6YDefiniteIntegrals

= The Fundamental Theory of Calculus =


 * Integral Calculus can be used to find the exact area under curves

Using a rectangle approximation for the area between the curve and the x-axis: If the width of the rectangle is h, then:

... ... Area of rectangle = f(x) × h

Approximate area between: will be: ... ... the **__sum__** of ** f(x) × h ** for all the rectangles between ** a ** and ** b **
 * the curve and
 * the x-axis and
 * ** x = a ** and
 * ** x = b **

The symbol Σ (** Sigma ==> Greek letter for capital "S") ** represents "Sum" so in notation, the area is:

math . \qquad \text {Area } = \displaystyle{\sum\limits_{x=a}^{b}{f(x)h}} \qquad. math

To make the approximation more accurate, make the rectangles thinner.

To make the rectangles very thin, take the limit as h --> 0.

math . \qquad \text {Area } = \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \displaystyle{\sum\limits_{x=a}^{b}{f(x)h}} \qquad. math

Then we replace h with dx. And we replace Σ with ò {the old English form of "S" for Sum}

math . \qquad \text {Area } = \displaystyle{ \int\limits_{x=a}^{b}{f(x) \; dx} } \qquad. math


 * Definite Integrals **

math . \qquad \displaystyle{\int} f(x) \; dx \qquad. math ... ... is called the ** indefinite integral ** and gives a function as the result

math . \qquad \displaystyle{\int\limits_{x=a}^{b}{f(x) \,dx}} \qquad. math ... ... is called the ** definite integral ** and gives a value (the area under the curve) as the result.

... ... a and b are called the ** terminals ** (or sometimes the ** limits **)


 * F(x) is defined as __an__ indefinite integral of f(x) **

The Fundamental Theorem of Integral Calculus

says that: Provided f(x) is __continuous__ and __smooth__ in the domain [a, b]

math . \qquad \displaystyle{\int\limits_a^b} f(x) \; dx = \Big[ F(x) \Big]_a^b = F(b) - F(a) \qquad. math

Example 1

math . \qquad \text{For } \; f(x) = 6x^2+3 \qquad. math

... ... Find the area between f(x), the x-axis, and between x = 1 and 3

math \\ . \qquad F(x) = \displaystyle{ \int{ 6x^2+3 }\;dx } \qquad. \\ . \\ . \qquad \qquad = \dfrac {6x^3}{3} + 3x \qquad \small{\textit{AN integral so don't need + c}} \qquad. \\ . \\ . \qquad \qquad = 2x^3 + 3x math

... ... ** Use the fundamental theorem of integral calculus **

math \\ . \qquad \text{Area } \; = \displaystyle{ \int\limits_1^3 } 6x^2+3 \,dx \qquad. \\ . \\ . \qquad \qquad \; = \Big[ 2x^3+3x \Big] _1^3 \qquad. \\ .\\ . \qquad \qquad \; = F(3) - F(1) math . math \\ . \qquad \qquad \; = (2 \times 3^3 + 3 \times 3) - (2 \times 1^3 + 3 \times 1) \qquad. \\ . \\ . \qquad \qquad \; = (54 + 9) - (2 + 3) \qquad. \\ . \\ . \qquad \qquad \; = 58 \; \small{ \text{ square units}} math


 * Example 2 **

math . \qquad \text{For } \; f(x) = \sin(x) \qquad. math

... ... Find the area between f(x), the x-axis, and between x = 0 and p /2

math \\ \text{Area } \; = \displaystyle {\int\limits_0^{\frac{\pi}{2}} } \sin (x) \; dx \qquad. \\ . \\ . \qquad \; = \Big[ - \cos(x) \Big]_0^{\frac{\pi}{2}} \qquad. math . math \\ . \qquad \; = \Big( -\cos \left( \dfrac{\pi}{2} \right) \Big) - \Big( - \cos \big(0\big) \Big) \qquad. \\ . \\ . \qquad \; = \Big( 0 \Big) - \Big( -1 \Big) \\. \\ . \qquad \; = 1 \; \small{ \text{ square units}} \qquad. math

NOTE
 * If you start with a gradient function and antidifferentiate, you end up with the original function.
 * If you start with the original function and antidifferentiate, you end up with the area under the curve.
 * The same process gives you two different results because the gradient, the curve and the area under the curve are linked mathematically.

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