# Thread: Unsplitting Limits of Integration

1. ## Unsplitting Limits of Integration

Suppose I was integrating some function $f(x)$ across the interval [a,d].

Suppose, though, that this function was a piecewise function defined as follows:

$f(x) = \left\{
\begin{array}{c l}
ug(x) & a \leq x \leq b \\
vg(x) & b \leq x \leq c
\end{array}
\right.
$

Where u and v are some constants.

So the integral looks something like this:

$I = \int_a^b u \, g(x) dx + \int_b^c v g(x) \,dx$

$= u\int_a^b \, g(x) dx + v\int_b^c g(x) \,dx$

Is there some way to express this as a single integral across the whole interval [a,c]?

I tried equating my general integral to $w\int_a^c g(x) \, dx$ and solved for w, but the answer would require that I'd have to compute the individiual integrals first before I could combine them. Which makes it rather pointless.

The reason I ask is that I'm integrating some pretty big, nasty functions, and it would save me a lot of time under exam conditions if I didn't have to write each integral out twice - it seems to me to be a perfect waste of time when they're both the same function being integrated over adjacent intervals!

2. If G is a primitive for g, then (obviously) $\int_a^c f = u\int_a^b g + v \int_b^c g = vG(c) - (v-u)G(b) - uG(a)$. Is this what you're looking for?

If G is a primitive for g, then (obviously) $\int_a^c f = u\int_a^b g + v \int_b^c g = vG(c) - (v-u)G(b) - uG(a)$. Is this what you're looking for?
Yes, this is the equation I derived earlier, and it does reduce the tricky integrations I am dealing with down to mere algebra.

However, what I was really asking was, given a split up integral as

$I = u\int_a^b g + v \int_b^c g$

Is there a way to write this such that, rather than having two integrals of the same thing over two adjacent intervals, you have one integral over one function through the whole interval?

4. Yes, obviously $\int_a^c f$......

Yes, obviously $\int_a^c f$......