# Definite and indefinite integrals

## Key Questions

• A definite integral looks like this:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

Definite integrals differ from indefinite integrals because of the $a$ lower limit and $b$ upper limits.

According to the first fundamental theorem of calculus, a definite integral can be evaluated if $f \left(x\right)$ is continuous on [$a , b$] by:

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$

If this notation is confusing, you can think of it in words as:

The integral of a function ($f \left(x\right)$) with limits $a$ and $b$ is the integral of that function evaluated at the upper limit ($F \left(b\right)$) minus the integral of that function evaluated at the lower limit ($F \left(a\right)$)
$F \left(x\right)$ just denotes the integral of the function.

Note that you will get a number and not a function when evaluating definite integrals. Also, you have to check whether the integral is defined at the given interval.

Let's look at an example .

${\int}_{-} {2}^{6} {x}^{3} + 2 \mathrm{dx}$

${x}^{3} + 2$ is defined for all real numbers, so the boundaries of $a$ and $b$ are defined. To evaluate this definite integral, we first find the integral function and then plug in the upper limit of 6 into the integral function, and subtract the integral function evaluated at the lower limit of -2.

${\int}_{-} {2}^{6} {x}^{3} + 2 \mathrm{dx} = {\left[\frac{1}{4} {x}^{4} + 2 x\right]}_{-} {2}^{6} = \left(\frac{1}{4} {\left(6\right)}^{4} + 2 \left(6\right)\right) - \left(\frac{1}{4} {\left(- 2\right)}^{4} + 2 \left(- 2\right)\right) = \left(336\right) - \left(0\right) = 336$

• Indefinite integrals are antiderivatives in general form.

$\int f \left(x\right) \mathrm{dx} = F \left(x\right) + C$,

where $F ' \left(x\right) = f \left(x\right)$.

I hope that this was helpful.

• Indefinite integrals have no lower/upper limits of integration. They are general antiderivatives, so they yield functions.

$\int f \left(x\right) \mathrm{dx} = F \left(x\right) + C$,

where $F ' \left(x\right) = f \left(x\right)$ and $C$ is any constant.

Definite integrals have lower and upper limits of integration ($a$ and $b$). They yield values.

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = F \left(b\right) - F \left(a\right)$,

where $F ' \left(x\right) = f \left(x\right)$.

I hope that this was helpful.