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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.

Solve by difficulty

The integral

\int_{0 }^{\pi }\sqrt{1+4\sin ^{2}\frac{x}{2}-4\sin \frac{x}{2}}dx

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<c_{1}<c_{2}<\ldots<c_{n-1}<c_{n}<b.

 

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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Piecewise Definite integration

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