# Thread: Fundamental Theorem of Calculus to evaluate the integrals

1. ## Fundamental Theorem of Calculus to evaluate the integrals

The integral sign has a 2 at the top and a 0 at the bottom

2/x^2+1 dx

2. Originally Posted by cara
The integral sign has a 2 at the top and a 0 at the bottom

2/x^2+1 dx
$\int_0^2 \frac{2}{x^2} + 1 \, dx$ or $\int_0^2 \frac{2}{x^2+1} \, dx$ ???

3. The second one you listed is the problem that I need

4. Originally Posted by cara
The second one you listed is the problem that I need
You have $\int_{0}^{2}\dfrac{2}{x^2+1}\;{dx}$. Let $x = \tan\theta$, then $\dfrac{dx}{d\theta} = \sec^2\theta \Rightarrow dx = \left(d\theta\right){\sec^2\theta}$. When $x = 2$, $\tan\theta = 2 \Rightarrow \theta = \arctan\left(2\right)$. When $x = 0$, $\tan{\theta} = 0 \Rightarrow \theta = 0$ . So we have $\int_{x=0}^{x=2}\dfrac{2}{x^2+1}\;{dx} = \int_{\theta=0}^{\theta=\arctan(2)}\dfrac{2}{\tan^ 2\theta+1}{\sec^2\theta}\;{d\theta} = \int_{0}^{\arctan(2)}{2}\;{d\theta}$ $= \bigg[2\theta\bigg]_{0}^{\arctan(2)} = \boxed{2\arctan(2)}.$

5. we are not using arc tan, but F(b)-F(a)

6. Originally Posted by cara
we are not using arc tan, but F(b)-F(a)
uhh ...

$F(2) - F(0) = 2[\arctan(2) - \arctan(0)] = 2\arctan(2)$

... as stated previously