Definite integral

A definite integral as the area under the function $f(x)$ between $a$ and $b$

The definite integral, when

$\int\limits_{a}^{b}f(x)dx$

is the area between the function $f(x)$ and the x-axis where $x$ ranges from $a$ to $b$ . According to the Fundamental theorem of calculus, if

$F'(x) = f(x)$

the definite integral can be calculated by:

$\int\limits_a^b f(x)dx=F(b)-F(a)$

To take the definite integral of this function, one would find the antiderivative of the function at $F(b)$ and subtract the value of $F(a)$ from this. This will be equal to the area under the function in $[a, b]$ .

A definite integral can be thought of as a Riemann sum of infinitely small rectangles, or

$\lim _{n\to \infty }\sum _{i=1}^{n}f(x_{i})\Delta x_{i}$

Integration can be thought of as a generalization of multiplication. If f(x)=constant then the definite integral reduces to ordinary multiplication:

$\int\limits_a^b k dx = k \int\limits_a^b 1 dx = k(b-a)$

Applications

Definite integrals have many applications in geometry, physics, and other fields. For example, volume is the integral of area, and velocity is the integral of acceleration. For instance, using integrals it is possible to calculate the volume of a solid object such as a solid of revolution or a pyramid.

Area under a curve

A definite integral in $[a, b]$ is equal to the area between the curve and the x-axis. For example, to calculate the area under the graph of $f(x)=\sqrt{x}$ on the interval $[0,4]$ , one would first take the integral as follows:

$\int \limits _{0}^{4}{\sqrt {x}}dx=\int \limits _{0}^{4}x^{\frac {1}{2}}dx={\dfrac {2}{3}}(4)^{\frac {3}{2}}-{\frac {2}{3}}(0)^{\frac {3}{2}}={\frac {2}{3}}(8)-0={\frac {16}{3}}$

Volume of a solid

See also: Solid of revolution, Volume by cross sections and Multiple integral

Suppose we want to find the volume of a cone with radius $r$ and height $h$ . We can do this by integrating the circular cross sections from 0 to $h$ . The radius will be equal to the distance between the x-axis and the function $y=r-\frac{r}{h}x$ , so the area of a cross section will be equal to $\pi\left(r-\frac{r}{h}x\right)^2$ or $\pi\left(r^2-\frac{2r^2}{h}x+\frac{r^2}{h^2}x^2\right)$ . Therefore, the volume of the cone will be equal to

$\begin{align}&\int\limits_0^h\pi\left(r^2-\dfrac{2r^2}{h}x+\frac{r^2}{h^2}x^2\right)dx=\pi\int\limits_0^h\left(r^2-\frac{2r^2}{h}x+\frac{r^2}{h^2}x^2\right)dx \\&=\pi\left(r^2x-\frac{r^2}{h}x^2+\frac{r^2}{3h^2}x^3\right)\bigg|^h_0 \\&=\pi\left(\left(r^2(h)-\frac{r^2}{h}(h)^2+\frac{r^2}{3h^2}(h)^3\right)-\left(r^2(0)-\frac{r^2}{h}(0)^2+\frac{r^2}{3h^2}(0)^3\right)\right) \\&=\pi\left(r^2h-r^2h+\frac{r^2h}{3}-0\right)=\frac{\pi}{3}r^2h\end{align}$

Area with polar coordinates

Given the polar function $r(\theta)$ , the area under the function as a Riemann sum is

$\lim_{n\to\infty}\sum_{i=1}^n\dfrac{1}{2}r(\theta_i)^2\Delta\theta_i$

As an integral, this would be

$\dfrac{1}{2}\int\limits_{\alpha}^{\beta}r(\theta)^2d\theta$

Arc length

See also: Arc length

Arc length can be calculated by summing an infinite number of infinitesimally small line segments along a curve. The formula for doing this with regular functions is

$\int\limits_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}dx$

With a parametric equation, this is equal to

$\int\limits_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2} dt$

In polar coordinates, arc length is equal to

$\int\limits_{\alpha}^{\beta}\sqrt{r'(\theta)^2+r(\theta)^2}d\theta$

See also

• Technique of integration
• Antiderivative
• Multiple integral
• Improper integral
• Integration by substitution