Conditions before computation
If u=g(x), carry the factor g′(x)dx into du. Constant factors may be adjusted, but missing variable factors cannot simply be invented. Either transform both bounds and evaluate entirely in u, or return to an antiderivative in x before using the original bounds. A decreasing g reverses the ordering of transformed endpoints and can introduce a negative differential factor. Logarithmic antiderivatives require absolute values generally; the particular positive intervals here allow ordinary logarithms.
Worked example with a method check
Evaluate ∫ from 0 to 1 of 2x(x²+3)² dx using substitution. Let u=x²+3; then du=2x dx and the bounds become 3 and 4. The integral becomes ∫u² du over those transformed bounds. The value is (4³−3³)/3=37/3.
Compare two methods
Compare ∫2x(x²+3)²dx with ∫(x²+3)²dx. The first has the differential factor for u=x²+3 and becomes ∫u²du. The second lacks 2x; pretending it has the same substitution would change the problem. It can instead be expanded as a polynomial. Choosing a recognizable inner expression is not sufficient without accounting for the entire integrand.
Transfer: solve before reading
Evaluate ∫₀¹ 2x(x²+2)²dx and check the result by expanding the polynomial. Then explain why using bounds 0 and 1 on ∫u²du would fail.
Transfer reasoning and answer
The u-bounds are 2 and 3, so the answer is (27−8)/3=19/3. Expansion gives 2x⁵+8x³+8x, whose integral on [0,1] is 1/3+2+4=19/3. Bounds 0 and 1 describe x, while the substituted u runs from 2 to 3.
Practice the next representation
Work all four task families linked to this lesson. For the illustrated items, state which numerical labels and relationships you used. For a written extension, compare each line of your solution with its original self-assessment criteria; these criteria are practice feedback, not an official AP score.
Evaluate ∫ from 0 to 1 of 2x(x²+3)² dx using substitution.
- Let u=x²+3; then du=2x dx and the bounds become 3 and 4.
- The integral becomes ∫u² du over those transformed bounds.
- The value is (4³−3³)/3=37/3.
Compare the other answer choices
Omit the division by 3 in the antiderivative of u².
Lower the integrand power before integrating.
Forget the transformed lower bound.
Explain your thinking.
Optional written extension. Self-assessed only; it does not change your automatic practice result. These are teaching criteria, not official awarded marks.
(a) State the relationship or theorem that makes this calculation valid, including its relevant conditions. (b) Show the intermediate computation and give the requested result. (c) Explain why one plausible competing method would fail for this problem.
This draft stays on this page only. Use paper if you want to keep your working.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
AP Calculus AB and BC Course and Exam DescriptionCalculus Volume 1: Fundamental Theorem of Calculus