Arc Length of Quadratic Graphs

Why quadratic graph lengths have an exact formula

A quadratic graph may bend substantially over an interval, so its length is generally greater than the horizontal distance between its endpoints. The arc-length formula turns that geometric question into an integral. For a quadratic, the derivative is linear, which makes a substitution possible and gives an exact antiderivative.

The key is to keep track of both the transformed endpoints and the coefficient in the substitution. In particular, the quadratic coefficient may be negative, but the resulting length must still be positive.

Definition: For a continuously differentiable graph $y=f(x)$ on an interval $[r,s]$, where $r<s$, its arc length is $$ L=\int_r^s \sqrt{1+\bigl(f'(x)\bigr)^2}\,dx. $$ For a quadratic $f(x)=ax^2+bx+c$, with $a\ne 0$, this becomes $$ L=\int_r^s \sqrt{1+(2ax+b)^2}\,dx. $$

The graph-length formula applies here because a quadratic has a continuous derivative on every finite interval. The integral measures the accumulated lengths of the small pieces of the curve; it is not simply the distance between the graph’s endpoints.

The antiderivative needed for a quadratic

After substituting $u=2ax+b$, the integrand takes the standard form $\sqrt{1+u^2}$. Define the inverse hyperbolic sine here by its logarithmic expression, and introduce a function that will serve as an antiderivative.

Definition: For every real $u$, let $$ \operatorname{arsinh}(u)=\ln\left(u+\sqrt{1+u^2}\right), \qquad H(u)=u\sqrt{1+u^2}+\operatorname{arsinh}(u). $$ Then $H'(u)=2\sqrt{1+u^2}$.

To verify the derivative, first differentiate the logarithmic expression. Since $\sqrt{1+u^2}>|u|$, the argument of the logarithm is positive for every real $u$. Differentiation gives

\begin{aligned} \frac{d}{du}\operatorname{arsinh}(u) &=\frac{1+\dfrac{u}{\sqrt{1+u^2}}}{u+\sqrt{1+u^2}}\\ &=\frac{1}{\sqrt{1+u^2}}. \end{aligned}

For the second equality, the numerator is $\bigl(\sqrt{1+u^2}+u\bigr)/\sqrt{1+u^2}$, so its factor $\sqrt{1+u^2}+u$ cancels with the denominator. Now differentiate the other term in $H$:

\begin{aligned} \frac{d}{du}\left(u\sqrt{1+u^2}\right) &=\sqrt{1+u^2}+\frac{u^2}{\sqrt{1+u^2}}\\ &=\frac{1+2u^2}{\sqrt{1+u^2}}. \end{aligned}

Adding the derivative of $\operatorname{arsinh}(u)$ yields

H'(u) =\frac{1+2u^2}{\sqrt{1+u^2}}+\frac{1}{\sqrt{1+u^2}} =2\sqrt{1+u^2}.

This proves the antiderivative fact we need. In particular, $\frac12H(u)$ is an antiderivative of $\sqrt{1+u^2}$.

Formula for the length of a quadratic graph

Let $f(x)=ax^2+bx+c$, with $a\ne 0$, and consider its graph for $r\le x\le s$. Its derivative is $f'(x)=2ax+b$, so the length integral is

L=\int_r^s \sqrt{1+(2ax+b)^2}\,dx.

Set $u=2ax+b$. Then $du=2a\,dx$, and the transformed endpoint values are $2ar+b$ and $2as+b$. The factor $a$ must remain in the calculation: if $a<0$, the transformed endpoints occur in decreasing order. Using $H'(u)=2\sqrt{1+u^2}$ gives the exact result.

Theorem: If $a\ne 0$ and $r<s$, the length of the graph $y=ax^2+bx+c$ from $x=r$ to $x=s$ is $$ L=\frac{H(2as+b)-H(2ar+b)}{4a}, $$ where $H(u)=u\sqrt{1+u^2}+\operatorname{arsinh}(u)$. The formula gives a positive length for either sign of $a$.

The formula can also be derived by differentiating its antiderivative. Define $F(x)=H(2ax+b)/(4a)$. The chain rule and the derivative of $H$ show that

F'(x) =\frac{H'(2ax+b)\,2a}{4a} =\frac{2\sqrt{1+(2ax+b)^2}\,2a}{4a} =\sqrt{1+(2ax+b)^2}.

Thus $F$ is an antiderivative of the length integrand, and evaluating $F(s)-F(r)$ gives the stated formula. If $a>0$, the transformed endpoint $2as+b$ is greater than $2ar+b$, and the numerator is positive. If $a<0$, their order is reversed, so the numerator and denominator are both negative. Their quotient is positive in either case.

Differentiate the quadratic.
Write $f'(x)=2ax+b$, then put this expression into the graph-length integral.

Transform the endpoints.
Calculate $u_r=2ar+b$ and $u_s=2as+b$ from the original interval endpoints.

Evaluate and simplify.
Use $L=[H(u_s)-H(u_r)]/(4a)$, retaining the sign of $a$ until the quotient is formed.

Worked examples

Worked Example: The graph of $y=x^2$ from $x=0$ to $x=1$

Here $a=1$, $b=0$, $r=0$, and $s=1$. The derivative is $2x$, so

L=\int_0^1\sqrt{1+4x^2}\,dx.

The transformed endpoints are $u_r=0$ and $u_s=2$. By the formula,

\begin{aligned} L &=\frac{H(2)-H(0)}{4}\\ &=\frac{2\sqrt{1+2^2}+\operatorname{arsinh}(2)-0}{4}\\ &=\frac{2\sqrt{5}+\operatorname{arsinh}(2)}{4}\\ &=\frac{\sqrt{5}}{2}+\frac{\operatorname{arsinh}(2)}{4}. \end{aligned}

To check, the antiderivative used is $F(x)=H(2x)/4$, and

F'(x)=\frac{H'(2x)\,2}{4} =\sqrt{1+4x^2},

which is exactly the integrand. The endpoint evaluation therefore gives the required length.

Worked Example: A shallow parabola on a symmetric interval

Find the length of $y=\frac14x^2$ from $x=-2$ to $x=2$. Here $a=\frac14$, $b=0$, so $4a=1$, and $u=x/2$. The transformed endpoints are $-1$ and $1$. Therefore,

\begin{aligned} L &=H(1)-H(-1)\\ &=2H(1)\\ &=2\left(\sqrt{2}+\operatorname{arsinh}(1)\right). \end{aligned}

The second line uses the fact that $H$ is odd: both $u\sqrt{1+u^2}$ and $\operatorname{arsinh}(u)$ change sign when $u$ is replaced by $-u$. For a direct check, an antiderivative in $x$ is $F(x)=H(x/2)$, and

F'(x)=H'(x/2)\frac12 =2\sqrt{1+\frac{x^2}{4}}\frac12 =\sqrt{1+\frac{x^2}{4}}.

This matches the length integrand, since $f'(x)=x/2$. The equal contributions on the two sides of the vertex are also reflected in the symmetric transformed endpoints.

Worked Example: A downward-opening parabola

Find the length of $y=-x^2+2x$ from $x=0$ to $x=2$. Here $a=-1$, $b=2$, and $f'(x)=2-2x$. The transformed endpoint values are $u_r=2$ and $u_s=-2$. Applying the formula, including its negative denominator, gives

\begin{aligned} L &=\frac{H(-2)-H(2)}{-4}\\ &=\frac{-2H(2)}{-4}\\ &=\frac{H(2)}{2}\\ &=\sqrt{5}+\frac{\operatorname{arsinh}(2)}{2}. \end{aligned}

Here $H(2)=2\sqrt5+\operatorname{arsinh}(2)$, and oddness gives $H(-2)=-H(2)$. To check the antiderivative, use $F(x)=H(2-2x)/(-4)$:

F'(x) =\frac{H'(2-2x)(-2)}{-4} =\sqrt{1+(2-2x)^2},

which equals the length integrand. Both the transformed endpoint difference and the denominator are negative, confirming that the computed length is positive.

A common sign mistake

For the downward-opening parabola in the previous example, a tempting but incorrect step is to keep the negative endpoint difference and divide by $4$ instead of by $4a=-4$. That would give

\frac{H(-2)-H(2)}{4} =\frac{-2H(2)}{4} =-\frac{H(2)}{2},

a negative “length.” The correction is to preserve the factor $4a$ from the substitution. Since $a=-1$, the correct denominator is $-4$, and the result is $H(2)/2>0$. Do not replace the formula’s denominator with $4$ or reorder transformed endpoints without also accounting for the change in sign.

Another useful check is to compare the answer with the horizontal distance $s-r$. The length of a graph over an interval cannot be less than that distance. This check can reveal an arithmetic or sign error, although it does not replace evaluating the integral correctly.

Using the result effectively

For a quadratic, the constant $c$ does not appear in the length formula. Adding a constant shifts the graph vertically but does not change its derivative, so it does not change the length. The coefficient $b$, by contrast, affects the derivative and therefore the transformed endpoints.

If $a=0$, the function is a line rather than a quadratic. Its derivative is the constant $b$, and on $[r,s]$ its length is $(s-r)\sqrt{1+b^2}$. For $a\ne 0$, use the quadratic formula above. In either case, specify the interval: the same graph can have different lengths over different intervals.

Key takeaway: For $y=ax^2+bx+c$, substitute $u=2ax+b$ into the arc-length integral. Evaluate the transformed endpoints in their original order and keep the signed factor $4a$; this produces the exact length even when the parabola opens downward.

Check Your Understanding

Use the quadratic length formula and show the endpoint values and sign of the denominator in each calculation.

For $y=x^2$ on $[0,2]$, what are the transformed endpoints?
Find the exact length of $y=\frac12x^2$ on $[-1,1]$.
For $y=-2x^2+x$ on $[0,1]$, identify $a$, the transformed endpoints, and the denominator in the formula.
Why does the constant term $c$ not affect the length of $y=ax^2+bx+c$?
What error occurs if the denominator $4a$ is replaced by $4$ when $a<0$?

What You'll Learn

Why quadratic graph lengths have an exact formula

The antiderivative needed for a quadratic

Formula for the length of a quadratic graph

Worked examples

Worked Example: The graph of \(y=x^2\) from \(x=0\) to \(x=1\)

Worked Example: A shallow parabola on a symmetric interval

Worked Example: A downward-opening parabola

A common sign mistake

Using the result effectively

Check Your Understanding

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Arc Length of Quadratic Graphs

What You'll Learn

Why quadratic graph lengths have an exact formula

The antiderivative needed for a quadratic

Formula for the length of a quadratic graph

Worked examples

Worked Example: The graph of \(y=x^2\) from \(x=0\) to \(x=1\)

Worked Example: A shallow parabola on a symmetric interval

Worked Example: A downward-opening parabola

A common sign mistake

Using the result effectively

Check Your Understanding

Test yourself: Calculus II final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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