Why Definite Integrals Need One More Step
A substitution can make a definite integral simpler by replacing a complicated expression with a new variable. For a definite integral, however, changing the variable also changes the endpoints. The lower and upper limits describe values of the integration variable, so they must be expressed in the new variable as well.
The guiding rule is simple: if \(u=g(x)\), then transform the integrand and differential using \(du=g'(x)\,dx\), and transform the limits using the same relation \(u=g(x)\). The resulting integral is entirely in \(u\); it should not retain limits that refer to \(x\).
When Is Substitution a Useful Signal?
Look for a composition in the integrand: an inner expression \(g(x)\) appears inside another function, and a factor proportional to \(g'(x)\) appears elsewhere. This is the definite-integral version of the inner-function fit considered in substitution. For instance, an expression involving \(x^2+1\) together with a factor \(2x\,dx\) suggests \(u=x^2+1\).
After choosing \(u=g(x)\), write down the changed differential and endpoints explicitly. If \(x=a\) is the lower endpoint, its new value is \(u=g(a)\); if \(x=b\) is the upper endpoint, its new value is \(u=g(b)\). The order is inherited from the original limits: it is not automatically reordered to make the lower limit numerically smaller.
Worked Example: A Trigonometric Composition
Worked Example: A Trigonometric Composition
Evaluate \(\displaystyle \int_0^1 2x\cos(x^2+1)\,dx\).
The inner expression is \(x^2+1\), whose derivative is \(2x\), matching the factor in the integrand. Set \(u=x^2+1\). Then \(du=2x\,dx\). Change the endpoints using this same substitution: at \(x=0\), \(u=0^2+1=1\); at \(x=1\), \(u=1^2+1=2\).
Check in the original variable: an antiderivative is \(\sin(x^2+1)\), since \(\frac{d}{dx}\sin(x^2+1)=\cos(x^2+1)\cdot 2x\). Evaluating it from \(0\) to \(1\) gives \(\sin 2-\sin 1\), as obtained by substitution.
Why the Formula Works
The substitution formula follows from the chain rule and the Fundamental Theorem of Calculus. The new limits are not an extra convention: they are what makes the transformed integral evaluate the same change in an antiderivative over the original interval.
Let \(F(t)=\int_{g(a)}^t f(u)\,du\). Since \(f\) is continuous, the Fundamental Theorem of Calculus gives \(F'(t)=f(t)\).
For \(H(x)=F(g(x))\), the chain rule gives \(H'(x)=F'(g(x))g'(x)=f(g(x))g'(x)\).
The Fundamental Theorem of Calculus gives \(\int_a^b f(g(x))g'(x)\,dx=H(b)-H(a)=F(g(b))-F(g(a))\). By the definition of \(F\), this difference is \(\int_{g(a)}^{g(b)}f(u)\,du\).
This argument also explains why the order of the new limits matters. The endpoint \(b\) contributes \(g(b)\), and the endpoint \(a\) contributes \(g(a)\). If \(g(b)<g(a)\), the transformed integral is oriented in the reverse direction and carries the corresponding sign.
Worked Example: Changing Limits in a Rational Integral
Worked Example: Changing Limits in a Rational Integral
Evaluate \(\displaystyle \int_0^2 \frac{x}{1+x^2}\,dx\).
The denominator contains \(1+x^2\), whose derivative is \(2x\). The numerator has \(x\), which is half of that derivative. Set \(u=1+x^2\). Then \(du=2x\,dx\), so \(x\,dx=\frac12\,du\). Transform the endpoints: when \(x=0\), \(u=1\); when \(x=2\), \(u=5\).
Check by differentiating an antiderivative in \(x\): \(\frac{d}{dx}\left(\frac12\ln(1+x^2)\right)=\frac12\cdot\frac{2x}{1+x^2}=\frac{x}{1+x^2}\). Its endpoint difference is \(\frac12\ln 5-\frac12\ln 1=\frac12\ln 5\), confirming the result.
A Common Mistake: Keeping the Old Limits
A frequent error is to change \(x\) to \(u\) in the integrand but leave the limits unchanged. In the rational example, \(u=1+x^2\) sends the original endpoints \(x=0\) and \(x=2\) to \(u=1\) and \(u=5\), not \(u=0\) and \(u=2\). Keeping \(0\) and \(2\) would make those limits refer to values of \(u\) that do not match the original interval.
Worked Example: Spotting the Bounds Error
For \(\displaystyle \int_0^2 \frac{x}{1+x^2}\,dx\), the substitution \(u=1+x^2\) gives \(x\,dx=\frac12\,du\). A wrong calculation might be written as
This equality is invalid because the limits were not converted: the correct \(u\)-limits are \(1\) and \(5\). In addition, the displayed integral on the right diverges at \(u=0\). It is an improper integral, and \(\int_0^2 \frac{1}{u}\,du\) does not have a finite value. It therefore cannot be evaluated as \(\ln 2\).
The correction is to change the endpoints together with the variable:
Checking the endpoints before evaluating helps prevent both the bounds mismatch and the resulting improper-integral error.
Reversed Limits and Orientation
Do not reverse the transformed limits merely because they appear in descending numerical order. A definite integral with upper limit less than lower limit is defined with a sign: \(\int_c^d f(u)\,du=-\int_d^c f(u)\,du\). This preserves the orientation of the original interval under substitution.
Worked Example: A Substitution with Reversed Limits
Evaluate \(\displaystyle \int_2^0 3x^2e^{x^3}\,dx\).
The inner expression \(x^3\) has derivative \(3x^2\). Set \(u=x^3\), so \(du=3x^2\,dx\). The original lower endpoint \(x=2\) becomes \(u=8\), and the upper endpoint \(x=0\) becomes \(u=0\). Keep that order:
The result is negative, consistent with integrating the positive function \(3x^2e^{x^3}\) from the larger \(x\)-value \(2\) down to \(0\). Check directly: \(\frac{d}{dx}e^{x^3}=3x^2e^{x^3}\), and evaluating \(e^{x^3}\) from \(2\) to \(0\) gives \(e^0-e^8=1-e^8\).
A Practical Check Before Evaluating
Before carrying out the final antiderivative calculation, pause to make sure the transformed integral is coherent. The variable in the integrand, differential, and limits should all be the new variable. A useful quick audit is:
- Does the chosen inner expression match a derivative factor in the integrand?
- Has the differential been rewritten correctly, including any constant factor?
- Were both endpoints converted using the substitution?
- Were the converted limits kept in their original order?
- Is the transformed integrand defined throughout the interval of integration?
This last check matters because a substitution does not make a divergent integral finite. It must also be possible to apply the integral and substitution rules on the interval in question, or any endpoint singularity must be treated as an improper integral under the appropriate conditions.
Check Your Understanding
For each question, focus on how the variable and limits change together.
- For \(\int_0^1 4x(x^2+3)^2\,dx\), choose a substitution and state the transformed integrand and limits.
- For \(u=2x-1\) on the interval \(x=1\) to \(x=3\), what are the new lower and upper limits?
- Why is it incorrect to keep the original limits after replacing \(x\) with \(u=1+x^2\) in an integral from \(x=0\) to \(x=2\)?
- If a substitution changes the limits from \(u=5\) to \(u=1\), should they be reordered before evaluating? Explain.
- What theorem and differentiation rule together justify the definite-integral substitution formula?
