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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To Australians at least, Sir Joseph Banks is an outstanding figure in botanical history, revealing to the world the rich diversity of antipodean flora and fauna.

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Nonno's explanations always did a round-robin circuit of science - history - philosophy, then back to science. Always back to science. As if it were that that drove the mechanism of time, and not the other way round. And philosophy? Well, that just went along for the ride.

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