If k∫01x f(3x)dx=∫03t f(t)dt then k=

I=k∫01x f(3x)dx=∫03t f(t)dtTake K∫01xf(3x)dx Put 3x=y 3dx=dy x=0⇒y=0x=1⇒y=3 I=K∫03y3f(y)dy3 =K9∫03yf(y)dy =K9∫03t f(t)dt We can replace y by 't' Now K9∫03t f(t)dt=∫03t f(t)dt K9=1⇒K=9

If k∫01x f(3x)dx=∫03t f(t)dt then k=

I=k∫01x f(3x)dx=∫03t f(t)dtTake K∫01xf(3x)dx Put 3x=y 3dx=dy x=0⇒y=0x=1⇒y=3 I=K∫03y3f(y)dy3 =K9∫03yf(y)dy =K9∫03t f(t)dt We can replace y by 't' Now K9∫03t f(t)dt=∫03t f(t)dt K9=1⇒K=9

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If $k \int_{0}^{1} x f (3 x) d x = \int_{0}^{3} t f (t) d t t h e n k =$

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$\frac{1}{9}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$I = k \int_{0}^{1} x f (3 x) d x = \int_{0}^{3} t f (t) d t$
$T a k e K \int_{0}^{1} x f (3 x) d x P u t 3 x = y 3 d x = d y x = 0 \Rightarrow y = 0$

$x = 1 \Rightarrow y = 3 I = K \int_{0}^{3} \frac{y}{3} f (y) \frac{d y}{3} = \frac{K}{9} \int_{0}^{3} y f (y) d y = \frac{K}{9} \int_{0}^{3} t f (t) d t [W e c a n r e p l a c e y b y' t'] N o w \frac{K}{9} \int_{0}^{3} t f (t) d t = \int_{0}^{3} t f (t) d t \frac{K}{9} = 1 \Rightarrow K = 9$