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Introductory integration

introductory integration

Integration can be viewed as one of two processes; either the process of undoing differentiation, or determining the area under a curve. The fundamental theorem of calculus tells us that, up to a constant, these processes are the same thing.

In this section we shall look at the integration of polynomial terms, $ax^n$ and the simple extension to rational powers, $bx^\nu$. (We require that $n,\nu \neq -1$ as $n,\nu=-1$ is a special case that will be dealt with separately.)

Like differentiation, integration is a linear operation, so 

$$\int f(x) + g(x)dx = \int f(x)dx + \int g(x)dx$$

and $$ \int af(x)dx = a\int f(x)dx,\;a\in \mathbb{R}$$

Viewed as the process of reversing differentiation, we get the following results

$$\int x^n dx = \frac{1}{n+1}x^n + c$$

$$\int 0dx = c, \; \int 1dx = x+ c$$

The final rule here is simply the case n = 0, but is explicitly included for clarity.

The appearance of the $c$ is needed since any constant differentiates to $0$ and so when integrating, a zero must return an arbitrary constant. For example, $x^2 + 5$ and $x^2 + 3$ both differentiate to give $2x$, so integrating $2x$ must give $x^2 + c$

From the fundamental theorem of calculus, if instead of wanting just the antiderivative, we want the area contained beneath a curve, we apply limits to our integral, so the area contained between the curve y = f(x) and the lines x= a and x = b is given by

$$ \mbox{Area} = \int_a^b f(x) dx = \left[F(x)\right]_a^b = F(b) - F(a)$$

where $F(x)$ is an antiderivative of $f(x)$

Introductory Integration 1

Determine the area enclosed by the lines $x= 2$, $x =4$ and the curve $y= 12 + 2x - x^2$, shown in the diagram

solution - press button to display

The area is given by $$ \mbox{Area} = \int_2^4 12 + 2x - x^2 dx $$ Evaluating the integral gives

$$ \begin{align} \mbox{Area} &= \int_2^4 12 + 2x - x^2 dx \\ &= \left[12x + x^2 - \frac{1}{3}x^3\right]_2^4\\ &= \left(12(4) + (4)^2 - \frac{1}{3}4^3\right) - \left(12(2) + (2)^2 - \frac{1}{3}2^3\right) \\ &= \frac{128}{3} - \frac{76}{3} \\ &= \frac{52}{3} \end{align}$$

Introductory integration 2

Determine the equation of the curve $C$ in the form $y=f(x)$ that passes through the point $(1,1)$, give that $$ \frac{dy}{dx} = \left(x^2 - 2\right)^3 $$
solution - press button to display
In this case, we must first expand the brackets and then integrate. The value of $c$ can then be determined by using the initial condition. $$ \left(x^2 -2\right)^3 = x^6 - 6x^4 + 12x^2 - 8 $$ From this it follows that $$ \begin{align} \int \left(x^2 -2\right)^3 dx &= \int x^6 - 6x^4 + 12x^2 - 8 dx \\ &= \frac{1}{7}x^7 - \frac{6}{5}x^5 +4x^3 - 8x + c \end{align} $$ The equation of the curve is therefrore $$y = \frac{1}{7}x^7 - \frac{6}{5}x^5 +4x^3 - 8x + c $$ Evaluating this equation at the point $(1,1)$ yields $$ \begin{align} 1 &= \frac{1}{7} - \frac{6}{5} + 4 - 8 + c\\ c &= \frac{212}{35} \end{align} $$

Hence the equation of the curve is $$y = \frac{1}{7}x^7 - \frac{6}{5}x^5 +4x^3 - 8x + \frac{212}{35} $$