Definite Substitution and Changing Bounds

Overview

Central question: Why must the interval change when the variable changes? You will use substitution in a definite integral by following the two endpoint values through the same function that changes the integrand. A definite integral is not merely an antiderivative with numbers attached. Its bounds identify the input interval over which the accumulation occurs. Once the variable changes, those endpoint descriptions must change as well.

Start with

$$\int_1^3 2x\left(x^2+1\right)^2\,dx.$$

The inner expression is $x^2+1$, and $du=2x\,dx$. Before integrating, predict where the endpoints move. When $x=1$, the new value is $u=1^2+1=2$. When $x=3$, the new value is $u=3^2+1=10$. The transformed integral is therefore $\int_2^{10}u^2\,du$.

Carry the interval through the substitution

The substitution changes every occurrence of the old variable. The expression $\left(x^2+1\right)^2$ becomes $u^2$, the differential $2x\,dx$ becomes $du$, and the two endpoint values become $2$ and $10$. These changes form one coherent statement:

$$\int_1^3 2x\left(x^2+1\right)^2\,dx =\int_2^{10}u^2\,du.$$

Now evaluate entirely in $u$:

$$\int_2^{10}u^2\,du =\left[\frac{u^3}{3}\right]_2^{10} =\frac{1000-8}{3} =\frac{992}{3}.$$

There is no constant $C$ because the endpoints select one numerical accumulation rather than a family of antiderivatives. The original integrand is positive over $1\le x\le3$, so a positive result is expected. This sign check does not find the exact number, but it does reject a conclusion that contradicts the graph and the integrand.

A second valid workflow

You may instead substitute, integrate in $u$, and then return to $x$ before using the original bounds. Begin with the indefinite result:

$$\int u^2\,du=\frac{u^3}{3}.$$

Replace $u$ by $x^2+1$ before evaluating:

$$\int_1^3 2x\left(x^2+1\right)^2\,dx =\left[\frac{\left(x^2+1\right)^3}{3}\right]_1^3.$$

This returns the same value, $\frac{992}{3}$. The logic is strict but simple: either use $u$ throughout, with $u$-bounds, or return completely to $x$ before using $x$-bounds. Never combine $u^2\,du$ with the old bounds $1$ and $3$. That expression mixes two coordinate systems and does not describe the original accumulation.

Direction is preserved by the differential

An endpoint map need not run left to right. Let $u=5-x$ in

$$\int_1^4\sqrt{5-x}\,dx.$$

Then $du=-dx$. The lower endpoint $x=1$ maps to $u=4$, and the upper endpoint $x=4$ maps to $u=1$. Keep both the reversed bounds and the negative differential:

$$\int_1^4\sqrt{5-x}\,dx =-\int_4^1\sqrt{u}\,du =\int_1^4\sqrt{u}\,du.$$

The final integral is positive because the original integrand is positive. The sign did not disappear by convention; reversing the bounds and carrying $du=-dx$ preserve the orientation of the substitution. This is a transfer case: it shows why endpoints must be mapped rather than rearranged by habit.

Stop and check

A plausible incorrect setup is $\int_1^3u^2\,du$. The underlying misconception is that bounds are decorative labels rather than values of the integration variable. To repair it, apply the substitution to each old endpoint before writing the transformed integral. Do not guess a new bound from its position or from a pattern in the numbers.

Interpret the endpoint values

It is tempting to describe a definite substitution as “changing the numbers on the integral.” That language misses the mathematical meaning. A bound answers the question “which input value starts or ends this accumulation?” When the variable changes from $x$ to $u=g(x)$, the old endpoint values must be described through the new input system. The interval $1\le x\le3$ has not become a different physical interval; it has acquired the equivalent description $2\le u\le10$ for this particular transformation.

This viewpoint helps with non-numerical endpoints as well. If $u=\sqrt{x+1}$ and an integral runs from $x=a$ to $x=b$, the transformed limits are $\sqrt{a+1}$ and $\sqrt{b+1}$, provided the square root is defined on the interval. Each endpoint must satisfy the same domain conditions as every interior point. The transformed integral cannot be more broadly defined than the original problem.

There is a further practical check when an application includes units. If $x$ is a distance measured in meters and $u=x^2$, then $u$ has units of square meters. A lower or upper bound must carry the units belonging to its variable. This is another reason mixed bounds are invalid: $u$ and $x$ can represent quantities with different dimensions, even when their numerical values happen to match.

For a definite integral, the final number should answer the original question in original units. Substitution is an internal change of representation; it does not change what has been accumulated. State the result in the original context after the symbolic work is complete.

Retrieval check

Evaluate $\int_0^2 3x^2e^{x^3}\,dx$ using changed bounds. Choose $u=x^3$, so $du=3x^2\,dx$, $x=0$ maps to $u=0$, and $x=2$ maps to $u=8$. Thus

$$\int_0^2 3x^2e^{x^3}\,dx =\int_0^8e^u\,du =e^8-1.$$

Explain why every symbol in the middle integral is written in $u$. The next lesson broadens the question from how to execute one substitution to how visible structure can justify a first integration move.