Integration skills

Linearity

sums of integral is integrals of sums

$$\begin{array}{rl}U(x)+V(x)& =\int u+vdx\\ U(x)& =\int udx\\ V(x)& =\int vdx\\ ⟹\int u+vdx& =\int udx+\int vdx\end{array}$$

Apply fundamental theory of calculus:

$$\begin{array}{rl}\left[U(b)+V(b)\right]-\left[U(a)+V(b)\right]& =[U(b)-U(a)]+[V(b)-V(a)]\\ {\int }_a^{b}u+vdx& ={\int }_a^{b}udx+{\int }_a^{b}vdx\end{array}$$

same as $aU(x)$

$$\begin{array}{rl}\int audx& =aU(x)\\ \int udx& =U(x)\\ ⟹\int audx& =a\int udx\end{array}$$

applying fundamental theorem:

$$\begin{array}{rl}aU(x_2)-aU(x_1)& =a\left(U(x_2)-U(x_1)\right)\\ {\int }_{x_1}^{x_2}audx& =a{\int }_{x_1}^{x_2}udx\end{array}$$

$u$ always $=u$

$u=u(x)=u(v)=u(v(x))=\dots$ $u=u$, no matter it’s written as $u(x)$, $u(v)$, or $u(v(x))$

abbreviation of integrals

$$\begin{array}{rl}\int ydx& =\int y(x)dx\\ \int udv& =\int u(v)dv\\ \dots & \end{array}$$

substituting variable

Composition of functions

differentiate w.r.t. $v$ or $x$:

$$\begin{array}{rl}U(v)& =\int u(v)dv\\ U(v(x))& =\int u(v)v’(x)dx\\ \int u(v)dv& =\int u(v)v’(x)dx\end{array}$$

or simply:

$$\int u{v}^{\prime }dx=\int udv$$

By applying Fundamental Theorem of Calculus:

$$\begin{array}{rl}U(v_2)-U(v_1)& ={\int }_{v_1}^{v_2}udv\\ u(v(x_2))-U(v(x_1))& ={\int }_{x_1}^{x_2}uv’dx\\ {\int }_{x_1}^{x_2}uv’dx& ={\int }_{v_1}^{v_2}udv\end{array}$$

integration by parts

multiplying

$$\begin{array}{rl}u(x)v(x)& =\int u’(x)v(x)+u(x)v’(x)dx\\ & =\int v(u)du+\int u(v)dv\\ \int u(v)dv& =u(x)v(x)-\int v(u)dv\end{array}$$

Or simply:

$$uv=\int udv+\int vdu$$

by using Fundamental Theorem of Calculus:

$$uv{|}_a^b={\int }_a^b{u}^{\prime }v+u{v}^{\prime }dx={\int }_{v(a)}^{v(b)}udv+{\int }_{u(a)}^{u(b)}vdu$$ $${\int }_a^{b}u{v}^{\prime }dx=(u\cdot v){|}_a^b-{\int }_a^{b}u’xdx$$

for example:

$${\int }_a^{b}udx=(ux{)}_a^b-{\int }_a^{b}u’xdx$$