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Differentiation under Integral Sign

The area boun...

Question

The area bounded by the curve $y = f (x)$ , x-axis and the ordinates $x = 1$ and $x = b$ is $(b - 1) sin (3 b + 4)$ , then $f (x)$ equals-

(x - 1) cos (3 x + 4)

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sin (3 x + 4)

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sin (3 x + 4) + 3 (x - 1) cos (3 x + 4)

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N o n e o f t h e s e

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Solution

The correct option is C $sin (3 x + 4) + 3 (x - 1) cos (3 x + 4)$
$Re q u i r e d a r e a$
$b \int 1 f (x) d x = (b - 1) sin (3 b + 4)$
$d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . b$
$f (b) = (b - 1) 3 cos (3 b + 4) + sin (3 b + 4)$
$= 3 (b - 1) cos (3 b + 4) + sin (3 b + 4)$
$a t b = x$
$f (x) = 3 (3 x - 1) cos (3 x + 4) + sin (3 x + 4)$
Option C is correct answer.

Suggest Corrections

Similar questions

(b - 1) s i n (3 b + 4)

y = f (x)

x -

x = 1

x = b

(b - 1) cos (3 b + 4)

f (x)

,

x = 1

x = b

(b - 1) sin (3 b + 4),

x = 4

y = f (x)

f (x) = x^{2}, 0 \leq x \leq 1

f (x) = \sqrt{x}, x \geq 1

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