Why Keep Irrational Bounds Exact?
The intersection points of two curves do not always have integer or rational coordinates. Solving their equations may produce bounds such as \(\sqrt{3}\) or \(\frac{1+\sqrt{5}}{2}\). Those values are exact coordinates, not inconvenient approximations.
When a region is bounded by two curves, its area is found by integrating the upper function minus the lower function with respect to \(x\), or by using an equivalent horizontal-slice setup when appropriate. If the intersection coordinates are irrational, the same method applies: use the exact coordinates as limits and evaluate the antiderivative at those limits. Rounding the bounds too early changes the calculation and can make the reported area less accurate.
From Intersections to Exact Limits
Suppose continuous curves \(y=f(x)\) and \(y=g(x)\) enclose a region from \(x=a\) to \(x=b\), with \(f(x)\ge g(x)\) throughout that interval. The area is
The limits \(a\) and \(b\) are the horizontal coordinates of the endpoints of the region. If the endpoints are intersections, find them by setting \(f(x)=g(x)\) and solving. When the solutions are irrational, retain their exact forms. Before integrating, check that the curves have the stated order throughout the interval. If they switch order inside the region, the interval must be split at the crossing or the area must be expressed using an absolute difference.
The definite integral is evaluated using an antiderivative \(F\) of the difference: \(\int_a^b(f-g)\,dx=F(b)-F(a)\). This evaluation rule applies to exact irrational limits just as it does to integer limits. The notation \(F(\sqrt{3})\), for example, means that \(\sqrt{3}\) is substituted into the antiderivative; it does not mean that the bound should first be replaced by a rounded decimal.
Why the Difference Gives Area
A short justification helps explain both the integrand and the importance of the correct bounds. Divide the interval from \(a\) to \(b\) into narrow subintervals. Over a typical subinterval of width \(\Delta x\), the vertical distance between the curves is approximately \(f(x)-g(x)\). The corresponding thin rectangle therefore has approximate area \((f(x)-g(x))\Delta x\). Adding these approximate areas and taking the limit as the widths approach zero gives the definite integral of the vertical distance.
This reasoning assumes the upper and lower curves are correctly identified on the interval and that the functions are continuous there. The intersection coordinates specify where the region begins and ends; their being irrational does not change the area principle.
Worked Example: A Symmetric Region with Bounds \(\pm\sqrt{2}\)
Worked Example: A Symmetric Region with Bounds \(\pm\sqrt{2}\)
Find the area enclosed by \(y=2\) and \(y=x^2\).
Step 1: Find the intersections. Set the two expressions for \(y\) equal:
At either coordinate, \(x^2=2\), so both points lie on both curves. The region therefore has bounds \(-\sqrt{2}\) and \(\sqrt{2}\).
Step 2: Identify the upper curve. At \(x=0\), the parabola has value \(0\), while the horizontal line has value \(2\). Thus \(2-x^2\) is positive between the intersection points, and the line is above the parabola on this interval.
Step 3: Set up and evaluate the area.
Check: Differentiating \(2x-\frac{x^3}{3}\) gives \(2-x^2\), the integrand. Also, this integrand is nonnegative between the intersections, as required for the chosen upper-minus-lower order. The exact area is \(\frac{8\sqrt{2}}{3}\) square units, approximately \(3.771\) square units to three decimal places.
Worked Example: Shifted Parabola and Bounds \(1\pm\sqrt{3}\)
Worked Example: Shifted Parabola and Bounds \(1\pm\sqrt{3}\)
Find the area enclosed by \(y=2x+2\) and \(y=x^2\).
Step 1: Find the intersections.
Thus the bounds are \(1-\sqrt{3}\) and \(1+\sqrt{3}\). Substituting either root into \(x^2-2x-2\) gives zero, confirming the intersection equation.
Step 2: Determine the curve order. At \(x=1\), the line gives \(4\) and the parabola gives \(1\), so the line is above the parabola between the roots. The difference can also be written \(2x+2-x^2=3-(x-1)^2\), which is positive when \((x-1)^2<3\), precisely between these bounds.
Step 3: Integrate. Let \(u=x-1\). Then the bounds become \(-\sqrt{3}\) and \(\sqrt{3}\), and the difference becomes \(3-u^2\).
Check: The derivative of \(3u-\frac{u^3}{3}\) is \(3-u^2\). The integrand is positive for \(-\sqrt{3}<u<\sqrt{3}\), so the result is a positive area. The exact area is \(4\sqrt{3}\) square units, approximately \(6.928\) square units to three decimal places.
Exact Evaluation at Irrational Limits
The preceding calculations use the Fundamental Theorem of Calculus: if \(h\) is continuous on \([a,b]\) and \(F'(x)=h(x)\) there, then \(\int_a^b h(x)\,dx=F(b)-F(a)\). The theorem places no requirement that \(a\) or \(b\) be rational. The exact endpoint is substituted into \(F\) using ordinary algebra.
When the limits contain radicals, simplifying powers and using relationships satisfied by the intersection coordinates can keep the arithmetic manageable. For instance, if \(r^2=3\), then \(r^3=3r\). That identity is exact; it is preferable to replacing \(r\) with a rounded decimal and then calculating powers approximately.
Set the curve expressions equal, and keep exact roots such as radicals in the result.
Check that the intersection coordinates delimit the intended region and that the chosen upper-minus-lower difference is nonnegative across it.
Find an antiderivative and substitute the exact lower and upper bounds before simplifying.
Give the exact area first, then provide a clearly rounded decimal approximation if one is requested.
Worked Example: Golden-Ratio Intersection Bounds
Worked Example: Golden-Ratio Intersection Bounds
Find the area enclosed by \(y=x+1\) and \(y=x^2\).
Step 1: Solve for the bounds.
Write \(\alpha=\frac{1-\sqrt{5}}{2}\) and \(\beta=\frac{1+\sqrt{5}}{2}\). These are the exact lower and upper bounds. At either endpoint, \(x^2=x+1\), confirming that each is an intersection.
Step 2: Choose the integrand. At \(x=0\), which lies between the two roots, the line has value \(1\) and the parabola has value \(0\). Therefore \(x+1-x^2\) is the upper-minus-lower difference on the interval.
Step 3: Evaluate exactly. Completing the square gives
Set \(u=x-\frac{1}{2}\). The bounds become \(-\frac{\sqrt{5}}{2}\) and \(\frac{\sqrt{5}}{2}\):
Check: The derivative of \(\frac{5u}{4}-\frac{u^3}{3}\) is \(\frac{5}{4}-u^2\), as required. The difference is positive between the bounds because \(|u|<\frac{\sqrt{5}}{2}\) implies \(u^2<\frac{5}{4}\). The exact area is \(\frac{5\sqrt{5}}{6}\) square units, approximately \(1.863\) square units to three decimal places.
A Common Mistake: Rounding the Bounds First
A tempting shortcut is to replace an exact root with a nearby decimal before setting up the integral. For example, \(\sqrt{2}\approx1.414\), so one might use \(-1.414\) and \(1.414\) as limits in the first example. That produces an approximation, not the exact area, and the endpoint rounding introduces an error before the integration is complete.
The corrected calculation keeps the bounds exact:
Another error is to report a decimal as if it were the exact value. The expression \(\frac{8\sqrt{2}}{3}\) is the exact area; \(3.771\) is its approximation to three decimal places. State which form is exact and how the decimal has been rounded.
Check Your Understanding
For each problem, find the exact area first. If you give a decimal, round only at the final step.
- Find the intersections of \(y=4\) and \(y=x^2\), then set up the area integral using exact bounds.
- For \(y=2x+1\) and \(y=x^2\), solve for the intersection coordinates and determine which curve is above between them.
- Explain why \(\sqrt{5}\) should remain exact while evaluating an integral with \(\sqrt{5}\) as a bound.
- In the golden-ratio example, differentiate the antiderivative \(\frac{5u}{4}-\frac{u^3}{3}\) to check the integrand.
- State the difference between an exact area and its decimal approximation for the region between \(y=2\) and \(y=x^2\).
