As inverse of Differentiation MCQ Quiz in मल्याळम - Objective Question with Answer for As inverse of Differentiation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 13, 2025
Latest As inverse of Differentiation MCQ Objective Questions
Top As inverse of Differentiation MCQ Objective Questions
As inverse of Differentiation Question 1:
The anti - derivative of the function u(x)v”(x) - u”(x)v(x) is:
Answer (Detailed Solution Below)
As inverse of Differentiation Question 1 Detailed Solution
\(\smallint \left\{ {{\rm{u}}\left( {\rm{x}} \right){\rm{v''}}\left( {\rm{x}} \right) - {\rm{u''}}\left( {\rm{x}} \right){\rm{v}}\left( {\rm{x}} \right)} \right\}dx = \smallint {\rm{u}}\left( {\rm{x}} \right){\rm{v''}}\left( {\rm{x}} \right){\rm{dx}} - \smallint {\rm{u''}}\left( {\rm{x}} \right){\rm{v}}\left( {\rm{x}} \right){\rm{dx}}\)
\(= \left\{ {u\left( x \right)v'\left( x \right) - \smallint u'\left( x \right)v'\left( x \right)dx} \right\} - \left\{ {u\left( x \right)v'\left( x \right) - \smallint u'\left( x \right)v'\left( x \right)dx} \right\}\; + \;C\)
= f(x)g’(x) – g(x)f’(x) + CAs inverse of Differentiation Question 2:
If \(f\left( x \right) = \frac{1}{{\sqrt {ax + b} }}\), then its antiderivative is
Answer (Detailed Solution Below)
As inverse of Differentiation Question 2 Detailed Solution
Given function is \(f\left( x \right) = \frac{1}{{\sqrt {ax + b} }}\)
Its antiderivative is \(F\left( x \right) = \smallint f\left( x \right)dx + C\)
\(F\left( x \right) = \smallint \frac{1}{{\sqrt {ax + b} }}dx = \smallint \frac{1}{{{{\left( {ax + b} \right)}^{1/2}}}}dx = \smallint {\left( {ax + b} \right)^{ - 1/2}}dx = \frac{{{{\left( {ax + b} \right)}^{\frac{1}{2}}}}}{{\frac{1}{2}a}} + C = \frac{2}{a}{\left( {ax + b} \right)^{\frac{1}{2}}} + C\)