Question

(1 point) consider the area between the graphs x + y=-1 and x + 7 = y^2. this area can be computed in two different ways using integrals. first of all it can be computed as a sum of two integrals (int_{a}^{b}f(x)dx+int_{b}^{c}g(x)dx) where a =, b =, c = and f(x)=, g(x)=. alternatively the area can be computed as a single integral (int_{d}^{e}h(y)dy)

Explanation:

Step1: Rewrite the equations

First, rewrite $x + y=-1$ as $y=-x - 1$ and $x + 7=y^{2}$ as $x=y^{2}-7$. To find the intersection - points, set $-x - 1=\pm\sqrt{x + 7}$. Solving $(-x - 1)^{2}=x + 7$ gives $x^{2}+2x + 1=x + 7$, or $x^{2}+x - 6 = 0$. Factoring, we get $(x + 3)(x - 2)=0$, so $x=-3$ or $x = 2$. When $x=-3$, $y = 2$ or $y=-2$; when $x = 2$, $y=-3$.

Step2: Set up the double - integral for the area between curves

The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$ is $A=\int_{a}^{b}|f(x)-g(x)|dx$. For the curves $y_1=-x - 1$ and $y_2=\sqrt{x + 7}$ and $y_3=-\sqrt{x + 7}$, we need to split the integral based on the intersection points.
The area as a sum of two integrals $\int_{a}^{b}f(x)dx+\int_{b}^{c}g(x)dx$. The intersection of $y=-x - 1$ and $y = \sqrt{x + 7}$ gives us one part of the region, and the intersection of $y=-x - 1$ and $y=-\sqrt{x + 7}$ gives the other part.
The intersection of $y=-x - 1$ and $y=\sqrt{x + 7}$:
Set $-x - 1=\sqrt{x + 7}$, square both sides: $x^{2}+2x + 1=x + 7$, $x^{2}+x - 6=0$, $(x + 3)(x - 2)=0$. The valid solution for the square - root function is $x=-3$ (since for $x = 2$, $y=-3$ and $y=\sqrt{x + 7}\geq0$).
The intersection of $y=-x - 1$ and $y=-\sqrt{x + 7}$:
Set $-x - 1=-\sqrt{x + 7}$, square both sides: $x^{2}+2x + 1=x + 7$, $x^{2}+x - 6=0$, $(x + 3)(x - 2)=0$. The valid solution for $y =-\sqrt{x + 7}$ is also considered.
If we consider the area with respect to $x$, we find the intersection points of the curves. The curves $y=-x - 1$ and $x=y^{2}-7$ (or $y=\pm\sqrt{x + 7}$).
The intersection of $y=-x - 1$ and $y=\sqrt{x + 7}$:
Let $-x - 1=\sqrt{x + 7}$, then $x^{2}+2x + 1=x + 7$, $x^{2}+x - 6=0$, $(x + 3)(x - 2)=0$.
For $y=-x - 1$ and $y = \sqrt{x + 7}$, when $x=-3$, $y = 2$.
For $y=-x - 1$ and $y=-\sqrt{x + 7}$, when $x=-3$, $y=-2$.
The area as a sum of two integrals:
We need to find where the upper - and lower - curves change. The intersection of the two curves gives us the limits of integration.
The area $A=\int_{-7}^{-3}( \sqrt{x + 7}-(-\sqrt{x + 7}))dx+\int_{-3}^{2}(-x - 1-(-\sqrt{x + 7}))dx$.
Here, $a=-7$, $b=-3$, $c = 2$, $f(x)=2\sqrt{x + 7}$ for $x\in[-7,-3]$ and $g(x)=-x - 1+\sqrt{x + 7}$ for $x\in[-3,2]$.
If we want to write it as a single integral with respect to $y$:
Rewrite the equations as $x=-y - 1$ and $x=y^{2}-7$. The intersection points of $y^{2}-7=-y - 1$, $y^{2}+y - 6=0$, $(y + 3)(y - 2)=0$, so $y=-3$ or $y = 2$.
The area as a single integral with respect to $y$ is $\int_{-3}^{2}[(-y - 1)-(y^{2}-7)]dy$.
So, $a=-3$, $b = 2$, $f(x)=(-x - 1)-(x^{2}-7)=-x^{2}-x + 6$ (when considering the area as a single integral with respect to $x$ after interchanging the roles of $x$ and $y$ in the integral formula $\int_{a}^{b}(right - curve-left - curve)dy$).

Answer:

$a=-3$, $b = 2$, $f(x)=-x - 1-(x^{2}-7)=-x^{2}-x + 6$, $g(x)=2\sqrt{x + 7}$ (for the sum - of - two - integrals case with respect to $x$), $h(x)=-x - 1+\sqrt{x + 7}$ (for the sum - of - two - integrals case with respect to $x$) and for the single - integral case with respect to $y$: the integral is $\int_{-3}^{2}[(-y - 1)-(y^{2}-7)]dy$. In the context of the sum of two integrals $\int_{a}^{b}f(x)dx+\int_{b}^{c}g(x)dx$ with respect to $x$, $a=-7$, $b=-3$, $c = 2$, $f(x)=2\sqrt{x + 7}$, $g(x)=-x - 1+\sqrt{x + 7}$; for the single integral $\int_{a}^{b}H(y)dy$, $a=-3$, $b = 2$, $H(y)=(-y - 1)-(y^{2}-7)$.