Definite integrals and areas between curves (AA SL)

Here are y = x and y = x². They cross at 0 and at 1, trapping a region between them. Slide the strip and watch its height: it is always the top curve minus the bottom one.

x = 0.50

0.500top curve, y = x

0.250bottom curve, y = x²

0.250the gap

The strip is widest in the middle and closes to nothing at both crossing points.

Definite integrals by hand

∫_a^b f(x) dx = g(b) − g(a), where g is an antiderivative of f

∫₀²(x² + 1) dx= [x³3 + x]₀² = (83 + 2) − 0 = 143 ≈ 4.667 no + C: it cancels between the limits

When the curve dips below the axis

∫₀^2π sin x dx = 0, because the hump above cancels the hump below. The area is 4.

So sketch it, find where the curve crosses the axis, split the integral there and take each piece positively. "Find the integral" and "find the area" are different instructions and the examiner means the one they wrote.

Areas between curves

Area = ∫_a^b (top − bottom) dx

One integral of the difference, not two areas subtracted afterwards. It works even when part of the region is below the x-axis, because the difference is still positive wherever the top curve is on top.

Worked example

x= x² gives x(1 − x) = 0, so they meet at x = 0 and x = 1 find the limits first, always Area= ∫₀¹(x − x²) dx the line is above the parabola between 0 and 1 = [x²2 − x³3]₀¹ = 12 − 13 = 16

Check which one is on top. Between 0 and 1, x is bigger than x². Outside that interval it is the other way round. Subtracting the wrong way gives the right size with a minus sign, and an area cannot be negative.

Your turn

1. Evaluate ∫₀²(x² + 1) dx to 3 decimal places.

2. Find the area enclosed between y = x and y = x². Give it to 4 decimal places.

3. ∫₀^2π sin x dx = 0. What is the area between the curve and the x-axis over that interval?

Where the marks go

Finding the limits by solving the two curves equal to each other. On a between-curves question that is usually the first mark and everything depends on it.

Writing the integral of the difference, with the right curve on top, before evaluating anything.

Splitting at a crossing point when the question says area. An unsplit integral answers a different question.

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