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Integral calculus

Integral calculus is the study of areas under curves. If the area under a curve is split into strips, each of length \( \delta x \), so the area of one of the strips is \( y_i \delta x \). Then the total area under the curve from \( x = a \) to \( x = b \) is

\begin{displaymath}
\lim_{\delta x \to 0} \sum_{a}^{b} y_i \delta x
\end{displaymath}

In calculus notation, this is written

\begin{displaymath}
\int_{a}^{b} y \mathrm{d}x
\end{displaymath}

The above is a definite integral which gives the area between the curve \( y = f(x) \) and the \( x \)-axis. Integration is, in fact, the opposite of differentiation, and an indefinite integral gives the antiderivative of the gradient function, i.e. \( \int \frac{\mathrm{d}y}{\mathrm{d}x} \mathrm{d}x
= y + c \), where \( c \) is a constant. The constant of integration is necessary since a constant always differentiates to \( 0 \). Also \( \frac{\mathrm{d}}{\mathrm{d}x} \int y \mathrm{d}x = y \). As differentiation, integration is linear, i.e.

\begin{displaymath}
\int \biggl( a f(x) + b g(x) \biggr) \mathrm{d}x = a \int f(x)
\mathrm{d}x + b \int g(x) \mathrm{d}x
\end{displaymath}

To evaluate a definite integral, we use the fundamental theorem of calculus which states that if \( F'(x)
= f(x) \),

\begin{displaymath}
\int_{a}^{b} f(x) \mathrm{d}x = \Biggl [ \int f(x) \mathrm{d}x \Biggr ]_{a}^{b} = F(b) - F(a)
\end{displaymath}

The indefinite integral evaluated at \( x = a \) gives the area between \( x = p \) and \( x = a \), where $p$ is a point depending on the integrand. Hence,

\begin{displaymath}
\int_{a}^{b} f(x) \mathrm{d}x = \int_{p}^{b} f(x) \mathrm{d}x - \int_{p}^{a} f(x) \mathrm{d}x =
F(b) - F(a)
\end{displaymath}

This justifies the previous statement. (This is obvious geometrically.)

To derive standard integrals, we simply reverse the differentiation process. For example

\begin{displaymath}
\frac{\mathrm{d}}{\mathrm{d}x} x^n = nx^{n-1}
\end{displaymath}

Integrating gives

\begin{displaymath}
x^n = n \int x^{n-1} \mathrm{d}x + c
\end{displaymath}

(since integration is also linear) so that

\begin{displaymath}
\int x^n \mathrm{d}x = \frac{x^{n+1}}{n+1} + c
\end{displaymath}

Other results can be derived in a similar way. To find the area between the curve \( y = \sin{x} \), the \( x \)-axis, and the lines \( x = 0 \) and \( x = \pi/2 \), we need to evaluate

\begin{displaymath}
\int_{0}^{\pi/2} \sin{x} \mathrm{d}x
\end{displaymath}

This is done as follows

\begin{eqnarray*}
\int_{0}^{\pi/2} \sin{x} \mathrm{d}x & = & \Biggl [ -\cos{x} \Biggr ]_{0}^{\pi/2}
 & = & -\cos{\frac{\pi}{2}} + \cos{0} = 1
\end{eqnarray*}





Subsections
next up previous
Next: Methods of integration Up: Calculus - a brief Previous: Reciprocal rule
Alexander Frolkin 2001-03-13