Integral Equation to Determine a Function Value

Question where an integral equation relating f, f', or f'' is given and one must solve for specific function values.

csat-suneung 2013 Q12 3 marks View
A continuous function $f ( x )$ satisfies
$$f ( x ) = e ^ { x ^ { 2 } } + \int _ { 0 } ^ { 1 } t f ( t ) d t$$
What is the value of $\int _ { 0 } ^ { 1 } x f ( x ) d x$? [3 points]
(1) $e - 2$
(2) $\frac { e - 1 } { 2 }$
(3) $\frac { e } { 2 }$
(4) $e - 1$
(5) $\frac { e + 1 } { 2 }$
jee-main 2019 Q82 View
If $\int e ^ { \sec x } \left( \sec x \tan x f ( x ) + \left( \sec x \tan x + \sec ^ { 2 } x \right) \right) d x = e ^ { \sec x } f ( x ) + C$, then a possible choice of $f ( x )$ is:
(1) $\sec x - \tan x - \frac { 1 } { 2 }$
(2) $\sec x + \tan x + \frac { 1 } { 2 }$
(3) $x \sec x + \tan x + \frac { 1 } { 2 }$
(4) $\sec x + x \tan x - \frac { 1 } { 2 }$
jee-main 2022 Q74 View
Let $g : ( 0 , \infty ) \rightarrow R$ be a differentiable function such that $\int \left( \frac { x \cos x - \sin x } { e ^ { x } + 1 } + \frac { g(x) ( e ^ { x } + 1 ) - x e ^ { x } } { ( e ^ { x } + 1 ) ^ { 2 } } \right) \mathrm { d } x = \frac { x g(x) } { e ^ { x } + 1 } + C$, for all $x > 0$, where $C$ is an arbitrary constant. Then
(1) $g$ is decreasing in $\left( 0 , \frac { \pi } { 4 } \right)$
(2) $g - g ^ { \prime }$ is increasing in $\left( 0 , \frac { \pi } { 2 } \right)$
(3) $g ^ { \prime }$ is increasing in $\left( 0 , \frac { \pi } { 4 } \right)$
(4) $g + g ^ { \prime }$ is increasing in $\left( 0 , \frac { \pi } { 2 } \right)$
jee-main 2024 Q74 View
If $\int \frac { 1 } { \mathrm { a } ^ { 2 } \sin ^ { 2 } x + \mathrm { b } ^ { 2 } \cos ^ { 2 } x } \mathrm {~d} x = \frac { 1 } { 12 } \tan ^ { - 1 } ( 3 \tan x ) +$ constant, then the maximum value of $\mathrm { a } \sin x + \mathrm { b } \cos x$, is:
(1) $\sqrt { 40 }$
(2) $\sqrt { 41 }$
(3) $\sqrt { 39 }$
(4) $\sqrt { 42 }$
jee-main 2025 Q6 View
If $\int \mathrm { e } ^ { x } \left( \frac { x \sin ^ { - 1 } x } { \sqrt { 1 - x ^ { 2 } } } + \frac { \sin ^ { - 1 } x } { \left( 1 - x ^ { 2 } \right) ^ { 3 / 2 } } + \frac { x } { 1 - x ^ { 2 } } \right) \mathrm { d } x = \mathrm { g } ( x ) + \mathrm { C }$, where C is the constant of integration, then $g \left( \frac { 1 } { 2 } \right)$ equals :
(1) $\frac { \pi } { 4 } \sqrt { \frac { e } { 3 } }$
(2) $\frac { \pi } { 6 } \sqrt { \frac { e } { 3 } }$
(3) $\frac { \pi } { 4 } \sqrt { \frac { e } { 2 } }$
(4) $\frac { \pi } { 6 } \sqrt { \frac { e } { 2 } }$