Introduction to integration
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What does the ∫ symbol mean?
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5.5.18 cards
What does the ∫ symbol mean?
"Integrate with respect to x." The integral symbol ∫ paired with dx means find the antiderivative — the reverse of differentiation.
It is the elongated S for "sum".
State the power rule for integration.
∫xⁿ dx = xⁿ⁺¹/(n+1) + C, provided n ≠ −1. Add 1 to the power, divide by the new power, add C.
Opposite of the power rule for differentiation.
Why must you always include +C in an indefinite integral?
Because constants disappear when you differentiate. Infinitely many functions have the same derivative — +C represents all of them.
Example: d/dx(x²+5) = d/dx(x²−7) = 2x.
∫(4x³ − 6x + 2) dx = ?
x⁴ − 3x² + 2x + C. Integrate each term: 4·x⁴/4 = x⁴, 6·x²/2 = 3x², 2·x = 2x.
Integrate term by term.
What is the first step when integrating a product like x(x+3)?
Expand the brackets first: x(x+3) = x² + 3x. Then integrate: x³/3 + 3x²/2 + C.
You cannot integrate products directly — expand first.
∫x^(1/2) dx = ?
(2/3)x^(3/2) + C. Add 1: 1/2 + 1 = 3/2. Divide by 3/2: divide by 3/2 = multiply by 2/3.
Don't panic with fractions — same rule applies.
How do you check an integral is correct?
Differentiate your answer. If you get back the original integrand, your integral is correct.
Differentiation and integration are inverse operations.
∫(x² − 3)/x dx = ?
Rewrite: x²/x − 3/x = x − 3x⁻¹. Integrate: x²/2 − 3ln|x| + C.
Split the fraction first, then use power rule.
5.5.28 cards
What is a definite integral?
An integral with limits [a, b] that gives a specific number — the signed area between the curve and the x-axis from x = a to x = b.
Unlike indefinite integrals, no +C is needed.
State the Fundamental Theorem of Calculus.
∫[a to b] f(x) dx = F(b) − F(a), where F is any antiderivative of f.
Evaluate F at b, then subtract F at a.
Evaluate ∫[1 to 3] 2x dx.
F(x) = x². F(3) − F(1) = 9 − 1 = 8.
Integrate to get F(x), then apply limits.
If f(x) < 0 on [a, b], what does the definite integral give?
A negative number. The integral gives signed area — negative when the curve is below the x-axis. For total area, take the absolute value.
Below x-axis = negative integral.
How do you find the area between two curves y = f(x) and y = g(x)?
1) Find intersections: solve f(x) = g(x) to get limits a and b. 2) Identify the top function. 3) Integrate [f(x) − g(x)] from a to b.
Always: top minus bottom.
Find the area under y = x² + 1 from x = 0 to x = 2.
∫[0 to 2] (x²+1) dx = [x³/3 + x] from 0 to 2 = (8/3 + 2) − 0 = 14/3 ≈ 4.67 square units.
Integrate then evaluate F(2) − F(0).
On IB Paper 2, how can you evaluate definite integrals?
Use your GDC. But always write the integral notation first (e.g., ∫[a to b] f(x) dx = ...). Marks are given for the setup, not just the answer.
GDC gives the number; marks need the setup.
Area between y = x and y = x² from x = 0 to x = 1.
∫[0 to 1] (x − x²) dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6 square units.
y=x is above y=x² on [0,1]. Integrate top − bottom.
5.5.38 cards
What is an initial condition in integration?
A specific point (x₀, y₀) that the function passes through. Used to find the exact value of the constant C.
Initial condition removes the ambiguity of +C.
f'(x) = 4x − 1, f(2) = 5. Find f(x).
Step 1: Integrate → f(x) = 2x² − x + C. Step 2: f(2) = 8 − 2 + C = 5 → C = −1. Answer: f(x) = 2x² − x − 1.
Substitute the point AFTER integrating.
If f'(x) = 6x and the curve passes through (0, 4), what is C?
Integrate: f(x) = 3x² + C. Substitute (0, 4): 3(0) + C = 4 → C = 4. So f(x) = 3x² + 4.
Easiest initial condition: use x = 0.
In kinematics, if v(t) = 3t², s(0) = 5, what is s(t)?
Integrate: s(t) = t³ + C. Use s(0) = 5: C = 5. So s(t) = t³ + 5.
v = ds/dt so s = ∫v dt + C.
How many initial conditions do you need to fully determine a function after integrating twice?
Two initial conditions — one for each integration, since each introduces a new constant (C₁ and C₂).
Each ∫ adds one unknown constant.
a(t) = 10, v(0) = 3, s(0) = 1. Find s(t).
v(t) = 10t + 3 (use v(0)=3 → C₁=3). s(t) = 5t² + 3t + C₂. Use s(0)=1 → C₂=1. s(t) = 5t² + 3t + 1.
Integrate twice with separate constants.
What is the "particular solution" vs "general solution" of an integral?
General solution: f(x) + C (all possible solutions). Particular solution: the specific function once C is found using an initial condition.
Initial condition converts general → particular.
dy/dx = 3x² + 2x, and y = 10 when x = 1. Find y.
Integrate: y = x³ + x² + C. Use (1, 10): 1 + 1 + C = 10 → C = 8. So y = x³ + x² + 8.
Substitute after integrating, not before.
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