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Substitution and Reversing the Chain Rule · Tutorial 1157 of 1495

The Unchanged Bounds Error

Learn why substitution requires matching bounds and integrand variables, and how to evaluate correctly using either of two consistent routes.

Intermediate 9 min read

What You'll Learn

  • Identify when transformed bounds no longer match the integration variable.
  • Convert the original endpoints using the substitution.
  • Keep the correct orientation when the inner function is decreasing.
  • Evaluate by changing bounds or by returning to the original variable.
  • Check a definite-integral result using an antiderivative.

Why Bounds Matter in Substitution

A definite integral records both an integrand and an interval. In \(\int_a^b f(x)\,dx\), the variable is \(x\), and the endpoints \(a\) and \(b\) are values of \(x\). If a substitution changes the variable from \(x\) to \(u\), the endpoints must also be expressed as values of \(u\), unless the calculation returns to \(x\) before evaluation.

The unchanged-bounds error occurs when a solution rewrites the integrand using \(u\) but keeps the original numerical endpoints without checking whether they are still the correct endpoints. This can produce a plausible-looking calculation with the wrong value. The issue is not the substitution itself; it is mixing two incompatible descriptions of the same interval.

Key rule: In a definite integral, the variable, integrand, differential, and bounds must belong to the same variable. After substituting \(u=g(x)\), either use the transformed bounds \(u=g(a)\) and \(u=g(b)\), or substitute back to \(x\) before using \(a\) and \(b\).

Two Consistent Ways to Evaluate

Suppose the original integral is over \(a\leq x\leq b\), and the substitution is \(u=g(x)\). There are two standard routes.

  1. Change the bounds: calculate \(u=g(a)\) and \(u=g(b)\), rewrite the entire integral in \(u\), and evaluate using those \(u\)-bounds.
  2. Return to \(x\): find an antiderivative expressed in \(u\), replace \(u\) by \(g(x)\), and then evaluate the resulting expression at \(x=a\) and \(x=b\).

Both routes give the same result when performed correctly. What is not valid is to use an integrand and differential in \(u\) while retaining bounds that describe \(x\), or to use \(x\)-bounds on a \(u\)-integral merely because those numbers are convenient.

Definition: For the substitution \(u=g(x)\) in an integral with \(x\)-bounds \(a\) and \(b\), the transformed bounds are \(g(a)\) and \(g(b)\). They are obtained by applying the substitution to each original endpoint, in its original order.

Why the Bounds Must Transform

The relationship between the bounds and the variable follows from the same rule that justifies substitution. Suppose the original integrand can be written as \(h(g(x))g'(x)\), where \(g\) is differentiable and \(h\) is continuous on the relevant range. Let \(H\) be an antiderivative of \(h\), so \(H'(u)=h(u)\). By the chain rule,

$$ \frac{d}{dx}H(g(x)) = H'(g(x))g'(x) = h(g(x))g'(x). $$

Therefore, by the Fundamental Theorem of Calculus, the original integral is

$$ \int_a^b h(g(x))g'(x)\,dx = H(g(b))-H(g(a)). $$

On the other hand, applying the Fundamental Theorem in the \(u\)-variable gives

$$ \int_{g(a)}^{g(b)}h(u)\,du = H(g(b))-H(g(a)). $$

Both expressions have the same endpoint values, so

$$ \int_a^b h(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)}h(u)\,du. $$

This derivation also explains why the order of the transformed bounds matters. The lower \(x\)-endpoint \(a\) maps to \(g(a)\), and the upper \(x\)-endpoint \(b\) maps to \(g(b)\). If \(g\) is decreasing, then \(g(a)\) may be greater than \(g(b)\). Keep that order; the resulting reversed orientation is part of the equality.

Worked Example: A Power of a Quadratic

Worked Example: A Power of a Quadratic

Evaluate \(\int_0^1 2x(x^2+1)^3\,dx\) using \(u=x^2+1\).

First, differentiate the substitution:

$$ u=x^2+1, \qquad du=2x\,dx. $$

The original endpoints are \(x=0\) and \(x=1\). Transform each one:

$$ x=0 \implies u=0^2+1=1, \qquad x=1 \implies u=1^2+1=2. $$

Now the integrand, differential, and bounds are all in \(u\):

$$ \int_0^1 2x(x^2+1)^3\,dx = \int_1^2 u^3\,du = \left[\frac{u^4}{4}\right]_1^2 = \frac{2^4}{4}-\frac{1^4}{4} = \frac{16}{4}-\frac{1}{4} = \frac{15}{4}. $$

To check by returning to \(x\), an antiderivative is \(\frac{(x^2+1)^4}{4}\). Its derivative is

$$ \frac{d}{dx}\left(\frac{(x^2+1)^4}{4}\right) = \frac{1}{4}\cdot 4(x^2+1)^3\cdot 2x = 2x(x^2+1)^3. $$

Evaluating this antiderivative at the original \(x\)-bounds gives

$$ \left[\frac{(x^2+1)^4}{4}\right]_0^1 = \frac{(1^2+1)^4}{4}-\frac{(0^2+1)^4}{4} = \frac{16}{4}-\frac{1}{4} = \frac{15}{4}. $$

Both consistent routes agree.

The Error in Concrete Form

In the example above, after replacing \(x^2+1\) by \(u\) and \(2x\,dx\) by \(du\), the transformed integral is \(\int u^3\,du\). The numbers \(0\) and \(1\) describe the original \(x\)-interval, not the corresponding \(u\)-interval. Keeping those numbers gives the incorrect calculation

$$ \int_0^1 u^3\,du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}. $$

This is wrong because \(x=0\) corresponds to \(u=1\), not \(u=0\), and \(x=1\) corresponds to \(u=2\), not \(u=1\). The transformed integral must therefore run from \(u=1\) to \(u=2\), giving \(\frac{15}{4}\). The incorrect result is not a small rounding discrepancy: it evaluates a different interval.

Common pitfall: A substitution changes more than the symbols in the integrand. It changes the variable of integration and therefore changes what the endpoints mean. Never carry numerical bounds across a substitution without transforming and checking them.

Worked Example: A Decreasing Substitution

Worked Example: A Decreasing Substitution

Evaluate \(\int_0^1 (2-2x)^2\,dx\) using \(u=2-2x\).

Differentiate and solve for the differential:

$$ u=2-2x, \qquad du=-2\,dx, \qquad dx=-\frac{1}{2}\,du. $$

Transform the endpoints in their original order:

$$ x=0 \implies u=2, \qquad x=1 \implies u=0. $$

The substitution is decreasing, so the transformed lower bound is larger than the transformed upper bound. Keep the order \(2\) to \(0\), and include the negative factor from \(dx\):

$$ \int_0^1 (2-2x)^2\,dx = -\frac{1}{2}\int_2^0 u^2\,du = -\frac{1}{2}\left[\frac{u^3}{3}\right]_2^0 = -\frac{1}{2}\left(0-\frac{8}{3}\right) = \frac{4}{3}. $$

Check by returning to \(x\). An antiderivative is \(-\frac{(2-2x)^3}{6}\), since

$$ \frac{d}{dx}\left(-\frac{(2-2x)^3}{6}\right) = -\frac{1}{6}\cdot 3(2-2x)^2(-2) = (2-2x)^2. $$

Its value at the upper endpoint minus its value at the lower endpoint is

$$ -\frac{(2-2\cdot1)^3}{6} - \left(-\frac{(2-2\cdot0)^3}{6}\right) = 0-\left(-\frac{8}{6}\right) = \frac{4}{3}. $$

The reversed \(u\)-bounds and the negative factor in \(dx\) work together to give the positive value of the original integral.

Worked Example: An Exponential Integrand

Worked Example: An Exponential Integrand

Evaluate \(\int_1^2 2x e^{x^2}\,dx\) using \(u=x^2\).

Differentiate the substitution and transform the endpoints:

$$ u=x^2, \qquad du=2x\,dx, \qquad x=1 \implies u=1, \qquad x=2 \implies u=4. $$

Rewrite and evaluate entirely in \(u\):

$$ \int_1^2 2x e^{x^2}\,dx = \int_1^4 e^u\,du = \left[e^u\right]_1^4 = e^4-e. $$

For a check in the original variable, an antiderivative is \(e^{x^2}\), because

$$ \frac{d}{dx}e^{x^2} = e^{x^2}\cdot 2x = 2x e^{x^2}. $$

Evaluating at the original endpoints gives

$$ \left[e^{x^2}\right]_1^2 = e^{2^2}-e^{1^2} = e^4-e. $$

If the original bounds \(1\) and \(2\) had been retained on \(\int e^u\,du\), the result would have been \(e^2-e\), which does not match the endpoint values of the original antiderivative.

A Practical Check Before Evaluating

When a definite-integral solution contains a substitution, pause before applying the evaluation brackets. Check that each part of the integral is expressed in the same variable. In particular, identify which variable the limits describe. If the integrand and differential are in \(u\), the limits must be \(u\)-values. If the limits are still \(x\)-values, first rewrite the antiderivative in \(x\).

1
Record the original interval.
For \(\int_a^b f(x)\,dx\), the endpoints are \(x=a\) and \(x=b\).
2
Apply the substitution to both endpoints.
If \(u=g(x)\), calculate \(g(a)\) and \(g(b)\) separately, preserving their order.
3
Choose one evaluation route.
Use the transformed bounds with the \(u\)-integral, or return to \(x\) before evaluating with the original bounds.
4
Check the result.
When practical, differentiate the antiderivative in \(x\) and evaluate it at the original endpoints.

This check is especially useful when the substitution is decreasing, when its endpoint values are not visually obvious, or when a calculation contains several changes of variable. A quick endpoint calculation can reveal a bounds mismatch before it affects the final answer.

Check Your Understanding

For each question, focus on whether the variable and bounds describe the same interval.

  1. For \(\int_0^2 2x(x^2+3)^2\,dx\), use \(u=x^2+3\). What are the transformed bounds?
  2. In a transformed integral \(\int_1^3 h(u)\,du\), what would it mean to keep original \(x\)-bounds instead?
  3. If \(u=5-x\) and the original bounds are \(x=1\) to \(x=4\), what is the correct order of the transformed bounds?
  4. State the two consistent ways to finish a definite integral after substitution.
  5. Why does a decreasing substitution not justify swapping transformed bounds without also accounting for the sign or orientation?

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