+0

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

0
566
1

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
Sort:

#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

### 31 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details