# Modification of second fundamental theorem of calculus

Let $$f:[a, b]\rightarrow \mathbb{R}$$ be continuous and let $$F:[a, b]\rightarrow \mathbb{R}$$ be defined by $$F(x)=\int_a^x f(t) \,dt$$. Then $$F$$ is differentiable whose derivative is $$f$$.

Now define $$G$$ on $$[a, b]$$ by $$G(x)=\int_x^b f(t)\, dt$$.

It can be shown, following the proof of above theorem, that $$G$$ is differentiable with derivative $$-f$$ (is this correct)? However, I wanted to see, how the assertion about $$G$$ follows from that of $$F$$ without following the proof of assertion for $$F$$? Any hint?

• Use $-G = \intop_b^x f(x) dx$ – Shai Deshe Jan 19 at 14:51

Note that $$A = \int_a^b f(t) dt$$ is a constant and $$G(x) = \int_x^b f(t) dt = \int_a^b f(t) dt - \int_a^x f(t) dt = A - F(x)$$
Here's a hint: start with the fact that if $$a \le x \le b$$ then $$\int_a^b f(t) \, dt = \int_a^x f(t) \, dt + \int_x^b f(t) \, dt$$ which is a simple consequence of the definition of definite integrals.