JC2 H2 Maths Bukit Timah | Substitution Limits

Did you know? When you substitute u=x²+1 into a definite integral from x=0 to x=2, the new limits are not 0 and 2. They are u=1 and u=5. A student can make the correct substitution and integrate accurately, yet lose the final answer by evaluating a function of u at the old x-limits. The quickest repair is a two-column limit table, written before the integration begins.

This JC2 H2 Mathematics Tutor Bukit Timah guide answers one precise A-Level 9758 question: when a definite integral is solved by substitution, how do the bounds, differential and any minus signs change? The method is taught through complete worked examples, a trigonometric reversal, a symmetry check and 40 original substitution clinics. The first rule is given immediately so readers can find the calculation they need without working through the whole chapter.

The official 2027 SEAB H2 Mathematics 9758 syllabus includes integration by a given substitution, definite-integral evaluation and interpreting a definite integral. It excludes reduction formulae. This guide stays with stated simple substitutions and exact bounds. It also distinguishes signed integrals from geometric area, so learners do not automatically interpret every negative definite integral as a physically negative area.

Quick answer: the variable and the bounds must change together

For a differentiable substitution u=g(x), with du=g′(x)dx, an integral of the form ∫ from x=a to x=b f(g(x))g′(x)dx becomes ∫ from u=g(a) to u=g(b) f(u)du. Replace both bounds with values of u. If you keep the original x-limits instead, you must first find an antiderivative and convert it back to an expression in x before applying those original limits. Either approach works; mixing them does not.

Identify u: use the substitution supplied by the problem, or one clearly matching the integrand.
Calculate du: include the coefficient and sign of g′(x).
Convert the lower bound: u(a)=g(a).
Convert the upper bound: u(b)=g(b).
Integrate in one variable: once in u, write only u-bounds.
Check orientation: if the substitution reverses the bounds, the resulting negative differential or reversed interval must be handled consistently.
Audit independently: differentiate your substituted antiderivative or compare with a numerical calculator value.

Worked example one: a polynomial substitution

Evaluate ∫₀² 2x(x²+1)³ dx. Use u=x²+1, so du=2x dx. At x=0, u=1; at x=2, u=5. Therefore the integral becomes ∫₁⁵ u³ du=[u⁴/4]₁⁵=(625−1)/4=156. The evaluation is exact, with no need to substitute back into x because the bounds have already changed.

The wrong mixed-variable calculation would evaluate [u⁴/4] between the old numbers 0 and 2, giving 4. This value does not represent the original integral. A second route confirms 156: first find the antiderivative (x²+1)⁴/4, then apply the original x-limits from 0 to 2. The result is again (5⁴−1⁴)/4=156. Differentiate that antiderivative to verify the original integrand.

A two-column limit table prevents the common error

Old x bound	Calculation under u=x²+1	New u bound
x=0	u=0²+1	1
x=2	u=2²+1	5
Orientation	u rises from 1 to 5	∫₁⁵

Write the table before entering any numbers into an antiderivative. If u involves a logarithm, exponential, trigonometric function or negative linear expression, the converted limits may not look like the old ones. That is expected. The upper number on the page is not “2” merely because the old integral once ended at x=2.

Worked example two: the logarithm emerges automatically

Evaluate ∫₀¹ 3x²/(1+x³) dx. Set u=1+x³, so du=3x² dx. The original lower bound x=0 gives u=1; the upper x=1 gives u=2. The integral becomes ∫₁² 1/u du=[ln u]₁²=ln 2. Because u is positive throughout, no absolute-value issue arises in this specific evaluation.

If you choose to integrate back in x, the antiderivative is ln(1+x³), which produces ln 2−ln 1=ln 2 at the original bounds. A student who writes ln u and immediately inserts x=0 and x=1 without converting the integrand back into x has skipped the logical identification of the variable. The result happens to look plausible, which is why the method audit matters.

Worked example three: a substitution reverses the limits

Consider ∫₀^(π/2) sin x cos x dx. Put u=cos x, so du=−sin x dx. The old lower bound x=0 gives u=1, while the old upper bound x=π/2 gives u=0. Thus the integral becomes −∫₁⁰ u du. Reversing the bounds gives +∫₀¹ u du, so the answer is 1/2. If you neglect the negative sign from du, the sign of the answer becomes wrong.

Try the alternative u=sin x. Now du=cos x dx and the bounds convert from u=0 to u=1 without reversal, giving ∫₀¹ u du=1/2 directly. Both substitutions are valid. Their agreement is a powerful independent check and shows why a reversed range is not an error if the differential sign is handled correctly.

Why orientation is part of the calculation

In a definite integral, ∫ᵦᵃ f(u)du=−∫ₐᵦ f(u)du. When u=g(x) decreases as x increases, the transformed upper value may be smaller than the transformed lower value. This is not a signal to silently sort the two bounds into ascending order. Either preserve their original transformed order and the differential sign, or explicitly reverse the order with a minus sign.

A short test is u=1−x on x∈[0,1]. The transformed bounds run from 1 down to 0, and du=−dx. Hence ∫₀¹ 1 dx can be written −∫₁⁰ 1 du=1. Writing ∫₀¹ 1du without showing how the minus signs cancel may reach the same number by accident, but such shortcuts become unreliable for more complicated integrands.

Even if u is not one-to-one, the derivative rule may still work

Evaluate ∫₋₁¹ 2x/(1+x²) dx using u=1+x², du=2x dx. Both old endpoints map to u=2, so the formal transformed integral is ∫₂² 1/u du=0. This is correct: the original integrand is odd in x, and its signed integral over the symmetric interval also equals zero. The substitution need not be globally one-to-one for this derivative-matched definite-integral identity to hold.

However, if an application asks for the geometric area under a signed curve, a zero definite integral does not necessarily mean zero area. Positive and negative contributions may cancel. Always distinguish the signed integral from the total geometric area, as explained in the separate Area Between Crossing Curves tutorial.

The given substitution might need a constant factor

Suppose the integrand is x(1+x²)² rather than 2x(1+x²)². Under u=1+x², du=2x dx, so x dx=du/2. On x from 0 to 1, the new bounds are 1 and 2, and the integral is (1/2)∫₁² u² du=(1/2)[u³/3]₁²=7/6. The half factor must be retained outside the integral.

Students sometimes focus so heavily on rewriting the bounds that they lose the scaling between dx and du. The complete substitution has three parts: changed variable, changed differential and changed limits. Omitting any one may produce a different integral. A 15-second line showing du=2x dx and x dx=du/2 prevents the error.

A trigonometric substitution with the other direction

For ∫₀^(π/2) sin³x cos x dx, set u=sin x, so du=cos x dx and the new bounds are 0 and 1. The resulting integral is ∫₀¹ u³ du=1/4. Using u=cos x would also work but require rewriting sin²x=1−cos²x and handling reversed bounds. Choose the substitution that leaves the fewest unnecessary transformations while respecting any substitution specified by the question.

The syllabus includes trigonometric integration, but this example’s focus is on the substitution limits. A different article in this batch owns the identities for integrating sin²x and cos²x. Keeping the two intents separate helps learners find the exact method they need instead of blending every integration technique into one ambiguous checklist.

A numerical or derivative check can catch an impossible value

Consider the original worked integral ∫₀² 2x(x²+1)³ dx. The integrand is nonnegative on [0,2], so its integral must be nonnegative. It also grows substantially near the upper end, which makes the wrong answer 4 suspicious. The exact antiderivative (x²+1)⁴/4 differentiates to 2x(x²+1)³, confirming the calculation.

A graphing calculator can approximate the original definite integral as a numerical check. It cannot prove that the substitution was stated consistently, and in questions where method marks are expected you should still show the transformed bounds. When the numerical and algebraic answers disagree, check the variable labels and differential factors before re-entering the same expression into the calculator.

One-minute Bukit Timah diagnostic

Give u=x²+1 for ∫₀² 2x(x²+1)³ dx. Ask for the new limits without integrating: 1 and 5. Then ask for the new integrand: u³. Next switch to u=cos x for ∫₀^(π/2) sin x cos x dx. The bounds become 1 to 0 and the differential adds a minus sign. Finally change the integrand to x(1+x²)² and ask for the missing factor: one-half. These checks diagnose three independent failure points in one minute.

Thirty-two original substitution clinics

The next 32 exercises are grouped by substitution family: polynomial powers, logarithmic reciprocals, exponentials and trigonometric powers. Each has a verified transformed bound and exact answer. Work a few from each family before mixing them. The numbers are original mathematical constructions, not official past-year examination questions or actual pupil outcomes.

1. Polynomial power with changed limits

Question. Evaluate ∫₀^1 2x(x²+1)^1 dx using the substitution u=x²+1. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=1 and upper bound u=2. The transformed definite integral is ∫_1^2 u^1 du. Evaluate it as [u^2/2]_1^2=1.5. The power rule gives (2^2−1^2)/2. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. The transformed bounds are values of u, not copies of the original x limits. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

2. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^1 [2x/(x²+1)] dx using the substitution u=x²+1. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=1 and upper bound u=2. The transformed definite integral is ∫_1^2 (1/u)du. Evaluate it as [ln u]_1^2=ln(2/1). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. Checking the antiderivative by differentiation is independent of the transformed-bound calculation. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

3. Exponential chain with changed limits

Question. Evaluate ∫₀^1 2x exp(x²+0) dx using the substitution u=x²+0. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=0 and upper bound u=1. The transformed definite integral is ∫_0^1 e^u du. Evaluate it as [e^u]_0^1=e^1−e^0. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. The du factor is part of the substitution; a missing constant produces a wrong exact answer. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

4. Sine-power substitution with changed limits

Question. Evaluate ∫₀^(π/2) (sin x)^1 cos x dx using the substitution u=sin x. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=cos x dx, with the lower bound becoming u=0 and upper bound u=1. The transformed definite integral is ∫₀¹ u^1 du. Evaluate it as [u^2/2]₀¹=1/2. The bounds are sin 0=0 and sin(π/2)=1; the transformed result has no π factor. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. The old upper x bound may map to a number much larger than x itself, which is perfectly valid. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

5. Polynomial power with changed limits

Question. Evaluate ∫₀^2 2x(x²+2)^2 dx using the substitution u=x²+2. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=2 and upper bound u=6. The transformed definite integral is ∫_2^6 u^2 du. Evaluate it as [u^3/3]_2^6=69.333333. The power rule gives (6^3−2^3)/3. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. If the original integrand is nonnegative on the interval, the definite result should not be negative. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

6. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^2 [2x/(x²+2)] dx using the substitution u=x²+2. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=2 and upper bound u=6. The transformed definite integral is ∫_2^6 (1/u)du. Evaluate it as [ln u]_2^6=ln(6/2). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. Logarithmic integrals require positive arguments or absolute values according to the actual domain. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

7. Exponential chain with changed limits

Question. Evaluate ∫₀^2 2x exp(x²+1) dx using the substitution u=x²+1. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=1 and upper bound u=5. The transformed definite integral is ∫_1^5 e^u du. Evaluate it as [e^u]_1^5=e^5−e^1. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. For a trigonometric substitution, the endpoint values may be zero or one even when the old bound contains π. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

8. Sine-power substitution with changed limits

Question. Evaluate ∫₀^(π/2) (sin x)^2 cos x dx using the substitution u=sin x. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=cos x dx, with the lower bound becoming u=0 and upper bound u=1. The transformed definite integral is ∫₀¹ u^2 du. Evaluate it as [u^3/3]₀¹=1/3. The bounds are sin 0=0 and sin(π/2)=1; the transformed result has no π factor. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

Independent check and transfer. A change of variable rewrites the entire integral, not merely its algebraic integrand. Try redoing this one with an antiderivative expressed in the original variable x, then apply the old x-limits instead; the answer must match. Next change the upper limit of the original integral and predict how the new u-bound should change before calculating. That direct endpoint audit is a stronger skill than copying the substituted expression from a worked example.

9. Polynomial power with changed limits

Question. Evaluate ∫₀^3 2x(x²+3)^3 dx using the substitution u=x²+3. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=3 and upper bound u=12. The transformed definite integral is ∫_3^12 u^3 du. Evaluate it as [u^4/4]_3^12=5163.75. The power rule gives (12^4−3^4)/4. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

10. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^3 [2x/(x²+3)] dx using the substitution u=x²+3. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=3 and upper bound u=12. The transformed definite integral is ∫_3^12 (1/u)du. Evaluate it as [ln u]_3^12=ln(12/3). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

11. Exponential chain with changed limits

Question. Evaluate ∫₀^3 2x exp(x²+2) dx using the substitution u=x²+2. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=2 and upper bound u=11. The transformed definite integral is ∫_2^11 e^u du. Evaluate it as [e^u]_2^11=e^11−e^2. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

12. Sine-power substitution with changed limits

Question. Evaluate ∫₀^(π/2) (sin x)^3 cos x dx using the substitution u=sin x. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=cos x dx, with the lower bound becoming u=0 and upper bound u=1. The transformed definite integral is ∫₀¹ u^3 du. Evaluate it as [u^4/4]₀¹=1/4. The bounds are sin 0=0 and sin(π/2)=1; the transformed result has no π factor. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

13. Polynomial power with changed limits

Question. Evaluate ∫₀^1 2x(x²+4)^4 dx using the substitution u=x²+4. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=4 and upper bound u=5. The transformed definite integral is ∫_4^5 u^4 du. Evaluate it as [u^5/5]_4^5=420.2. The power rule gives (5^5−4^5)/5. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

14. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^1 [2x/(x²+4)] dx using the substitution u=x²+4. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=4 and upper bound u=5. The transformed definite integral is ∫_4^5 (1/u)du. Evaluate it as [ln u]_4^5=ln(5/4). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

15. Exponential chain with changed limits

Question. Evaluate ∫₀^1 2x exp(x²+3) dx using the substitution u=x²+3. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=3 and upper bound u=4. The transformed definite integral is ∫_3^4 e^u du. Evaluate it as [e^u]_3^4=e^4−e^3. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

16. Sine-power substitution with changed limits

Question. Evaluate ∫₀^(π/2) (sin x)^4 cos x dx using the substitution u=sin x. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=cos x dx, with the lower bound becoming u=0 and upper bound u=1. The transformed definite integral is ∫₀¹ u^4 du. Evaluate it as [u^5/5]₀¹=1/5. The bounds are sin 0=0 and sin(π/2)=1; the transformed result has no π factor. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

17. Polynomial power with changed limits

Question. Evaluate ∫₀^2 2x(x²+1)^1 dx using the substitution u=x²+1. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=1 and upper bound u=5. The transformed definite integral is ∫_1^5 u^1 du. Evaluate it as [u^2/2]_1^5=12. The power rule gives (5^2−1^2)/2. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

18. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^2 [2x/(x²+5)] dx using the substitution u=x²+5. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=5 and upper bound u=9. The transformed definite integral is ∫_5^9 (1/u)du. Evaluate it as [ln u]_5^9=ln(9/5). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

19. Exponential chain with changed limits

Question. Evaluate ∫₀^2 2x exp(x²+0) dx using the substitution u=x²+0. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=0 and upper bound u=4. The transformed definite integral is ∫_0^4 e^u du. Evaluate it as [e^u]_0^4=e^4−e^0. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

20. Sine-power substitution with changed limits

Question. Evaluate ∫₀^(π/2) (sin x)^5 cos x dx using the substitution u=sin x. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=cos x dx, with the lower bound becoming u=0 and upper bound u=1. The transformed definite integral is ∫₀¹ u^5 du. Evaluate it as [u^6/6]₀¹=1/6. The bounds are sin 0=0 and sin(π/2)=1; the transformed result has no π factor. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

21. Polynomial power with changed limits

Question. Evaluate ∫₀^3 2x(x²+2)^2 dx using the substitution u=x²+2. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=2 and upper bound u=11. The transformed definite integral is ∫_2^11 u^2 du. Evaluate it as [u^3/3]_2^11=441. The power rule gives (11^3−2^3)/3. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

22. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^3 [2x/(x²+1)] dx using the substitution u=x²+1. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=1 and upper bound u=10. The transformed definite integral is ∫_1^10 (1/u)du. Evaluate it as [ln u]_1^10=ln(10/1). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

23. Exponential chain with changed limits

Question. Evaluate ∫₀^3 2x exp(x²+1) dx using the substitution u=x²+1. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=1 and upper bound u=10. The transformed definite integral is ∫_1^10 e^u du. Evaluate it as [e^u]_1^10=e^10−e^1. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

24. Sine-power substitution with changed limits

Question. Evaluate ∫₀^(π/2) (sin x)^6 cos x dx using the substitution u=sin x. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=cos x dx, with the lower bound becoming u=0 and upper bound u=1. The transformed definite integral is ∫₀¹ u^6 du. Evaluate it as [u^7/7]₀¹=1/7. The bounds are sin 0=0 and sin(π/2)=1; the transformed result has no π factor. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

25. Polynomial power with changed limits

Question. Evaluate ∫₀^1 2x(x²+3)^3 dx using the substitution u=x²+3. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=3 and upper bound u=4. The transformed definite integral is ∫_3^4 u^3 du. Evaluate it as [u^4/4]_3^4=43.75. The power rule gives (4^4−3^4)/4. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

26. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^1 [2x/(x²+2)] dx using the substitution u=x²+2. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=2 and upper bound u=3. The transformed definite integral is ∫_2^3 (1/u)du. Evaluate it as [ln u]_2^3=ln(3/2). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

27. Exponential chain with changed limits

Question. Evaluate ∫₀^1 2x exp(x²+2) dx using the substitution u=x²+2. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=2 and upper bound u=3. The transformed definite integral is ∫_2^3 e^u du. Evaluate it as [e^u]_2^3=e^3−e^2. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

28. Sine-power substitution with changed limits

29. Polynomial power with changed limits

Question. Evaluate ∫₀^2 2x(x²+4)^4 dx using the substitution u=x²+4. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=4 and upper bound u=8. The transformed definite integral is ∫_4^8 u^4 du. Evaluate it as [u^5/5]_4^8=6348.8. The power rule gives (8^5−4^5)/5. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

30. Logarithmic reciprocal with changed limits

Question. Evaluate ∫₀^2 [2x/(x²+3)] dx using the substitution u=x²+3. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=3 and upper bound u=7. The transformed definite integral is ∫_3^7 (1/u)du. Evaluate it as [ln u]_3^7=ln(7/3). The denominator u stays positive on the integration range, so the natural logarithm is well-defined here. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

31. Exponential chain with changed limits

Question. Evaluate ∫₀^2 2x exp(x²+3) dx using the substitution u=x²+3. Before integrating, write its differential and convert the old lower and upper bounds separately. The two new numbers must describe the same endpoints in the new variable; the name of the dummy integration variable is not interchangeable with its original limits.

Complete substitution. We have du=2x dx, with the lower bound becoming u=3 and upper bound u=7. The transformed definite integral is ∫_3^7 e^u du. Evaluate it as [e^u]_3^7=e^7−e^3. Unlike a logarithmic reciprocal, the antiderivative of e^u is e^u itself. The calculation stays entirely in u after the new bounds have been introduced, which is why there is no need to substitute x back before numerical evaluation.

32. Sine-power substitution with changed limits

Eight quick practice answers

∫₀² 2x(x²+1)³ dx=156, with u limits 1 to 5.
∫₀¹ 3x²/(1+x³) dx=ln 2, with u limits 1 to 2.
∫₀^(π/2) sin x cos x dx=1/2; u=cos x reverses the limits.
∫₀¹ x(1+x²)² dx=7/6; the differential contributes 1/2.
∫₀¹ 2x e^(x²) dx=e−1.
∫₀¹ 2x/(1+x²) dx=ln 2.
∫₋₁¹ 2x/(1+x²) dx=0; the signed integrand is odd.
∫₀^(π/2) sin³x cos x dx=1/4.

Five-day revision and three-learner audit

Day one: practise converting bounds under u=x²+a without doing any integration. Day two: include the differential and missing constant factors, then calculate basic polynomial powers. Day three: apply the same rules to logarithmic reciprocals and exponentials. Day four: practise trigonometric substitutions whose bounds reverse, including one method using u=sin x and another using u=cos x for the same integral. Day five: mix all types and verify answers by differentiating a proposed antiderivative or using a suitable graphing-calculator numerical check.

In a three-student tutorial, one person can choose and verify u, another can transform the differential, and a third can convert the bounds and evaluate. Rotating these roles makes it easier to identify whether a wrong final answer came from the variable substitution, a missing coefficient or incorrect endpoint conversion. All three decisions matter before calculating the antiderivative.

Frequently asked questions

Must I always change the limits in substitution?

If you evaluate in the new variable u, yes. Alternatively, integrate in u, substitute back to x, and use the original x-limits. Do not mix the two methods.

What happens if the new upper limit is smaller?

Keep the transformed endpoint order and differential sign. Reversing limits explicitly introduces an additional minus sign.

Why do I need the du coefficient?

A substitution replaces the differential as well as the expression. A mismatch by a constant factor changes the integral’s value.

Can u map both endpoints to the same value?

Yes. If the integrand matches a derivative-based substitution, the transformed definite integral may have identical endpoint values and evaluate to zero, as with an odd integrand on a symmetric interval.

Does a definite integral always equal geometric area?

No. It is a signed integral. Regions below the axis can contribute negative values, so total geometric area may require splitting and absolute values.

Can a calculator replace showing u-bounds?

It can provide a numerical check, but when mathematical working is required the substitution and correct converted endpoints still need justification.

Is integration by given substitution in the 2027 H2 syllabus?

Yes. The integration techniques and definite integrals sections include this method, while excluding reduction formulae.

What is the fastest final audit?

Check the derivative factor, converted bounds and the consistency of the integration variable. An independent antiderivative in x should give the same value.

Official syllabus and editorial note

Source: SEAB Singapore–Cambridge H2 Mathematics 9758 (2027), Integration Techniques §5.3 and Definite Integrals §5.4. All 32 clinics are original mathematical exercises, not actual examination questions, observed student outcomes or guarantees of tuition results.

JC2 H2 Mathematics Tutor Bukit Timah | Definite Integration by Substitution: Change the Limits