Returning To The Original Variable Before Evaluation

Introduction

A substitution changes the variable used to describe an integral. For a definite integral, the limits must be handled consistently with that change. One valid route is to change the limits into the new variable and evaluate there. Another is to find an antiderivative in the new variable, substitute back to the original variable, and then use the original limits.

This tutorial develops the second route. Its central rule is simple: if the limits are still values of $x$, the expression being evaluated must also be written in terms of $x$. Returning to the original variable makes the connection between the antiderivative and the original limits explicit.

The Method and Its Conditions

Suppose the integrand has the form $f(g(x))g'(x)$, and use the substitution $u=g(x)$. Then $du=g'(x)\,dx$, so the integral in the new variable has integrand $f(u)$. If $F$ is an antiderivative of $f$, then $F(u)$ is an antiderivative in the $u$-variable. To use the original $x$-limits, replace $u$ by $g(x)$ before evaluating.

Method: For an integral $\int_a^b f(g(x))g'(x)\,dx$, let $u=g(x)$, find an antiderivative $F(u)$ of $f(u)$, and return to $x$ by writing $F(g(x))$. Then evaluate $F(g(b))-F(g(a))$. This method applies when the functions involved are defined and continuous on the interval and the chain rule and Fundamental Theorem of Calculus apply there.

The limits $a$ and $b$ are original $x$-values. They are not values of $u$. Thus, the expression $F(u)$ cannot be evaluated at $x=a$ and $x=b$ until $u$ has been replaced by $g(x)$. The returned expression $F(g(x))$ is in the same variable as the original limits.

Worked Examples

Worked Example: A Polynomial Inner Function

Evaluate $\int_0^2 2x(x^2+1)^3\,dx$ by returning to $x$ before applying the limits.

Set $u=x^2+1$. Then $du=2x\,dx$, so the integrand and differential become $u^3\,du$. An antiderivative in $u$ is

\int u^3\,du=\frac{u^4}{4}.

Return to $x$ by substituting $u=x^2+1$:

\frac{u^4}{4}=\frac{(x^2+1)^4}{4}.

Now apply the original limits $x=0$ and $x=2$:

\int_0^2 2x(x^2+1)^3\,dx = \left.\frac{(x^2+1)^4}{4}\right|_0^2 = \frac{(2^2+1)^4}{4}-\frac{(0^2+1)^4}{4} = \frac{5^4-1^4}{4} = \frac{625-1}{4} = 156.

Check the returned antiderivative by differentiating:

\frac{d}{dx}\left(\frac{(x^2+1)^4}{4}\right) = \frac14\cdot4(x^2+1)^3\cdot2x = 2x(x^2+1)^3.

This matches the integrand. Substituting the endpoints into the returned antiderivative gives $156$, as required.

Worked Example: An Exponential with a Cubic Inner Function

Evaluate $\int_0^1 3x^2e^{x^3+2}\,dx$.

Let $u=x^3+2$. Then $du=3x^2\,dx$, and the integral in $u$ has integrand $e^u$. Since an antiderivative of $e^u$ is $e^u$, return to the original variable:

e^u=e^{x^3+2}.

Use the original limits $x=0$ and $x=1$:

\int_0^1 3x^2e^{x^3+2}\,dx = \left.e^{x^3+2}\right|_0^1 = e^{1^3+2}-e^{0^3+2} = e^3-e^2.

Differentiate the returned expression to check it:

\frac{d}{dx}\left(e^{x^3+2}\right) = e^{x^3+2}\cdot3x^2 = 3x^2e^{x^3+2}.

The derivative agrees with the original integrand, and the endpoint evaluation gives $e^3-e^2$.

Worked Example: A Trigonometric Inner Function

Evaluate $\int_0^{\pi/3}\cos x\sqrt{1+\sin x}\,dx$.

Set $u=1+\sin x$, so $du=\cos x\,dx$. The integral in $u$ becomes $\int u^{1/2}\,du$, whose antiderivative is $\frac23u^{3/2}$. Substitute back:

\frac23u^{3/2}=\frac23(1+\sin x)^{3/2}.

Evaluate using the original $x$-limits:

\begin{aligned} \int_0^{\pi/3}\cos x\sqrt{1+\sin x}\,dx &= \left.\frac23(1+\sin x)^{3/2}\right|_0^{\pi/3}\\ &= \frac23\left(1+\sin\frac{\pi}{3}\right)^{3/2} -\frac23(1+\sin0)^{3/2}\\ &= \frac23\left(1+\frac{\sqrt3}{2}\right)^{3/2}-\frac23. \end{aligned}

Check the antiderivative by the chain rule:

\frac{d}{dx}\left(\frac23(1+\sin x)^{3/2}\right) = \frac23\cdot\frac32(1+\sin x)^{1/2}\cos x = \cos x\sqrt{1+\sin x}.

The derivative is the original integrand. The evaluated result is therefore $\frac23\left(1+\frac{\sqrt3}{2}\right)^{3/2}-\frac23$.

Why Returning to $x$ Works

The method is justified by the chain rule. If $F'(u)=f(u)$ and $u=g(x)$, then differentiating the composite expression $F(g(x))$ gives

\frac{d}{dx}F(g(x))=F'(g(x))g'(x)=f(g(x))g'(x).

Thus $F(g(x))$ is an antiderivative, in the original variable, of the original integrand. The Fundamental Theorem of Calculus then says that the integral from $a$ to $b$ is the value of this antiderivative at $b$ minus its value at $a$.

Justification: If $F'=f$, $g$ is differentiable, and $f(g(x))g'(x)$ is continuous on $[a,b]$, then $$ \int_a^b f(g(x))g'(x)\,dx = F(g(b))-F(g(a)). $$ Indeed, the chain rule shows that $\frac{d}{dx}F(g(x))=f(g(x))g'(x)$. Applying the Fundamental Theorem of Calculus to this antiderivative gives the stated equality.

This explains why returning to $x$ is not an extra computational trick. It constructs the antiderivative to which the original $x$-limits apply. If instead the integral is kept entirely in $u$, its limits must be expressed as $u$-values; the two approaches agree when each variable is paired with its own limits.

A Common Variable-Mismatch Error

A frequent mistake is to find an antiderivative in $u$ but then substitute the original $x$-limits into that expression as though they were $u$-values. For the first example, the antiderivative in $u$ was $\frac{u^4}{4}$, while the original limits were $x=0$ and $x=2$. Incorrectly inserting $0$ and $2$ directly gives

\frac{2^4}{4}-\frac{0^4}{4}=4.

This is wrong because $0$ and $2$ are $x$-values, not the corresponding $u$-values. The substitution is $u=x^2+1$, so returning to $x$ gives $\frac{(x^2+1)^4}{4}$. Using the original bounds with that expression gives

\frac{(2^2+1)^4}{4}-\frac{(0^2+1)^4}{4} = \frac{625-1}{4} = 156.

The correction is to check the variable before evaluating: an expression in $u$ requires $u$-limits; an expression in $x$ requires $x$-limits. Do not combine an antiderivative in one variable with endpoint values from another.

A Reliable Evaluation Sequence

When choosing the return-to-$x$ route, keep the calculation organized in this order:

Choose the substitution.
Write $u=g(x)$ and compute $du=g'(x)\,dx$.

Find the antiderivative in $u$.
Integrate the transformed integrand and obtain $F(u)$.

Return to the original variable.
Replace each $u$ in the antiderivative by $g(x)$, giving $F(g(x))$.

Apply the original bounds.
Evaluate $F(g(b))-F(g(a))$ for the original interval $[a,b]$.

Check the result.
Differentiate $F(g(x))$ to confirm it reproduces the original integrand, then verify the endpoint subtraction.

Why This Route Is Useful

Returning to the original variable can make the role of the original interval easier to see. It also provides a direct check: differentiating the returned expression must recover the integrand in $x$. This is especially helpful when the transformed antiderivative is simple but the original limits are already convenient to use.

The method depends on matching variables correctly, not on whether one route is always better than another. For any definite integral evaluated by substitution, keep track of which variable appears in the antiderivative and which variable the limits describe. Once these agree, the endpoint calculation follows directly from the Fundamental Theorem of Calculus.

Check Your Understanding

For each question, keep the variable in the antiderivative matched to the variable used for evaluation.

For $\int_0^1 2x(x^2+3)^2\,dx$, what antiderivative do you obtain after returning to $x$, and what value results from the original bounds?
If $u=g(x)$ and $F'(u)=f(u)$, what is the derivative of $F(g(x))$?
Why is it invalid to insert original $x$-limits directly into an antiderivative written only in $u$?
What condition must hold between the returned antiderivative and the original integrand for the endpoint evaluation to be valid?
For $\int_0^{\pi/2}\sin x\,e^{\cos x}\,dx$, name a substitution and describe how you would return to $x$ before using the original limits.

Returning To The Original Variable Before Evaluation

What You'll Learn

Introduction

The Method and Its Conditions

Worked Examples

Worked Example: A Polynomial Inner Function

Worked Example: An Exponential with a Cubic Inner Function

Worked Example: A Trigonometric Inner Function

Why Returning to \(x\) Works

A Common Variable-Mismatch Error

A Reliable Evaluation Sequence

Why This Route Is Useful

Check Your Understanding

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Returning To The Original Variable Before Evaluation

What You'll Learn

Introduction

The Method and Its Conditions

Worked Examples

Worked Example: A Polynomial Inner Function

Worked Example: An Exponential with a Cubic Inner Function

Worked Example: A Trigonometric Inner Function

Why Returning to \(x\) Works

A Common Variable-Mismatch Error

A Reliable Evaluation Sequence

Why This Route Is Useful

Check Your Understanding

Test yourself: Calculus I 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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