Let f:[1,∞] be a differentiable function such that f(1)=2. If 6∫1xf(t)dt=3xf(x)−x3 for all x≥1, then the value of f(2) is

Solution: 6∫1xf(t)dx=3xf(x)−x3 differentiate wrt x 6f(x)=3f(x)+3xf′(x)−3x2 3f(x)=3xf′(x)−3x2 y=x(dxdy)−x2 ydx−xdy=−x2dx nowx2xdy−ydx=dx ⇒∫d(xy)=∫dx xy=x+c if x=1 , theny(1)=2 12=1+c soc=1 xy=x+1 y=x2+x f(2)=22+2=6 f(x)=6 answer

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Let $f : [1, \infty]$ be a differentiable function such that $f (1) = 2 .$ If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1,$ then the value of $f (2)$ is

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Harper Discussed

Let

f : [1, \infty]

be a differentiable function such that

f (1) = 2 .

6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}

for all

x \geq 1,

then the value of

f (2)

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Text solutionVerified

Solution:

6 \int_{1}^{x} f (t) d x = 3 x f (x) - x^{3}

differentiate wrt x

6 f (x) = 3 f (x) + 3 x f^{'} (x) - 3 x^{2}

3 f (x) = 3 x f^{'} (x) - 3 x^{2}

y = x (\frac{d y}{d x}) - x^{2}

y d x - x d y = - x^{2} d x

now

\frac{x d y - y d x}{x ^{2}} = d x

\Rightarrow \int d (\frac{y}{x}) = \int d x

\frac{y}{x} = x + c

if x=1 , then

y (1) = 2

\frac{2}{1} = 1 + c

c = 1

\frac{y}{x} = x + 1

y = x^{2} + x

f (2) = 2^{2} + 2 = 6

f (x) = 6

answer

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Question Text	Let $f : [1, \infty]$ be a differentiable function such that $f (1) = 2 .$ If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1,$ then the value of $f (2)$ is
Answer Type	Text solution:1
Upvotes	150

Let f:[1,∞] be a differentiable function such that f(1)=2. If 6∫1x​f(t)dt=3xf(x)−x3 for all x≥1, then the value of f(2) is

Text solutionVerified

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Let $f : [1, \infty]$ be a differentiable function such that $f (1) = 2 .$ If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1,$ then the value of $f (2)$ is