08 2intcalc

# 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 each rectangle is h, then:

• Area of one rectangle $= f\big(x\big) \times h$
• Approximate area contained by:
• the curve $f\big(x\big)$ and
• the x-axis and
• $x = a$ and
• $x = b$
• the area of all the rectangles will be:
• the sum of $f\big(x\big) \times h$ for all the rectangles between a and b

.

• The symbol $\Sigma$ … (Sigma ==> Greek letter for capital "S")
• represents "Sum" so in notation, the area is:

… … Area $= \displaystyle{\sum\limits_{x=a}^{b}{f(x)h}}$
.

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

• To make the approximation more accurate, make the rectangles thinner and have more of them.

.

• To make the rectangles very thin, take the limit as $h \rightarrow 0$.

… … Area $= \begin{matrix} \lim \\ h \to 0 \\ \end{matrix} \; \displaystyle{\sum\limits_{x=a}^{b} f\big(x\big)h}$

.

• Then we replace h with dx ….. where dx represents an extremely small change in x.
• And we replace $\Sigma \text{ with } \displaystyle{\int}$ {this is the old English form of "S" for Sum}

… … Area $= \displaystyle{ \int\limits_{x=a}^{b}{f\big(x\big) \; dx} }$

.

## Definite Integrals

• $\displaystyle{\int} f(x) \; dx$ … … is called the indefinite integral and gives a function as the result

.

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

.

• $F(x)$ is defined as an indefinite integral of $f(x)$

.

## The Fundamental Theorem of Integral Calculus

.

The Fundamental Theorem of Integral Calculus says that:

… … Provided $f(x)$ is continuous and smooth in the domain $\big[a,\; b\big]$
.

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

.
.

### Example 1

For $\; f(x) = 6x^2+3$

Find the area between $f(x)$, the x-axis, and between $x = 1 \text{ and } x = 3$
.

Solution

… … $f(x) = 6x^2+3$ … … terminals are 1 and 3
.

… … $F(x) = \displaystyle{ \int{ 6x^2+3 }\;dx }$
.

… … $F(x) = \dfrac {6x^3}{3} + 3x$

… … Remember $F(x)$ is AN integral so don't need +c
.

… … $F(x) = 2x^3 + 3x$

.

… … Use the fundamental theorem of integral calculus
.

… … Area $= \displaystyle{ \int\limits_1^3 } 6x^2+3 \,dx$
.

… … … … $= \Big[ 2x^3+3x \Big] _1^3$
.

… … … … $= F(3) - F(1)$
.

… … … … $= \Big(2 \times 3^3 + 3 \times 3\Big) - \Big(2 \times 1^3 + 3 \times 1\Big)$
.

… … … … $= \Big(54 + 9\Big) - \Big(2 + 3\Big)$
.

… … … … $= 58$ square units
.

### Example 2

For $\; f(x) = \sin\big(x\big)$

Find the area between $f(x)$, the x-axis, and between $x = 0$ and $x = \dfrac{\pi}{2}$

Solution:

… … Area $= \displaystyle {\int\limits_0^{\frac{\pi}{2}} } \sin (x) \; dx$
.

… … … … $= \Big[ - \cos(x) \Big]_0^{\frac{\pi}{2}}$
.

… … … … $= \Big( -\cos \big( \dfrac{\pi}{2} \big) \Big) - \Big( - \cos \big(0\big) \Big)$
.

… … … … $= \Big( 0 \Big) - \Big( -1 \Big)$
.

… … … … $= 1$ square units

.

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.

.