
This is not generally true. However, it is true if you have definite integrals. Therefore I will change the question slightly to proving that: \begin{align*} \int_a^bf(x)-g(x)\ \mathrm{d}x = \int_a^bf(x)\ \mathrm{d}x - \int_a^b g(x)\ \mathrm{d}x \end{align*} First, draw a "random" function for $f(x)$, between $a < x < b$, along with a "random" function for $g(x)$, between $a < x < b$, and $f(x)-g(x)$ between $a$ and $b$.
Then $\int_{a}^{b}f(x)$ is the "area under the graph of $f(x)$ between $a$ and $b$", while $\int_{a}^{b}g(x)$ is the "area under the graph of $g(x)$ between $a$ and $b$". By looking at your pictures you should be able to convince yourself that for any well behaved function defined between $a$ and $b$, the difference between these two areas is equal to the area under the graph of $f(x) - g(x)$. This "proves" your question.