Evaluate F, the integral from y = -1.65 to 0 of the integral from x = -y/1.1015 to 0 of (14.1 - 11.515y) dx dy.

Step-by-step solution

1. 1

   Notice the integrand does not involve x. Integrating a constant (in $x$) over an interval just multiplies it by the signed width: $\int_{-y/1.1015}^{0}(14.1-11.515y)\,dx = (14.1-11.515y)\left(0-\left(-\frac{y}{1.1015}\right)\right) = (14.1-11.515y)\frac{y}{1.1015}.$

2. 2

   Check the sign of that width before going further. For $y\in[-1.65,0]$ we have $-\tfrac{y}{1.1015}\ge 0$, so the lower limit is above the upper limit $0$ and the width $\tfrac{y}{1.1015}$ is negative. Since $14.1-11.515y>0$ throughout, the inner integral is negative and $F$ must come out negative - this is the step where a sign is most often lost.

3. 3

   Reduce to a single integral and expand. $F = \frac{1}{1.1015}\int_{-1.65}^{0}\left(14.1y-11.515y^2\right)dy.$

4. 4

   Find the antiderivative. $\int\left(14.1y-11.515y^2\right)dy = 7.05y^2-\frac{11.515}{3}y^3.$

5. 5

   Apply the limits in the right order. At $y=0$ the antiderivative is $0$, so $F=\tfrac{1}{1.1015}\left[0 - G(-1.65)\right]$ where, with $(-1.65)^2 = 2.7225$ and $(-1.65)^3 = -4.492125$, $G(-1.65) = 7.05(2.7225)-\frac{11.515}{3}(-4.492125) = 19.193625+17.242273 = 36.435898.$

6. 6

   Divide and state the signed result. $F = \frac{-36.435898}{1.1015} \approx -33.0784.$ Direct numerical quadrature of $(14.1-11.515y)\tfrac{y}{1.1015}$ over $[-1.65,0]$ returns $-33.0784368$, confirming both the magnitude and the sign. If the intended region is the triangle between $x=0$ and $x=-\tfrac{y}{1.1015}$ with limits written in increasing order, the magnitude $33.0784$ is the value to quote.