As inverse of Differentiation MCQ Quiz in मराठी - Objective Question with Answer for As inverse of Differentiation - मोफत PDF डाउनलोड करा
Last updated on Mar 18, 2025
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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 \(F\left( x \right) = \smallint f\left( x \right)dx + C\)
\(F\left( x \right) = \smallint \frac{1}{{\sqrt {ax + b} }}dx = \smallint \frac{1}{{{{\left( {ax + b} \right)}^{1/2}}}}dx = \smallint {\left( {ax + b} \right)^{ - 1/2}}dx = \frac{{{{\left( {ax + b} \right)}^{\frac{1}{2}}}}}{{\frac{1}{2}a}} + C = \frac{2}{a}{\left( {ax + b} \right)^{\frac{1}{2}}} + C\)