Scalar Line Integrals on Graphs y Equals g of x

From a Graph to an Ordinary Integral

A curve given by $y=g(x)$ can look like a familiar graph, but integrating a scalar field along that curve is not usually the same as integrating the field with respect to $x$. The curve has its own length: moving a small amount horizontally also moves vertically whenever $g'(x)$ is nonzero. The line integral must account for this extra distance.

This tutorial develops a direct formula for a graph over an interval $[a,b]$. The method is to substitute the graph into the scalar field and multiply by the graph’s arc-length factor. The resulting expression is an ordinary single-variable integral.

Key formula: If $g$ is continuously differentiable on $[a,b]$, and $f$ is a continuous scalar field along the graph, then the scalar line integral over the graph segment is $$ \int_C f\,ds = \int_a^b f(x,g(x))\sqrt{1+(g'(x))^2}\,dx. $$

Why the Arc-Length Factor Appears

Let the graph segment run from $x=a$ to $x=b$, where $a<b$. A parameterization that follows the graph is $\mathbf{r}(t)=(t,g(t))$ for $a\leq t\leq b$. Its velocity is $\mathbf{r}'(t)=(1,g'(t))$, so its speed is the length of that velocity vector.

\|\mathbf{r}'(t)\| = \sqrt{1+(g'(t))^2}.

By the definition of a scalar line integral along a smooth parameterized curve, the field is evaluated at the curve point and multiplied by the speed. Thus $f(\mathbf{r}(t))=f(t,g(t))$, and the speed supplies the factor $\sqrt{1+(g'(t))^2}$. Renaming $t$ as $x$ gives the formula above.

Conditions for the formula: The graph is considered over a finite interval $[a,b]$; $g$ is continuously differentiable there; and $f(x,g(x))$ is continuous on that interval. Under these conditions the displayed ordinary integral exists. If the graph or field has a discontinuity, the integral requires separate treatment and the formula alone does not guarantee existence.

The factor has a geometric interpretation. Over a small horizontal change $dx$, the graph changes vertically by approximately $g'(x)\,dx$. The small length along the graph is therefore approximately $$ ds=\sqrt{(dx)^2+(g'(x)\,dx)^2} =\sqrt{1+(g'(x))^2}\,dx $$ when $dx$ is taken positive along the interval. A steeper graph has more arc length per unit horizontal distance, which is why the factor is at least $1$.

A Reliable Setup

For each graph integral, three actions make the setup clear: specify the interval, substitute $y=g(x)$ into the field, and include the arc-length factor. Only after these steps should the resulting one-variable integral be simplified or evaluated.

Record the graph and interval.
Write $y=g(x)$ for $a\leq x\leq b$, with $a<b$.

Differentiate the graph function.
Find $g'(x)$, then form $\sqrt{1+(g'(x))^2}$.

Restrict the field to the graph.
Replace $y$ by $g(x)$ in $f(x,y)$, giving $f(x,g(x))$.

Integrate and check.
Evaluate the resulting ordinary integral, then differentiate an antiderivative or verify any substitution and endpoint values.

Worked Polynomial Examples

Worked Example: A Quadratic Graph with a Convenient Substitution

Evaluate the scalar line integral of $f(x,y)=2x$ along $y=x^2$, for $0\leq x\leq 1$.

Here $g(x)=x^2$, so $g'(x)=2x$. The field restricted to the graph remains $2x$. Therefore, $$ \int_C f\,ds = \int_0^1 2x\sqrt{1+4x^2}\,dx. $$ Set $u=1+4x^2$. Then $du=8x\,dx$, so $2x\,dx=\frac14\,du$. The endpoints become $u=1$ at $x=0$ and $u=5$ at $x=1$. Thus $$ \int_0^1 2x\sqrt{1+4x^2}\,dx = \frac14\int_1^5 u^{1/2}\,du = \frac14\left[\frac23u^{3/2}\right]_1^5 = \frac{5\sqrt5-1}{6}. $$ Check the antiderivative in the original variable: $$ \frac{d}{dx}\left(\frac{(1+4x^2)^{3/2}}{6}\right) = \frac16\cdot\frac32(1+4x^2)^{1/2}\cdot8x = 2x\sqrt{1+4x^2}. $$ It differentiates to the integrand, and its endpoint difference is $(5\sqrt5-1)/6$.

Worked Example: A Cubic Graph and a Matching Field

Evaluate the scalar line integral of $f(x,y)=3x^3$ along $y=x^3$, for $0\leq x\leq 1$.

For $g(x)=x^3$, $g'(x)=3x^2$, so the arc-length factor is $\sqrt{1+9x^4}$. The field is already a function of $x$, and hence $$ \int_C f\,ds = \int_0^1 3x^3\sqrt{1+9x^4}\,dx. $$ Let $u=1+9x^4$. Then $du=36x^3\,dx$, which means $3x^3\,dx=\frac1{12}\,du$. The transformed endpoints are $1$ and $10$. Consequently, $$ \int_0^1 3x^3\sqrt{1+9x^4}\,dx = \frac1{12}\int_1^{10}u^{1/2}\,du = \frac1{12}\cdot\frac23\left(10^{3/2}-1\right) = \frac{10\sqrt{10}-1}{18}. $$ For a check, differentiate $F(x)=\frac1{18}(1+9x^4)^{3/2}$: $$ F'(x) = \frac1{18}\cdot\frac32(1+9x^4)^{1/2}\cdot36x^3 = 3x^3\sqrt{1+9x^4}. $$ This is exactly the integrand, and $F(1)-F(0)=(10\sqrt{10}-1)/18$.

Worked Example: A Constant Field on a Curved Graph

Evaluate the integral of the constant scalar field $f(x,y)=1$ along $y=x^2$, for $0\leq x\leq 1$.

The graph derivative is $2x$, so $$ \int_C 1\,ds = \int_0^1\sqrt{1+4x^2}\,dx. $$ An antiderivative is $$ F(x)=\frac{x}{2}\sqrt{1+4x^2}+\frac14\ln\left(2x+\sqrt{1+4x^2}\right). $$ To check, differentiating the first term gives $$ \frac12\sqrt{1+4x^2}+\frac{2x^2}{\sqrt{1+4x^2}}, $$ and differentiating the logarithmic term gives $1/(2\sqrt{1+4x^2})$. Combining these terms over the common denominator $\sqrt{1+4x^2}$ gives $\sqrt{1+4x^2}$, as required. Evaluating at the endpoints yields $$ F(1)-F(0) = \frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5). $$ The field equals $1$, but the answer is greater than $1$: the graph segment has more length than its horizontal interval.

Polynomial Data Do Not Guarantee a Polynomial Integral

In the examples, the graph functions and fields are polynomials, but the arc-length factor contains a square root. The ordinary integrand therefore need not be a polynomial, and it need not have a simple polynomial antiderivative. For instance, the constant field on $y=x^2$ produces $\sqrt{1+4x^2}$, not a polynomial. The method still gives the correct setup; the remaining one-variable integral may require a substitution or a logarithmic expression.

When a substitution is used, its differential and transformed bounds must agree. In the first example, $u=1+4x^2$ increases from $1$ to $5$. Keeping the original bounds $0$ and $1$ after changing to $u$ would mix variables and produce an invalid integral. Either transform the bounds, as shown, or convert the antiderivative back to $x$ before evaluating.

A Common Mistake: Integrating with Respect to Horizontal Distance

A frequent error is to replace $ds$ by $dx$, as though every small horizontal step had the same length as the corresponding piece of the curve. For the constant field on $y=x^2$ from $0$ to $1$, this mistaken calculation gives $$ \int_0^1 1\,dx=1. $$ That is the length of the horizontal interval, not the length of the graph. Since $g'(x)=2x$, the correct calculation is $$ \int_C 1\,ds = \int_0^1\sqrt{1+4x^2}\,dx = \frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5). $$ The missing factor would only equal $1$ everywhere if $g'(x)=0$ throughout the interval, meaning the graph is horizontal. More generally, check that the square root contains $1+(g'(x))^2$, not merely $1+g'(x)$ and not just $1$.

Why This Formula Is Useful

Many curves in applications are naturally described as graphs, and many scalar quantities are distributed along those curves. The formula converts a geometric integral into a familiar single-variable calculation while preserving the distance measured along the curve. It also separates two ingredients that should not be confused: the scalar field’s value at each point and the amount of curve length represented by each horizontal increment.

A useful quick check is that $\sqrt{1+(g'(x))^2}\geq 1$. For a nonnegative field, the graph integral should therefore be at least as large as the integral of the restricted field with respect to $x$, whenever that comparison is meaningful. If a calculation on a sloping graph produces a factor smaller than $1$, revisit the derivative or the arc-length calculation.

Check Your Understanding

For each question, identify the graph, field restriction, and arc-length factor before evaluating.

What is the arc-length factor for the graph $y=4x^2$?
Set up, but do not evaluate, the scalar line integral of $f(x,y)=x+y$ along $y=x^3$, for $0\leq x\leq 1$.
For $y=2x+1$, what constant arc-length factor appears when integrating with respect to $x$?
Why is $\int_a^b f(x,g(x))\,dx$ generally not the scalar line integral along the graph?
In a substitution $u=1+9x^4$ over $0\leq x\leq1$, what are the transformed endpoints?