Introduction
A line and a parabola can intersect at two points, enclosing a finite region between their graphs. Finding that region's area requires three connected steps: solve for the intersections, determine which curve is above the other, and integrate the vertical distance between them.
For a line and a quadratic function, these steps can often be made especially efficient. Once the intersections are known, the difference of the functions factors into a simple expression, and the area has a compact formula.
Finding the Intersection Bounds
Write the parabola as \(y=ax^2+cx+d\), where \(a\ne 0\), and the line as \(y=mx+b\). At an intersection, their \(y\)-values agree, so set the functions equal:
Rearranging gives a quadratic equation:
If this equation has two distinct real roots \(r_1<r_2\), the graphs intersect at \(x=r_1\) and \(x=r_2\). These are the vertical bounds of the region between the curves. The discriminant of the quadratic is \(\Delta=(c-m)^2-4a(d-b)\): two distinct intersections occur when \(\Delta>0\), one repeated intersection when \(\Delta=0\), and no real intersections when \(\Delta<0\).
With two distinct intersections of a line and a nondegenerate parabola, the graphs bound a region between those intersection points. To determine the correct subtraction order, compare the functions at any test value strictly between the roots. A positive value of line minus parabola means the line is above; a negative value means the parabola is above.
Worked Example: Find the Bounds and Integrate
Worked Example: Find the Bounds and Integrate
Find the area enclosed by \(y=x+1\) and \(y=x^2-3x+1\).
Step 1: Find the intersections. Set the functions equal:
Move everything to one side and factor:
Thus the intersection values are \(x=0\) and \(x=4\). These give the bounds \([0,4]\).
Step 2: Identify the upper curve. Subtract the parabola from the line:
At the test value \(x=2\), the difference is \(-4+8=4>0\). The line is above the parabola between the intersections.
Step 3: Integrate top minus bottom.
Check: Differentiating \(-x^3/3+2x^2\) gives \(-x^2+4x\), the integrand. Also, the difference is zero at both bounds: at \(x=0\), it is \(0\), and at \(x=4\), it is \(-16+16=0\). The enclosed area is \(\frac{32}{3}\) square units.
Why the Difference Has a Predictable Shape
The line-minus-parabola difference is a quadratic whose roots are the intersection values. Therefore, if the distinct intersections are \(r_1\) and \(r_2\), that difference factors in terms of those roots. This factorization explains both which graph is upper and why a shortcut area formula exists.
Justification. The line-minus-parabola difference is \(-ax^2+(m-c)x+(b-d)\). It has roots \(r_1\) and \(r_2\), and its leading coefficient is \(-a\). Thus it equals \(-a(x-r_1)(x-r_2)\), which is the same as \(a(x-r_1)(r_2-x)\).
For every \(x\) strictly between \(r_1\) and \(r_2\), both \(x-r_1\) and \(r_2-x\) are positive. The sign of the difference is therefore the sign of \(a\): if \(a>0\), the line is above; if \(a<0\), the parabola is above. In either case, the vertical distance between the graphs is \(|a|(x-r_1)(r_2-x)\). Integrating this distance gives the area.
Let \(L=r_2-r_1\) and set \(u=x-r_1\). Then \(x=r_1\) corresponds to \(u=0\), \(x=r_2\) corresponds to \(u=L\), and \(r_2-x=L-u\). Therefore,
This proves the formula. It applies when the two intersections are distinct and \(a\ne 0\), so that the graphs are a genuine line and parabola with a nonzero-width enclosed region.
Worked Example: The Parabola Is Above the Line
Worked Example: The Parabola Is Above the Line
Find the area enclosed by \(y=x\) and \(y=-2x^2-x+4\).
Step 1: Solve for the intersections.
The roots are \(x=-2\) and \(x=1\), so the bounds are \([-2,1]\).
Step 2: Choose the upper function. The line-minus-parabola difference is
At the midpoint \(x=-\frac12\), this difference is \(\frac12-1-4=-\frac92<0\). The parabola is above the line, so subtract the line from the parabola.
Step 3: Calculate the area.
At the upper bound, the antiderivative is \(-\frac23-1+4=\frac73\). At the lower bound, it is \(\frac{16}{3}-4-8=\frac{16}{3}-12=-\frac{20}{3}\). Hence
Check: The derivative of the antiderivative is \(-2x^2-2x+4\), as required. The line-minus-parabola difference is zero at \(x=-2\) and \(x=1\). The shortcut formula gives \(\frac{|{-2}|(1-(-2))^3}{6}=\frac{2\cdot27}{6}=9\), agreeing with the integral. The area is \(9\) square units.
A Shortcut When the Intersections Are Known
The formula can save time once the intersection values have been found. The coefficient \(a\) is the coefficient of \(x^2\) in the parabola, and the interval width is \(r_2-r_1\). The absolute value matters because area is positive regardless of which curve lies above.
Set the line and parabola equal, solve the quadratic, and order the roots as \(r_1<r_2\).
Evaluate line minus parabola at a test point between the roots to identify the upper graph.
Integrate upper minus lower, or use \(\frac{|a|(r_2-r_1)^3}{6}\) after confirming the two distinct intersections.
Worked Example: Use the Area Formula and Confirm It
Worked Example: Use the Area Formula and Confirm It
Find the area enclosed by \(y=2x-1\) and \(y=3x^2-13x+11\).
Step 1: Find the intersection values.
Thus \(r_1=1\) and \(r_2=4\), giving width \(r_2-r_1=3\).
Step 2: Determine which curve is above.
At \(x=2\), this difference is \(-12+30-12=6>0\), so the line is above the parabola.
Step 3: Use the formula. The parabola's quadratic coefficient is \(a=3\), so
Check by integration: The line-minus-parabola difference is the correct top-minus-bottom integrand.
At \(x=4\), the antiderivative is \(-64+120-48=8\). At \(x=1\), it is \(-1+\frac{15}{2}-12=-\frac{11}{2}\). The difference is \(8-(-\frac{11}{2})=\frac{16}{2}+\frac{11}{2}=\frac{27}{2}\). Differentiating the antiderivative returns \(-3x^2+15x-12\), confirming the integral. The area is \(\frac{27}{2}\) square units.
A Common Sign Mistake
Do not assume that the line is always above the parabola, or use the absolute value of the final signed integral as a substitute for checking the graph order. For example, in the second worked example the line-minus-parabola difference at the midpoint is \(-\frac92\), so the parabola is on top. Integrating line minus parabola over \([-2,1]\) gives \(-9\), not the area. The correction is to reverse the subtraction and integrate parabola minus line, giving \(9\).
A repeated intersection also requires care. If the discriminant is zero, the line and parabola touch at one \(x\)-value instead of crossing at two distinct bounds. A single tangency does not, by itself, enclose a finite region between these two graphs. The area formula requires two distinct intersection values.
Check Your Understanding
For each problem, find the intersections, identify the upper graph between them, and determine the enclosed area.
- Find the area between \(y=2x+1\) and \(y=x^2+1\).
- For \(y=-x+2\) and \(y=x^2-3x\), what test point can you use between the intersection bounds, and which graph is above there?
- A parabola has quadratic coefficient \(a=-4\) and intersects a line at \(x=-1\) and \(x=2\). Use the shortcut formula to find the enclosed area.
- What does a zero discriminant tell you about the intersections of a line and a parabola, and why is that not a two-bound enclosure?
- Explain why the area formula contains \(|a|\), rather than \(a\).
