# Differential equation True/ False

Every continuous function has an antiderivative

I thought this statement was false, but it seems that it is true.

I thought that it suppose to be every antiderivative is a continuous but the other way around is wrong. Similarly, when you say every continuous functions is not a derivative, but every derivative is a continuous.

If $F(x) = \int_0^x f(t) \, dt$, then $F(5) - F(3) = \int_3^5 f(t) \, dt$

True.

If $F(x) = \int_0^x f(t) \, dt$ and $G(x) = \int_2^x f(t) \, dt$ then $F(x) = G(x) + C$

False, it seems weird to me. I do not have a logical reason.

Could anyone check my answers and tell me if I am right or wrong?

• Your pictures are slightly chopped off at the bottom. Are the lower bounds of integration 0 and 2 respectively? – Semiclassical Jul 15 '14 at 6:11
• @Semiclassical yes it is, and sorry because I don't know how to enter the symbols. – user157908 Jul 15 '14 at 6:13
• Since there are no given bounds on C, (C could be +ve, -ve, or zero), the answer (true or false) does not depend on what the limits of integration are for F(x) and G(x). – Ozzah Jul 15 '14 at 6:19
• Regarding continuous functions: intuitively, if you can draw a curve with one progressing line (perhaps jaggedly, but not broken) can you associate an area under it? If so, there should be some function which describes how the area increases as you move left to right. – Semiclassical Jul 15 '14 at 6:20
• Also I would reread over the fundamental theorem of calculus - this shows that there exists an antiderivative of $f$ given $f$ is continuous and how to solve the other problems. – DanZimm Jul 15 '14 at 6:34

Hint: $$\int_0^xf(t)dt = \int_0^2f(t) dt + \int_2^x f(t)dt.$$
$\int_0^2f(t) dt$ is just a number, a constant number not depending on $x$.
When $f$ is continuous on an interval $I\subset{\mathbb R}$ then for any three points $a$, $b$, $c\in I$ (in any order) one has $$\int_a^c f(t)\ dt=\int_a^b f(t)\ dt+\int_b^c f(t)\ dt\ .$$