definite-integration

Important Points

• 1st fundamental theorem of calculus $\frac{d}{dx}{\int }_a^{x}f(x)dx=f(x)\forall x\ge a$
• 2nd fundamental theorem of calculus ${\int }_a^{b}f(x)dx=li{m}_{x\to b^{-}}F(x)-li{m}_{x\to a^{+}}F(x)$
• Substitutiion must be continuous in the interval/limits of integration ${\int }_a^{b}f(x)dx$ Here, if t = g(x) is not continuous in $(a,b)$ then we cannot use it as a substitution

  What we can do is instead of integration from a to b we can integration firstly from a to c then c to b here $c\in (a,b)$ and g(x) is continuous in both $(a,c)$ and $(c,b)$

• If the function f(x) is going up and down the x-axis then the area is given by ${\int }_a^{b}f(x)dx=\text{All the areas above x-axis}-\text{All the areas below x-axis}$
• if f(x) is continuous then the following implies that f(x) = 0 has atleast 1 real root in $(a,b)$ ${\int }_a^{b}f(x)dx=0$

Properties of Definite Integration

1. If the limits are correct, we can replace x with any variable like ${\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt$
2. ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$
3. ${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$
4. ${\int }_a^{b}f(x)dx={\int }_a^{b}f((a+b)-x)dx$
5. $${\int }_{-a}^{a}f(x)dx=\left\{\begin{array}{lr}0,& \text{if }f(-x)=-f(x)\\ 2{\int }_0^{a}f(x)dx,& \text{if }f(-x)=f(x)\end{array}\right\}$$
6. ${\int }_0^{2a}f(x)dx={\int }_0^{a}f(x)dx+{\int }_0^{a}f(2a-x)dx$ $$=\left\{\begin{array}{lr}0,& \text{if }f(2a-x)=-f(x)\\ 2{\int }_0^{a}f(x)dx,& \text{if }f(2a-x)=f(x)\end{array}\right\}$$
7. If (a+x) = f(x) then ${\int }_0^{na}f(x)dx=n{\int }_0^{a}f(x)dx$
8. If m is thel east vavlue and M is the greatest value of the function f(x) on the interval $[a,b]$ $m(b-a)\le {\int }_a^{b}f(x)dx\le M(b-a)$
9. If f is continuous on $[a,b]$ and u(x) and v(x) are differentiable functions of x whose value lie in $[a,b]$ $\frac{d}{dx}{\int }_{u(x)}^{v(x)}f(t)dt=f(v(x))\frac{dv}{dx}-f(u(x))\frac{du}{dx}$

Manipulating Limits

NOTE

If you want to do something like add, subtract or multiply limits by a constant, then it is adviced to just do that with x and put that equal to t and apply substitution

• ${\int }_a^{b}f(x)dx={\int }_{a+c}^{b+c}f(x-c)dx={\int }_{a-c}^{b-c}f(x+c)dx$
• ${\int }_a^{b}f(x)dx=k{\int }_{\frac{a}{k}}^{\frac{b}{k}}f(kx)dx=\frac{1}{k}{\int }_{ak}^{bk}f(\frac{x}{k})dx$

Periodic Functions

let’s consider f(x) to be a periodic function with period T

• ${\int }_0^{nT}f(x)dx=n{\int }_0^{T}f(x)dx$
• ${\int }_{mT}^{nT}f(x)dx=(n-m){\int }_0^{T}f(x)dx$
• ${\int }_{a+nT}^{b+nT}f(x)dx={\int }_a^{b}f(x)dx$
• ${\int }_a^{a+nT}f(x)dx=n{\int }_0^{T}f(x)dx$

Summation of infinite series

${\lim }_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^{n}f(\frac{r}{n})={\int }_0^{1}f(x)dx$

in the above equation

• $\frac{1}{n}=dx\frac{r}{n}=x$
• $\text{lower limit}=li{m}_{n\to \infty }\frac{r_{min}}{n}$
• $\text{upper limit}=li{m}_{n\to \infty }\frac{r_{max}}{n}$