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

#1**+5 **

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

#1**+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