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Topic 5.16Math AA HL16 flashcards

Integration by parts (HL only)

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Card 1 of 165.16.1
5.16.1
Question

What is integration by substitution undoing?

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All Flashcards in Topic 5.16

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5.16.18 cards

Card 1concept
Question

What is integration by substitution undoing?

Answer

The chain rule — it's the reverse chain rule. You let u = the inner function so f'(g)·g' dx becomes f'(u) du.

Card 2concept
Question

How do you choose u in a substitution?

Answer

Let u be the INNER function whose derivative (up to a constant) also appears in the integrand.

Card 3concept
Question

After choosing u, how do you replace dx?

Answer

Differentiate: du = u' dx, then rewrite u' dx (or dx) in terms of du.

Card 4concept
Question

For a DEFINITE integral, what's the clean way to finish?

Answer

Change the limits to u-values (put each x-limit into u), then evaluate in u — no switching back.

Card 5concept
Question

Find ∫ 2x(x² + 1)⁴ dx.

Answer

u = x² + 1, du = 2x dx → ∫ u⁴ du = (x² + 1)⁵/5 + C.

Card 6concept
Question

Evaluate ∫₀^(π/2) sin³x cos x dx.

Answer

u = sin x, limits 0→1 → ∫₀¹ u³ du = 1/4.

Card 7concept
Question

Only a constant factor is missing from u' — what do you do?

Answer

Balance it: e.g. if du = 2x dx but you have x dx, then x dx = ½ du. You can pull constants out, never variables.

Card 8concept
Question

Find ∫ x√(x² + 3) dx.

Answer

u = x² + 3, x dx = ½ du → ½∫ u^(1/2) du = ⅓(x² + 3)^(3/2) + C.

5.16.28 cards

Card 9formula
Question

State the integration by parts formula.

Answer

∫ u dv = uv − ∫ v du. You differentiate u and integrate dv.

Card 10concept
Question

What does LIATE help you decide?

Answer

Which factor to call u: Log, Inverse-trig, Algebra (powers of x), Trig, Exponential — first listed becomes u.

Card 11concept
Question

When do you use integration by parts (not substitution)?

Answer

For a PRODUCT of two different kinds of function (e.g. x·eˣ, x·sin x, x·ln x) where no inner-derivative pair is present.

Card 12concept
Question

Find ∫ x cos x dx.

Answer

u = x, dv = cos x dx → x sin x − ∫ sin x dx = x sin x + cos x + C.

Card 13concept
Question

How is ∫ ln x dx found by parts?

Answer

u = ln x, dv = dx (v = x): x ln x − ∫ 1 dx = x ln x − x + C.

Card 14concept
Question

How many times must you apply parts for ∫ x² eˣ dx?

Answer

Twice — each round drops the power of x by one (x² → x → constant).

Card 15concept
Question

Find ∫ x ln x dx.

Answer

u = ln x, dv = x dx → (x²/2)ln x − ∫ x/2 dx = (x²/2)ln x − x²/4 + C.

Card 16concept
Question

Find ∫ x² eˣ dx.

Answer

Parts twice: eˣ(x² − 2x + 2) + C.

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