If $f$ is a continuous function on the interval $a ≤ x ≤ b$ and $F$ is an antiderivative of $f$ on $a ≤ x ≤ b$, then:

$$∫↙{a}↖{b} f(x) dx = [F(x)]_a^b = F(b) - F(a)$$- $∫↙{a}↖{b} kf(x)dx = k∫↙{a}↖{b} f(x)dx$
- $∫↙{a}↖{b} (f(x)±g(x))dx = ∫↙{a}↖{b} f(x)dx ± ∫↙{a}↖{b} g(x)dx$
- $∫↙{a}↖{a} f(x)dx = 0$
- $∫↙{a}↖{b} f(x)dx = -∫↙{b}↖{a} f(x)dx$
- $∫↙{a}↖{b} f(x)dx = ∫↙{a}↖{c} f(x)dx + ∫↙{c}↖{b} f(x)dx$
- $∫u⋅dv = uv - ∫v⋅du$

- Constant Rule: $∫ kdx = kx + C$
- Power Rule: $∫x^n dx = 1/(n + 1) x^{n+1} + C$, $n ≠ 1$
- Constant multiple rule: $∫kf(x)dx = k ∫f(x) dx$
- Sum of difference rule: $∫(f(x) ± g(x))dx = ∫f(x)dx ± ∫g(x)dx$
- $e^x$ and $1/x$ integrals: $1/{x}dx = lnx + C$, $x > 0$, $∫e^xdx = e^x + C$
- Linear composition: $∫f(ax + b)dx = 1/{a}F(ax + b) + C$, where $F'(x) = f(x)$

Two common methods for solving integrals are:

Substitution: ${dy}/{dx} = {dy}/{du}⋅{du}/{dx}$

Rearranging the equation to obtain the form $∫f(g(x))g'(x)dx$.

An indefinite integral is a family of functions that differ by the constant C. A definite integral is of the form:

$$∫↙a↖b f(x) dx$$Leibniz invented the ∫ symbol for integration by the 1680s.

$∫↙a↖b f(x) dx$ is read as 'the integral from a to b of f(x) with respect to x'.

If a function $f$ is defined for a ≤ x ≤ b and ${lim}↙{n→∞}∑↙{i=1}↖n f(x_i)Δx_i$ exists, then $f$ is integrable on a ≤ x ≤ b.

This is the definite integral and it is denoted as $∫↙a↖b f(x) dx$, where a and b are the lower and upper limits of integration.

If $f$ is a non-negative function for a ≤ x ≤ b, then $∫↙a↖b f(x) dx$ gives the area under the curve from x = a to x = b.

Definite integrals resolve to a number. This is because the constant C which is in the indefinite integral solution can be cancelled during the procedure of quantifying the definite integral.

From the graph of the function $f(x) = 4 - x^2$, determine the integral and use this to calculate the area under the curve from x = -2 to x = +2.

$A = ∫↙{-2}↖{2} f(x) dx = ∫↙{-2}↖{2} (4 - x^2) dx $ $= [4x - {x^3}/3 + C]_{-2}^{2} $ $= (8 - 8/3 + C) - (-8 + 8/3 + C) $ $= 16 - {16}/3 + C - C $ $= {32}/3 $ $= 10.67$

If $y_1$ and $y_2$ are continuous on $a ≤ x ≤ b$ for all x in $a ≤ x ≤ b$, then the area between $y_1$ and $y_2$ from x = a to x = b is given by:

$$∫↙{a}↖{b} (y_1 - y_2)dx $$

The area of a curve is never negative. Therefore, the part of a function which graphs to below the x-axis needs to be calculated separately from the part above the x-axis, and made positive.

If $f(x)$ is negative between $n$ and $p$, the area between the curve of $f(x)$, the x-axis, and the lines $x = n$ and $x = p$ is:

$$|∫_n^pf(x)dx|$$If functions $f$ and $g$ are continuous in the interval [a, b], and $f(x) ≥ g(x) for the region $a ≤ x /le; b$, the the area between the two graphs is:

$$A = ∫_a^bf(x)dx - ∫_a^bg(x)dx = ∫_a^b(f(x)- g(x)dx$$Associated Mathematicians:

- Isaac Newton
- Gottfried Leibnitz
- Jakob Bernoulli
- Johann Bernoulli (I)

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