Polynomial Curves With One Enclosed Region

From polynomial intersections to area

Two polynomial graphs can meet at points that form the left and right boundaries of an enclosed region. The intersections tell us where the area calculation begins and ends. Between those points, the vertical height of the region is the value of the upper polynomial minus the value of the lower polynomial.

The key is to keep the order of the curves consistent across the interval. Once that order is established, the area is a definite integral. For polynomials, the intersections are found by solving a polynomial equation, and the integral can be evaluated using polynomial antiderivatives.

Intersections and the area formula

Let $f$ and $g$ be polynomial functions. Their graphs intersect at an $x$-value $x$ precisely when $f(x)=g(x)$. Thus, to find the intersections, solve $f(x)-g(x)=0$, then use each solution in either function to find the corresponding $y$-coordinate.

Suppose the two distinct intersections that bound the region have $x$-coordinates $a$ and $b$, where $a<b$. If $f(x)>g(x)$ for every $x$ strictly between $a$ and $b$, then $f$ is the upper curve and $g$ is the lower curve throughout that interval. The area is given by the following result.

Theorem: If $f$ and $g$ are polynomials, $f(a)=g(a)$, $f(b)=g(b)$, $a<b$, and $f(x)>g(x)$ for every $x\in(a,b)$, then the area of the region between their graphs from $x=a$ to $x=b$ is $$ A=\int_a^b \bigl(f(x)-g(x)\bigr)\,dx. $$ If instead $g(x)>f(x)$ throughout the interval, the area is $\int_a^b(g(x)-f(x))\,dx$.

The strict inequality in the interior ensures that the same curve stays above the other throughout the region. At the endpoints the curves meet, so their vertical separation is zero. In this tutorial, we focus on an interval bounded by intersections where the curves do not switch order inside the interval.

Why upper minus lower gives area

At a fixed $x$-value between $a$ and $b$, a thin vertical slice of the region has height $f(x)-g(x)$ when $f$ is above $g$. Its approximate area is therefore its height multiplied by a small width $\Delta x$. Adding many such slice areas gives an approximation to the total area.

More precisely, divide $[a,b]$ into subintervals and choose a sample point $x_i^*$ in each subinterval. The sum of the slice areas is approximately

\sum_i \bigl(f(x_i^*)-g(x_i^*)\bigr)\Delta x_i.

As the widths of the subintervals shrink, these sums approach the definite integral of the height function. Polynomials are continuous, so their difference is continuous and integrable on $[a,b]$. The limit of the sums is therefore

\int_a^b \bigl(f(x)-g(x)\bigr)\,dx.

This argument explains both the formula and its assumptions: the integrand is the vertical height, and it must represent that height consistently on the interval. The definite integral then gives the area. Since the integrand is positive in the interior, the result is positive as expected.

A reliable method

For two polynomial graphs that enclose one region, use these steps:

Find the intersections.
Set the polynomial expressions equal, solve for the $x$-coordinates, and find the corresponding $y$-coordinates if needed.

Choose the interval.
Use the left and right intersection $x$-coordinates as the integration limits.

Determine which graph is on top.
Compare the function values at an interior point, or factor their difference and check its sign throughout the interval.

Integrate upper minus lower.
Evaluate the definite integral and check that the final area is positive.

Factoring the difference is often especially useful. Once the intersections are known, the factors corresponding to the intersection $x$-values may make the sign between them clear. The sign check is about the whole interval, not merely the endpoint values, where the difference is zero.

Worked examples

Worked Example: A line and a parabola

Find the area enclosed by $y=x+2$ and $y=x^2$.

Step 1: Find the intersections. Set the expressions equal:

x+2=x^2 \quad\Longrightarrow\quad x^2-x-2=0 \quad\Longrightarrow\quad (x-2)(x+1)=0.

Thus $x=-1$ or $x=2$. At $x=-1$, both expressions give $y=1$; at $x=2$, both give $y=4$. The intersections are $(-1,1)$ and $(2,4)$.

Step 2: Identify the upper curve. The line minus the parabola is

$$ (x+2)-x^2=-(x-2)(x+1). $$

For $-1<x<2$, we have $x+1>0$ and $x-2<0$, so $-(x-2)(x+1)>0$. Therefore the line is above the parabola throughout the interior.

Step 3: Integrate upper minus lower.

A=\int_{-1}^{2}(x+2-x^2)\,dx =\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}.

At the upper endpoint, the antiderivative is

\frac{2^2}{2}+2(2)-\frac{2^3}{3} =2+4-\frac{8}{3} =\frac{10}{3}.

At the lower endpoint, it is

\frac{(-1)^2}{2}+2(-1)-\frac{(-1)^3}{3} =\frac12-2+\frac13 =-\frac76.

Subtract the lower-endpoint value from the upper-endpoint value:

A=\frac{10}{3}-\left(-\frac76\right) =\frac{20}{6}+\frac76 =\frac{27}{6} =\frac92.

Check: Differentiating $\frac{x^2}{2}+2x-\frac{x^3}{3}$ gives $x+2-x^2$, the stated integrand. The area $\frac92$ is positive, and substituting $x=-1$ and $x=2$ into the original curves confirms that both are intersection points.

Worked Example: A horizontal line and a shifted parabola

Find the area enclosed by $y=3$ and $y=x^2-2x$.

Step 1: Find the intersections.

3=x^2-2x \quad\Longrightarrow\quad x^2-2x-3=0 \quad\Longrightarrow\quad (x-3)(x+1)=0.

The intersection $x$-coordinates are $-1$ and $3$. At each, the parabola has value $3$, so the intersection points are $(-1,3)$ and $(3,3)$.

Step 2: Check the order. The horizontal line minus the parabola factors as

$$ 3-(x^2-2x)=-(x-3)(x+1). $$

For $-1<x<3$, $x+1>0$ and $x-3<0$, so this difference is positive. The line is above the parabola on the interval.

Step 3: Evaluate the area.

\begin{aligned} A&=\int_{-1}^{3}\bigl(3-(x^2-2x)\bigr)\,dx\\ &=\int_{-1}^{3}(3-x^2+2x)\,dx\\ &=\left[3x-\frac{x^3}{3}+x^2\right]_{-1}^{3}. \end{aligned}

At $x=3$, the antiderivative equals

3(3)-\frac{3^3}{3}+3^2=9-9+9=9.

At $x=-1$, it equals

3(-1)-\frac{(-1)^3}{3}+(-1)^2=-3+\frac13+1=-\frac53.

Consequently,

A=9-\left(-\frac53\right)=\frac{27}{3}+\frac53=\frac{32}{3}.

Check: The derivative of $3x-\frac{x^3}{3}+x^2$ is $3-x^2+2x$, which equals the upper-minus-lower expression. Substitution into both curves gives $3$ at each endpoint, as required.

Worked Example: A quartic and a constant

Find the area enclosed by $y=1$ and $y=x^4$.

Step 1: Find all intersections.

1=x^4 \quad\Longrightarrow\quad x^4-1=0 \quad\Longrightarrow\quad (x^2-1)(x^2+1)=0.

For real $x$, $x^2+1$ is never zero. The real solutions are $x=-1$ and $x=1$. Both curves have value $1$ at these points, giving intersections $(-1,1)$ and $(1,1)$.

Step 2: Check which curve is above. For $-1<x<1$, $|x|<1$, so $x^4<1$. Thus the constant function is above $x^4$ throughout the interval.

Step 3: Calculate and check the area.

A=\int_{-1}^{1}(1-x^4)\,dx =\left[x-\frac{x^5}{5}\right]_{-1}^{1}.

At $x=1$, the antiderivative is $1-\frac15=\frac45$. At $x=-1$, it is $-1+\frac15=-\frac45$. Therefore,

A=\frac45-\left(-\frac45\right)=\frac85.

Check: Differentiating $x-\frac{x^5}{5}$ gives $1-x^4$. Substitution confirms both endpoints satisfy $1=x^4$, and the positive area $\frac85$ agrees with the positive vertical height on the interior.

A common mistake: reversing the curves

The definite integral records signed accumulation. If the lower curve is mistakenly subtracted from the upper curve in the wrong order, the integral is negative. That negative result does not represent a negative geometric area; it signals that the integrand was not the region's positive vertical height.

For the line and parabola in the first example, reversing the order gives

\int_{-1}^{2}\bigl(x^2-(x+2)\bigr)\,dx =-\frac92.

The curves have not changed, so their geometric region still has area $\frac92$. The correction is to use line minus parabola, because the line is above the parabola throughout this interval. Checking the order before integrating prevents this error.

Why the intersection and sign checks matter

The bounds must describe the horizontal extent of the region being measured. Using incorrect intersection values includes too much or too little of the region. Likewise, the correct bounds alone do not determine the integrand: the integrand must be the vertical distance between the two graphs.

For polynomial curves, a useful check is to combine these facts: verify that the chosen endpoints solve $f(x)=g(x)$, then establish the sign of $f(x)-g(x)$ between them. If it is positive throughout, integrate $f-g$; if it is negative throughout, integrate $g-f$. Finally, differentiate the antiderivative used in the evaluation and confirm that the resulting area is positive.

When the area represents a physical quantity, its units are square units: for example, metres squared if both coordinate axes represent metres. The integral combines vertical lengths with horizontal widths, giving the corresponding area units.

Check Your Understanding

For each question, identify the intersections and consider how the order of the two curves determines the area integrand.

For $y=2x+3$ and $y=x^2$, find the intersection $x$-coordinates and determine which curve is above between them.
For $y=4$ and $y=x^2$, write, but do not evaluate, the definite integral for the enclosed area.
Why is it not enough to check which curve is higher at an intersection point?
If an integral using upper minus lower evaluates to a negative number, what should be checked?
Explain why the definite integral of the vertical height gives area rather than just a collection of slice heights.

Polynomial Curves With One Enclosed Region

What You'll Learn

From polynomial intersections to area

Intersections and the area formula

Why upper minus lower gives area

A reliable method