1.3 KiB
1.3 KiB
Formal Definition
Let f be a continuous function on an interval [a, b].
Divide [a, b] into n equal parts of width \Delta x = \dfrac{b-a}{n}.
Let x_0, x_1, x_2, \cdots, x_n be the endpoints of this subdivision. x_0 = a and x_n = b.
Define \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^nf(x_i)\Delta x
\Delta xrefers to the width of each sub-interval, or\frac{b-a}{n}.f(x_i)refers to the height of each subinterval, and can be found with the equationx_i = \Delta xi + a- Or, the width of each interval times the interval index, plus the starting offset.
Then let f be a continous function on [a, b] and let F be the antiderivative of f (i.e F'(x) = f(x)).
Then \int_a^b f(x) dx = F(b) - F(a).
Examples
- Find the area under the curve between 0 and 1 of the function
f(x) = x^2
\int_0^1 x^2 dx = \frac{1}{3} x^3 \Big |_0^1 = 1/3(1^3)- 1/3 (0)^3 = 1/3
- Find the Riemann Sum under the curve between -2 and 2 of the function
2x + 2.
\int_{-2}^2 (2x + 2)dx = \lim_{n \to \infty} \sum_{i = 1}^nf(x_i)\Delta x
Using the fact that
x_i = \Delta x + a, $
= lim_{n \to \infty} \sum_{i=1}^n(2x_i + 2)\frac{4}{n}
= \lim_{n \to \infty} \sum_{i = 1}^n (2(-2 +\frac{4i}{n}) + 2)\frac{4}{n}
= \lim_{n \to \infty} \sum_{i = 1}^n(-4 + \frac{8i}{n} + 2)\frac{4}{n}