Back to Topic 5.5 — Introduction to integration
5.5.2Math AI SL SL8 flashcards

Definite Integration and Area Under a Curve

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Card 1 of 85.5.2
5.5.2
Question

What is a definite integral?

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All 8 Flashcards — Definite Integration and Area Under a Curve

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Card 1definition

Question

What is a definite integral?

Answer

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.

💡 Hint

Unlike indefinite integrals, no +C is needed.

Card 2formula

Question

State the Fundamental Theorem of Calculus.

Answer

∫[a to b] f(x) dx = F(b) − F(a), where F is any antiderivative of f.

💡 Hint

Evaluate F at b, then subtract F at a.

Card 3example

Question

Evaluate ∫[1 to 3] 2x dx.

Answer

F(x) = x². F(3) − F(1) = 9 − 1 = 8.

💡 Hint

Integrate to get F(x), then apply limits.

Card 4concept

Question

If f(x) < 0 on [a, b], what does the definite integral give?

Answer

A negative number. The integral gives signed area — negative when the curve is below the x-axis. For total area, take the absolute value.

💡 Hint

Below x-axis = negative integral.

Card 5process

Question

How do you find the area between two curves y = f(x) and y = g(x)?

Answer

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.

💡 Hint

Always: top minus bottom.

Card 6example

Question

Find the area under y = x² + 1 from x = 0 to x = 2.

Answer

∫[0 to 2] (x²+1) dx = [x³/3 + x] from 0 to 2 = (8/3 + 2) − 0 = 14/3 ≈ 4.67 square units.

💡 Hint

Integrate then evaluate F(2) − F(0).

Card 7concept

Question

On IB Paper 2, how can you evaluate definite integrals?

Answer

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.

💡 Hint

GDC gives the number; marks need the setup.

Card 8example

Question

Area between y = x and y = x² from x = 0 to x = 1.

Answer

∫[0 to 1] (x − x²) dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6 square units.

💡 Hint

y=x is above y=x² on [0,1]. Integrate top − bottom.

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