Question

find \\(\int_{-4}^{-1} g(x) \\, dx\\). \\(\int_{-4}^{-1} g(x) \\, dx = \square\\)

Explanation:

Step1: Split integral into two parts

The interval $[-4, -1]$ can be split into $[-4, -2]$ and $[-2, -1]$, so:
$$\int_{-4}^{-1} g(x) dx = \int_{-4}^{-2} g(x) dx + \int_{-2}^{-1} g(x) dx$$

Step2: Calculate area for $[-4,-2]$

This region is a trapezoid with bases $1$ and $4$, height $2$. Area of trapezoid: $\frac{1}{2}(b_1+b_2)h$. Since it's below the x-axis, the integral is negative:
$$\int_{-4}^{-2} g(x) dx = -\frac{1}{2}(1+4)\times2 = -5$$

Step3: Calculate area for $[-2,-1]$

This region is a rectangle with length $1$ and width $4$. Since it's below the x-axis, the integral is negative:
$$\int_{-2}^{-1} g(x) dx = -4\times1 = -4$$

Step4: Sum the two results

Add the values from Step2 and Step3:
$$-5 + (-4) = -9$$

Answer:

$-9$