Definite integrals of polynomials

How to evaluate a definite integral

If $F(x)$ is an antiderivative of $f(x)$, then

The constant of integration cancels, so it is not written.

Example: $\displaystyle\int_1^2 (3x^2 - 2x)\,dx = \Bigl[x^3 - x^2\Bigr]_1^2 = (8 - 4) - (1 - 1) = 4$

Properties of definite integrals

The third property lets you join or split intervals. Sums and differences of integrals of the same function are easier to combine into one integral first. It is also how you split an interval for absolute values.

The 1/6 formula

Use it when both limits are roots of the quadratic integrand. It is a huge time saver for areas between a parabola and a line.

Example: $\displaystyle\int_{-1}^{3}(x^2 - 2x - 3)\,dx = \int_{-1}^{3}(x+1)(x-3)\,dx = -\frac{1}{6}\{3 - (-1)\}^3 = -\frac{32}{3}$

To prove it, write $(x-\alpha)(x-\beta) = (x-\alpha)^2 - (\beta-\alpha)(x-\alpha)$ and integrate.

Even and odd functions

A function with $f(-x) = f(x)$, like $x^2$ or a constant, is even; one with $f(-x) = -f(x)$, like $x$ or $x^3$, is odd. On a symmetric interval from $-a$ to $a$:

Example: $\displaystyle\int_{-2}^{2}(x^3 + 3x^2 - x + 1)\,dx = 2\int_0^2 (3x^2 + 1)\,dx = 2\Bigl[x^3 + x\Bigr]_0^2 = 20$

Integrals with absolute values

Remove an absolute value such as $|x - 2|$ by cases on the sign of the inside. Split the interval where the inside is zero, integrate each piece and add.

A sketch shows this as the sum of the areas of two triangles. For a quadratic like $|x^2 - 4x + 3|$, factor it to find where the sign changes.

Functions defined by integrals

In an equation such as $f(x) = 3x^2 - 2\displaystyle\int_0^1 f(t)\,dt$, notice that $\displaystyle\int_0^1 f(t)\,dt$ is a constant; call it $k$. Then $f(x) = 3x^2 - 2k$ and

so $k = \frac{1}{3}$ and $f(x) = 3x^2 - \frac{2}{3}$.

An integral with $x$ in the upper limit, $\displaystyle\int_a^x f(t)\,dt$, is a function of $x$ with the following derivative, which is often used by differentiating both sides.