If $f'\left( x \right) = \frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}\,\,{\rm{and}}\;\,\,f\left( 0 \right) = 0,\;\,{\rm{then}}\;\,f\left( 1 \right)$ is equal to

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

f'\left( x \right) = \frac{{dx}}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}\,\,\mathrm{and}\;\,\,f\left( 0 \right) = 0,\;\,\mathrm{then}\;\,f\left( 1 \right)

is equal to

View answer and explanation

Correct option: C

\frac{1}{{\sqrt 2 }}

Option C (

\frac{1}{{\sqrt 2 }}

) is correct because integrating

f'(x)

and using the initial condition

f(0) = 0

, we find

f(1) = \frac{1}{{\sqrt 2 }}

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