# When may we ignore the limits of integration?

When we try to evaluate an integral such as, say

$$\int_a^b{f(x)dx}$$

there is often the case that we can analytically find

$$\int{f(x)dx}$$

a little faster (imagine leaving away the evaluation for specific terms for partial integrations, if you lack an example. Partially integrating forces you to evaluate two terms instead of one all of the sudden).

Therefore, I feel tempted to leave away the limits of integration in general and evaluating it at the very end, to save some time.

Unfortunately, I'm not sure if this holds for every scenario. Think of integration by substitution, where the limits might change due to substituting.

And so, I want to ask: When exactly can I just ignore the limits of integration and apply them to an indefinite integral instead?

If you're having trouble to understand the difference, just give $\int_0^{\pi}{e^x\cdot cos(x)dx}$ a try. Finding $\int{e^x\cdot cos(x)dx}$ is easy, but trying to find $\int_0^{\pi}{e^x\cdot cos(x)dx}$ directly (using partial integration and evaluating all the terms mid-way) is somewhat cumbersome.

• To the contrary, the definite integral expressions in your example will be simpler. If we let $u=e^x$ and $dv=\cos x\,dx$ then $uv$ dies at both ends. Commented Aug 26, 2014 at 6:27
• You can always do this---this is the content of the Fundamental Theorem of Calculus: More precisely, if $F$ is any antiderivative of $f$, then $\int_a^b f(x) \,dx = F(b) - F(a)$, provided that $f$ is well-behaved on the interval $[a, b]$; here, continuity is sufficient for being well-behaved. In practice people very often do exactly this, though, NB that there are some definite integrals whose values can be evaluated, but for which the corresponding indefinite integrals have no antiderivative in terms of elementary functions. Commented Aug 26, 2014 at 6:28
• @AndréNicolas Could you elaborate on that? I'm not entirely sure if I understood your line of argument. Commented Aug 26, 2014 at 6:36
• In doing the integration, you will do integration by parts twice. The first integration by parts gives $\left.uv\right|_0^{\pi}uv-\int_0^\pi v\,du$. Here $u$ and $v$ are as above. Because $\sin 0=\sin\pi=0$, the first term $\left.uv\right|_0^{\pi}uv$ is $0$. Commented Aug 26, 2014 at 6:40
• @AndréNicolas I see now, thank you! If anyone wants to write an answer to this question, I'll mark it as solved. Commented Aug 26, 2014 at 6:59

$$\int_a^bf(x)\,dx = F(b) - F(a) = \left[ F(x) \right]_a^b = \left[ \int f(x)\,dx \right]_a^b$$
given that $f$ is well-behaved on $[a, b]$.