Definitions [1]
Let ( f(x) ) be a continuous function in the closed interval [a, b] and (h) be the length of (n) equal subintervals, then
\[\int_{a}^{b}f(x)dx=\lim_{n\to\infty}h\sum_{r=0}^{n}f(a+rh)\]
Theorems and Laws [1]
Let ( f(x) ) be a continuous function on a closed interval ([a, b]) and let \[\int\mathrm{f}(x)\mathrm{d}x=\mathrm{F}(x)+\mathrm{c},\] Then, \[\int_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(x\right)\mathrm{d}x=\left[\mathrm{F}(x)+\mathrm{c}\right]_{\mathrm{a}}^{\mathrm{b}}\] \[=\mathrm{F(b)-F(a)}\]
i.e., the definite integral of a function over ([a, b]) is equal to the difference of the values of its antiderivative at the upper and lower limits.
Key Points
(i)\[\int_{a}^{b}\mathrm{f}\left(x\right)\mathrm{d}x=\int_{a}^{b}\mathrm{f}\left(t\right)\mathrm{d}t\]
(ii)\[\int_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(x\right)\mathrm{d}x=-\int_{\mathrm{b}}^{\mathrm{a}}\mathrm{f}\left(x\right)\mathrm{d}x\]
(iii)\[\int_{a}^{b}f\left(x\right)\mathrm{d}x=\int_{a}^{c}f\left(x\right)\mathrm{d}x+\int_{c}^{b}f\left(x\right)\mathrm{d}x,a<c<b\]
(iv)\[\int_{0}^{a}\mathrm{f}\left(x\right)\mathrm{d}x=\int_{0}^{a}\mathrm{f}\left(\mathrm{a}-x\right)\mathrm{d}x\]
(v)\[\int_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(x\right)\mathrm{d}x=\int_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{a}+\mathrm{b}-x\right)\mathrm{d}x\]
(vi)\[\int_{0}^{2a}\mathrm{f}\left(x\right)\mathrm{d}x=\int_{0}^{a}\left[\mathrm{f}\left(x\right)+\mathrm{f}\left(2\mathrm{a}-x\right)\right]\mathrm{d}x\]
(vii)\[\int_{0}^{2a}\mathrm{f}\left(x\right)\mathrm{d}x=2\int_{0}^{a}\mathrm{f}\left(x\right)\mathrm{d}x,\] if f(2a − x) = f(x)
= 0, if f(2a − x) = −f(x)
(viii)$$\int_{-a}^{a} f(x) dx = \begin{cases} 2 \int_{0}^{a} f(x) dx, & \text{if } f(x) \text{ is an even function} \\ & \text{i.e., } f(-x) = f(x) \\ 0, & \text{if } f(x) \text{ is an odd function} \\ & \text{i.e., } f(-x) = -f(x) \end{cases}$$
- If f(t) is an odd function → its integral is an even function
- If f(t) is an even function → its integral is an odd function
