Area enclosed by a parabola line and vertical boundary

Key takeaway: This is an area-between-curves problem with a vertical boundary. The key steps are finding the intersection points, identifying the top and bottom functions, and integrating over the correct interval.

Find the intersection point

We are given

$f(x)=x^2+2x+11$, \quad $g(x)=-4x+2$, \quad x=0$.

First, find where the parabola and line intersect:

$x^2+2x+11=-4x+2$

$x^2+6x+9=0$

$(x+3)^2=0$

So the graphs intersect at

$x=-3$.

The enclosed region is bounded on the right by $x=0$ and on the left by $x=-3$.

Determine top minus bottom

On the interval $[-3,0]$, compare the functions. At $x=0$,

$f(0)=11$, \quad $g(0)=2$,

so the parabola is above the line.

The area is

$\int_{-3}^{0}\big(f(x)-g(x)\big)\,dx$.

Substitute:

$\int_{-3}^{0}\big((x^2+2x+11)-(-4x+2)\big)\,dx$

$=\int_{-3}^{0}(x^2+6x+9)\,dx$.

Integrate

Since

$x^2+6x+9=(x+3)^2$,

the integral becomes

$\int_{-3}^{0}(x+3)^2\,dx$.

An antiderivative is

$\frac{(x+3)^3}{3}$.

Evaluate from $-3$ to $0$:

$\frac{(0+3)^3}{3}-\frac{(-3+3)^3}{3}=\frac{27}{3}-0=9$.

Final answer

$\boxed{9}$

The correct choice is B.

Key idea

When a region is enclosed by a curve, a line, and a vertical boundary, the first step is often to find the intersection point that closes the region. Then integrate top minus bottom over that interval.

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Pitfalls the pros know 👇 A common error is integrating from $0$ to $3$ because the equation $(x+3)^2=0$ may tempt students to think the boundary is on the positive side. The actual intersection is at $x=-3$, so the interval is $[-3,0]$. Another mistake is subtracting in the wrong order. If you do $g(x)-f(x)$ here, the integral becomes negative, which signals the functions were reversed. In area problems, the absolute geometric area must be positive, so the top function must always be identified before integrating.

What if the problem changes? If the vertical boundary were $x=1$ instead of $x=0$, the enclosed region would change and the area would no longer be 9. A variant prompt could ask for the area enclosed by the same graphs and $x=1$. Then you would need to check whether the curves still form a closed region on the new interval and possibly split the integral if the top/bottom order changes. Small changes to a boundary line can completely alter the geometry of the region.

Tags: area between curves, intersection point, definite integral