Why Approximate Intersections Matter
To find the area enclosed by two curves, we first need to know where their boundaries meet. If the intersection equation has no convenient exact solution, an exact endpoint may be unavailable even though the region is easy to describe and its area is well-defined.
In that situation, numerical root-finding gives a usable approximation to the intersection. That approximation can then serve as an integration bound. The procedure requires care: a numerical root should be supported by evidence such as a sign-changing interval or a small residual, and the resulting area should be reported as an estimate.
Bracketing an Intersection
A useful first step is to find an interval \([a,b]\) where the continuous function \(h\) changes sign: \(h(a)\) and \(h(b)\) have opposite signs. The Intermediate Value Theorem then guarantees at least one root between \(a\) and \(b\). This does not guarantee that the root is unique; uniqueness requires additional information, such as \(h\) being strictly increasing or strictly decreasing on the interval.
Bisection repeatedly halves a sign-changing interval. If \(h(a)<0<h(b)\), evaluate \(h\) at the midpoint \(m=(a+b)/2\). If \(h(m)>0\), retain \([a,m]\); if \(h(m)<0\), retain \([m,b]\). If \(h(m)=0\), the root has been found exactly. When the signs are reversed, the same rule applies: retain the half whose endpoint values have opposite signs.
For the error bound, observe that each step retains an interval containing a root and halves its width. After \(n\) steps, the interval width is therefore \((b-a)/2^n\). Any root in that interval is at most half its width from the midpoint, giving the stated bound. If the sign change might enclose more than one root, bisection still retains at least one, but the particular root is not identified without further analysis.
Worked Example: Bisection for an Exponential Intersection
Worked Example: Bisection for an Exponential Intersection
Find an approximate intersection of \(y=e^x\) and \(y=x+2\), and use it to estimate the area between the curves and the vertical line \(x=0\).
Set \(h(x)=e^x-(x+2)\). At the endpoints \(x=1\) and \(x=1.5\),
The values have opposite signs, so continuity guarantees a root in \([1,1.5]\). The following bisection steps show the midpoint and its approximate function value. Each row retains the half-interval containing the sign change.
| Midpoint tested | Approximate \(h\) value | Retained interval |
|---|---|---|
| 1.25 | \(e^{1.25}-3.25\approx 3.490343-3.25=0.240343\) | \([1,1.25]\) |
| 1.125 | \(e^{1.125}-3.125\approx 3.080217-3.125=-0.044783\) | \([1.125,1.25]\) |
| 1.1875 | \(e^{1.1875}-3.1875\approx 3.279-3.1875\approx0.0915\) | \([1.125,1.1875]\) |
| 1.15625 | \(e^{1.15625}-3.15625\approx3.177-3.15625\approx0.0208\) | \([1.125,1.15625]\) |
| 1.140625 | \(e^{1.140625}-3.140625\approx3.129-3.140625\approx-0.0116\) | \([1.140625,1.15625]\) |
Continuing to halve the bracket gives \([1.14618,1.14620]\): the function is negative at the left endpoint and positive at the right. Use the midpoint \(r\approx1.14619\) as the intersection estimate. Substitution gives \(h(1.14619)\approx-0.000007\), a small residual consistent with the bracket.
On \(0\le x\le r\), the line is above the exponential: at \(x=0\), \(2>1\), and there is no intervening intersection. The area is therefore estimated using line minus exponential:
To check the numerical simplification, at the estimated root \(e^r\approx r+2\), so the area expression reduces to \(r^2/2+r-1\). With \(r=1.14619\), \(r^2\approx1.31375\), giving \(1.31375/2+1.14619-1\approx0.80307\). This is an estimated area, not an exact value.
Newton’s Method
Bisection is dependable when a sign-changing bracket is known, but it can require many steps to obtain several accurate digits. Newton’s method can converge faster when a suitable starting value is available. For an equation \(h(x)=0\), its update is
The method uses the tangent line to approximate the curve near the current estimate. The tangent line at \(x_n\) has equation \(y=h(x_n)+h'(x_n)(x-x_n)\). Its \(x\)-intercept solves \(0=h(x_n)+h'(x_n)(x-x_n)\). If \(h'(x_n)\ne0\), solving for \(x\) gives the Newton update above.
This derivation explains the method but does not guarantee convergence. The function must be differentiable near the root, the derivative used in the update must be nonzero, and a suitable starting value is needed. Some starting values produce steps that move away from the root or fail because the derivative is zero. A small residual \(\lvert h(x_n)\rvert\) is useful evidence, but does not alone prove that the estimate is close to a root if the function changes very slowly nearby.
Worked Example: Newton’s Method for a Logarithmic Intersection
Worked Example: Newton’s Method for a Logarithmic Intersection
Find the intersection of \(y=\ln x\) and \(y=2-x\) for \(x\ge1\), then estimate the area between the curves and the line \(x=1\). Define \(h(x)=\ln x+x-2\). Its derivative is \(h'(x)=1/x+1\), which is positive for \(x\ge1\); thus \(h\) is strictly increasing there and can have at most one root. Also, \(h(1)=-1\) and \(h(2)=\ln2>0\), so a root lies between 1 and 2.
Start Newton’s method at \(x_0=1.5\). The function and derivative values are
The first update is \(x_1=1.5-(-0.094535)/1.666667\approx1.556721\). At this estimate, \(h(1.556721)\approx-0.000698\) and \(h'(1.556721)\approx1.642382\), so
Substitution gives \(h(1.557146)\) approximately \(0.000001\), so we use \(r\approx1.557146\). At \(x=1\), the line is above the logarithm, and the curves meet at \(r\). The area is
At the intersection, \(\ln r=2-r\). Substituting this relation into the antiderivative at \(r\) gives \(3r-r^2/2-r(2-r)=r+r^2/2\). The antiderivative at 1 is \(3-1/2-0=2.5\). Since \(r^2\approx2.42470\), the area check is \(1.557146+2.42470/2-2.5\approx0.26950\).
Worked Example: A Trigonometric Intersection
Worked Example: A Trigonometric Intersection
Find the positive intersection of \(y=\sin x\) and \(y=x/2\), and estimate the area between the curves from \(x=0\) to that intersection. Set \(h(x)=\sin x-x/2\). The endpoint \(x=0\) is an exact intersection. For the positive one, Newton’s method gives \(h'(x)=\cos x-1/2\).
Starting from \(x_0=2\), use \(\sin2\approx0.909297\) and \(\cos2\approx-0.416147\):
At \(x_1\approx1.900995\), \(h(x_1)\approx-0.00454\) and \(h'(x_1)\approx-0.824\). Thus the next update is approximately \(x_2=1.900995-(-0.00454)/(-0.824)\approx1.89549\). Substitution in the original equation gives \(\sin(1.89549)\approx0.94775\) and \(1.89549/2\approx0.94775\), agreeing to the shown precision.
On the interval between the two intersections, \(\sin x\) is above \(x/2\). Therefore, with \(r\approx1.89549\),
For a numerical check, \(r^2\approx3.5929\), so \(r^2/4\approx0.8982\), and \(\cos r\approx-0.3190\). Hence \(1-(-0.3190)-0.8982\approx0.4208\). Because the endpoint is approximate, the reported area is approximate as well.
Choosing and Checking a Numerical Method
The methods complement one another. Bisection is appropriate when continuity and a sign-changing interval are established; its error bound follows directly from the interval width. Newton’s method is useful when a good starting value is available and evaluating the derivative is practical, but its steps should be checked rather than assumed to converge. A common strategy is to bracket the root first, then use Newton’s method while monitoring whether the estimates remain sensible.
For curves \(y=f(x)\) and \(y=g(x)\), define \(h(x)=f(x)-g(x)\).
Use function values, a graph, or a table to identify a plausible interval and check for a sign change.
Apply bisection or Newton’s method, then substitute the estimate into \(h\) and report the residual or bracket.
On the interval, determine which curve is above the other and integrate upper minus lower using the approximate bounds.
A Common Pitfall: Treating an Estimate as Exact
Suppose a calculation gives a root estimate \(r\approx1.14619\). Writing an area integral with upper bound \(1.14619\) does not make that bound exact. The integral and its evaluated value inherit the root approximation. Report the result with an approximation sign, and avoid claiming more precision than the root-finding supports.
Another error is to solve \(f(x)=g(x)\) numerically but then use \(f-g\) without checking its sign across the interval. The area is the integral of the upper curve minus the lower curve. Test an interior point, or otherwise establish the curves’ order, before choosing the integrand. If the order changes inside the region, the interval must be split at the crossing points.
Check Your Understanding
Answer each question using the definitions and methods in this tutorial.
- What equation should be solved to find intersections of \(y=f(x)\) and \(y=g(x)\)?
- Why does a sign change across an interval guarantee a root when the function is continuous?
- After four bisection steps, what is the interval width in terms of the original width \(b-a\)?
- What tangent-line idea produces the Newton update, and why does that derivation not guarantee convergence?
- Why should an area calculated using a numerical intersection be reported as an estimate?
