Find the integral from -∞ to -1 of 1/(1+x²) dx.

I started working the problem out, and so far I got the lim as z→-∞ of [arctan (-1) - arctan (z)].

I'm a little bit confused exactly what is means when you plug -∞ into z, and how to solve the problem from here.

Thank you!

The indefinite integral is arctan x, as you have already indicated.

arctan(-infinity), the angle that has a tangent of -infinity, is (3/2)*pi

arctan(-1) = (7/8)*(2 pi) = (7/4)pi

That makes the answer pi/4

To find the integral from -∞ to -1 of 1/(1+x²), you correctly started by evaluating the antiderivative of the function. However, it seems that there might be a slight error in your calculations.

The antiderivative of 1/(1+x²) is arctan(x), so the integral becomes arctan(x)| from -∞ to -1.

When you evaluate the limits, you need to consider what happens as x approaches -∞ and -1.

As x approaches -∞, the arctan(x) term becomes -π/2. This is because the arctan function approaches -π/2 as the argument approaches -∞.

As x approaches -1, the arctan(x) term becomes -π/4. This is because the arctan function approaches -π/4 as the argument approaches -1.

Therefore, the integral from -∞ to -1 of 1/(1+x²) is:

[arctan(x)]| from -∞ to -1 = [arctan(-1) - arctan(-∞)] = -π/4 - (-π/2) = -π/4 + π/2 = π/4.

Hence, the value of the integral is π/4.

Note: To evaluate the limit as z approaches -∞, we usually use the concept of limits. In this case, we can simplify it by using the known limit of arctan(-∞) = -π/2.