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# Riemann integral

In the simplest settings we deal with $$f: [ a, b ] → ℝ$$ continuous. Then Riemann integral defines "the area under the graph of $f$".

Note: This makes sense in dimension 3 - volume under the surface and in dimension n:

• let $f: n-dim$ box $→ ℝ$ be scalar function, then
• graph $f ⊂ ℝ^{n+1}$.

Then it makes sense to talk about the $(n+1)$-dimensional volume "under the graph" of $f$.

The idea is to approximate the area with rectangles.

• Pick $k$, the number of intervals.
• Cut $[a,b]$ into $k$ intervals of equal length:
• $a = x_0 < x_1 < ... < x_k = l$.
• Pick a point in each: $c_1, ..., c_k$.
• Evaluate $f$ at $c_1, ..., c_k$. These numbers are the heights of rectangles, while the bases are $[ x_i, x_{i+1}], the intervals. • Calculate the areas of the rectangles and add them up: • widths of rectangles$∆x = ( b - a ) / k$, • heights of rectangles$f( c_1), ..., f( c_k)$. • Then we get total area of rectangles$= f(c_1) ∆x + f(c_2) ∆x + ... + f(c_k) ∆x$, called the Riemann sum of$f$for this partition of$[ a, b ]$. Consider $$\lim _{k→∞} ∑_{i=1}^k f( c_i ) ∆x$$ (or$lim_{∆x→0} ...), where $c_i$ denotes the midpoint of the respective intervals. If this limit exists, it is called the Riemann integral of $f$ from $a$ to $b$. We write $$∫_a^b f(x) dx,$$ and call $f$ an integrable function on $[ a, b ]$.

The relation of the sets of functions that are

• differentiable functions
• continuous functions
• integrable functions,

is shown on the right. The absolute value and the sign functions are the counterexamples.

Example. Let's compute something. Let

• $f(x) = x, a = 0, b = 1$, and
• $x_i = (1 / k) i$ the endpoints for $i = 1, ..., k$ ($k$ intervals).
• $c_i = (1 / k) i$ chosen points for $i = 0, ..., k-1$.

Then $$\lim _{k→∞} ∑ _{i=1}^k f( c_i ) ∆x$$ $$= \lim_{k→∞} ∑_{i=1}^k c_i ∆x$$ $$= \lim_{k→∞} ( c_1 + ... + c_k ) 1 / k$$ $$= \lim_{k→∞} ( 0 + (1 / k) + (2 / k ) + ... + (( k - 1 ) / k ) k$$ $$= \lim_{k→∞} ( (1 / k^2) + (2 / k^2) + ... + (( k - 1 ) / k^2) ) \text{ (arithmetic progression)}$$ $$= ( k ( k - 1 ) / 2 ) / k^2$$ $$= 1 / 2.$$ So, $$∫_{[0,1]} x dx$$ $$= ∫_0^1 x dx$$ $$= ( x^2 / 2 )|_0^1$$ $$= 1 / 2.$$