Piecewise Definite integration - Practice Questions & MCQ

Property 5

${\int }_{\mathrm{a}}^{\mathrm{b}}\mathbf{f}(\mathrm{x})\mathrm{d}\mathbf{x}={\int }_{\mathrm{a}}^{\mathrm{c}}\mathbf{f}(\mathrm{x})\mathrm{d}\mathbf{x}+{\int }_{\mathbf{c}}^{\mathrm{b}}\mathbf{f}(\mathbf{x})\mathrm{dx}\text{ where }c\in \mathbb{R}$

This property is useful when the function is in the form of piecewise or discontinuous or non-differentiable at $x=c$ in $(a,b)$.
Let
$\begin{array}{rl}& \frac{d}{dx}(F(x))=f(x)\\ & \begin{array}{rl}& {\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx\\ & ={F(x)|}_a^c+{F(x)|}_c^b\\ & =F(c)-F(a)+F(b)-F(c)\\ & =F(b)-F(a)\\ & ={\int }_a^{b}f(x)dx\end{array}\end{array}$
$\therefore {\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$

The above property can also be generalized into the following form

${\int }_a^{b}f(x)dx={\int }_a^{c_1}f(x)dx+{\int }_{c_1}^{c_2}f(x)dx+\dots +{\int }_{c_n}^{b}f(x)dx$
where, $a<c_1<c_2<\dots <c_{n-1}<c_n<b$.

Property 6

${\int }_0^{a}f(x)dx={\int }_0^{a/2}f(x)dx+{\int }_0^{a/2}f(a-x)dx$

Proof:
From the previous property,
${\int }_0^{a}f(x)dx={\int }_0^{a/2}f(x)dx+{\int }_{a/2}^{a}f(x)dx$

Put $x=a-t\Rightarrow dx=-dt$ in the second integral, when $x=a/2$, then $t=a/2$ and when $x=a$, then $t=0$
$\therefore \begin{array}{rl}{\int }_0^{a}f(x)dx& ={\int }_0^{a/2}f(x)dx+{\int }_{a/2}^{0}f(a-t)(-dt)\\ & ={\int }_0^{\mathrm{a}/2}\mathrm{f}(\mathrm{x})\mathrm{dx}+{\int }_0^{\mathrm{a}/2}\mathrm{f}(\mathrm{a}-\mathrm{t})\mathrm{dt}\\ {\int }_0^{a}f(x)dx& ={\int }_0^{a/2}f(x)dx+{\int }_0^{a/2}f(a-x)dx\end{array}$