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# Piecewise Definite integration - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Piecewise Definite integration is considered one the most difficult concept.

• 30 Questions around this concept.

The integral

equals:

## Concepts Covered - 1

Piecewise Definite integration

Property 5

$\mathbf{\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\;\;}\text{where }c\in\mathbb{R}$

This property is useful when function is in the form of piecewise or discontinuous or non-differentiable at x  = c in (a, b).

$\\\mathrm{Let\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{d}{dx}(F(x))=f(x)}\\\\\mathrm{\therefore \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_{a}^{c}f(x)\;dx+\int_{c}^{b}f(x)\;dx}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\left.F(x)\right|_{a} ^{c}+\left.F(x)\right|_{c} ^{b}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=F(c)-F(a)+F(b)-F(c)}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=F(b)-F(a)}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\int_{a}^{b}f(x)\;dx}$

The above property can be also generalized into the following form

$\\\mathbf{\int_{a}^{b} f(x) d x=\int_{a}^{c_{1}} f(x) d x+\int_{c_{1}}^{c_{2}} f(x) d x+\ldots+\int_{c_{n}}^{b} f(x) d x}\\\text { where, } \quad a

Property 6

$\mathbf{\int_{0}^{a} f(x)\; d x=\int_{0}^{a / 2} f(x)\; d x+\int_{0}^{a / 2} f(a-x) \;d x}$

Proof:

$\\\text{From the previous property,}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}\int_{0}^{a} f(x) d x=\int_{0}^{a / 2} f(x) d x+\int_{a / 2}^{a} f(x) d x\\ {\text { Put } x=a-t \Rightarrow d x=-d t \text { in the second integral, }} \\ {\text { when } x=a / 2, \text { then } t=a / 2 \text { and when } x=a, \text { then } t=0}\\\\\therefore \;\;\;\;\;\;\quad \int_{0}^{a} f(x) d x=\int_{0}^{a / 2} f(x) d x+\int_{a / 2}^{0} f(a-t)(-d t)\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\int_{0}^{a / 2} f(x) d x+\int_{0}^{a / 2} f(a-t) d t}\\\\ \mathrm{\;\;}\;\;\;\;\;\;\quad\int_{0}^{a} f(x) d x=\int_{0}^{a / 2} f(x) d x+\int_{0}^{a / 2} f(a-x) d x$

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## Books

### Reference Books

#### Piecewise Definite integration

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 8.11

Line : 49

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