Integration skills

Linearity

sums of integral is integrals of sums

$$\begin{array}{rl}U(x)+V(x)& =\int u+v,dx\\ U(x)& =\int u,dx\\ V(x)& =\int v,dx\\ ⟹\int u+v,dx& =\int u,dx+\int v,dx\end{array}$$

Apply fundamental theory of calculus:

\begin{align} \left\[U(b)+V(b) \right] - \left\[ U(a)+V(b) \right] &= \[U(b)-U(a)] + \[V(b)-V(a)] \\ \int\_{a}^{b} u+v ,dx&=\int\_{a}^{b} u ,dx+\int\_{a}^{b}v,dx \end{align}

same as $aU(x)$

$$\begin{array}{rl}\int au,dx& =aU(x)\\ \int u,dx& =U(x)\\ ⟹\int au,dx& =a\int u,dx\end{array}$$

applying fundamental theorem:

$$\begin{array}{rl}aU(x\_2)-aU(x\_1)& =a\left(U(x\_2)-U(x\_1)\right)\\ \int \_{x\_1}^{x\_2}au,dx& =a\int \_{x\_1}^{x\_2}u,dx\end{array}$$

$u$ 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

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

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:

$$\int u{v}^{\prime },dx=\int u,dv$$

By applying Fundamental Theorem of Calculus:

$$\begin{array}{rl}U(v\_2)-U(v\_1)& =\int \_{v\_1}^{v\_2}u,dv\\ u(v(x\_2))-U(v(x\_1))& =\int \_{x\_1}^{x\_2}uv’,dx\\ \int \_{x\_1}^{x\_2}uv’,dx& =\int \_{v\_1}^{v\_2}u,dv\end{array}$$

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:

$$uv=\int u,dv+\int v,du$$

by using Fundamental Theorem of Calculus:

$$uv{|}_a^b={\int }_a^b{u}^{\prime }v+u{v}^{\prime },dx=\int \_{v(a)}^{v(b)}u,dv+\int \_{u(a)}^{u(b)}v,du$$ $$\int \_a^{b}u{v}^{\prime },dx=(u\cdot v){|}_a^b-{\int }_a^{b}u’x,dx$$

for example:

$$\int \_a^{b}u,dx=(ux{)}_a^b-{\int }_a^{b}u’x,dx$$