# What is the integral of (x^2 - 4)dx from 0 to 4?

May 2, 2015

The answer for ${\int}_{0}^{4} \left({x}^{2} - 4\right) \mathrm{dx}$ is $\frac{16}{3}$.

When it comes to integrals, probably the best way is to think to yourself, "If I solved for an anti-derivative or integral, does taking the derivative of my answer give me the original function in the problem?"

To solve this integral, what you can do is split it up as follows:

(1) ${\int}_{0}^{4} \left({x}^{2} - 4\right) \mathrm{dx} = {\int}_{0}^{4} \left({x}^{2}\right) \mathrm{dx} - {\int}_{0}^{4} \left(4\right) \mathrm{dx}$

To integrate $4$ as an indefinite integral, a good rule of thumb is by an equation below for polynomials;

(2) $\int {x}^{a} = {x}^{a + 1} / \left(a + 1\right) + C$, where a is any number.

C denotes some constant that is crucial when doing indefinite integrals for all functions! By doing $4$ similar to Eq. 2,

$\int \left(4\right) \mathrm{dx} = 4 {x}^{0 + 1} / \left(0 + 1\right) + C = 4 x + C$

This makes sense since taking the derivative of $4 x + C$ gives us $4$.

If I take the definite integral of it however, I won't need the C since they would cancel. In addition, I need to evaluate from $0 \to 4$ the anti-derivative to compute the sum. Here is a full setup:

(3) int_0^4(4)dx=[4x+C~|_0^4
$= \left(4 \left(4\right) + \cancel{C}\right) - \left(\cancel{4 \left(0\right)} + \cancel{C}\right) = 16$

Thus, keep in mind that the constant $C$'s are not necessary for definite integrals and sometimes a zeroth term in the integral can be ignored while evaluating (unlike ${e}^{x}$, which ${e}^{0}$ becomes $1$).

For ${x}^{2}$, we'll first follow Eq. 2 again (works out by differentiating our anti-derivative):

int_0^4(x^2)dx=[x^(2+1)/(2+1)~|_0^4=[1/3x^3~|_0^4

And then we can solve for the sum of ${x}^{2}$ from $0 \to 4$:

(4) ${\int}_{0}^{4} \left({x}^{2}\right) \mathrm{dx} = \left[\frac{1}{3} {\left(4\right)}^{3} - 0\right] = \frac{64}{3}$

So now, we can combine Equations 3 and 4 to get the answer to Eq. 1:

${\int}_{0}^{4} \left({x}^{2} - 4\right) \mathrm{dx} = {\int}_{0}^{4} \left({x}^{2}\right) \mathrm{dx} - {\int}_{0}^{4} \left(4\right) \mathrm{dx} = \frac{64}{3} - 16$
$= \frac{64}{3} - \frac{48}{3} = \frac{16}{3}$

You can also just stick with the original function and integrate the sums inside knowing that they are separate, but the procedures above is a good way for beginners on definite integrals.

Remember that on some functions, they behave differently than polynomials, so just be aware of the derivatives.