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f(x)=2*(integral from x=0 to x=x of  1/(1+3f^2(t)) dt ))

 

 

Find integral from x=0 to x=1 of (f(t) dt)

Guest Feb 5, 2016

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 #1
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f(x)=2*(integral from x=0 to x=x of  1/(1+3f^2(t)) dt ))=

f(x) = (2 x)/(3 f^2 t+1)

 

Find integral from x=0 to x=1 of (f(t) dt)

Compute the definite integral: integral_0^1 f(t) dx Apply the fundamental theorem of calculus. The antiderivative of f(t) is x f(t): = x f(t)|_(x = 0)^1 Evaluate the antiderivative at the limits and subtract. x f(t)|_(x = 0)^1 = 1 f(t)-0 f(t) = f(t): Answer: | | = f(t)

Guest Feb 5, 2016
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 #1
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+5
Best Answer

f(x)=2*(integral from x=0 to x=x of  1/(1+3f^2(t)) dt ))=

f(x) = (2 x)/(3 f^2 t+1)

 

Find integral from x=0 to x=1 of (f(t) dt)

Compute the definite integral: integral_0^1 f(t) dx Apply the fundamental theorem of calculus. The antiderivative of f(t) is x f(t): = x f(t)|_(x = 0)^1 Evaluate the antiderivative at the limits and subtract. x f(t)|_(x = 0)^1 = 1 f(t)-0 f(t) = f(t): Answer: | | = f(t)

Guest Feb 5, 2016

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