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The Definite Integral and the Area Under a Curve

Estimate the area under a curve with rectangles, then evaluate it exactly by substituting the limits into the antiderivative. Areas below the axis come out negative and have to be handled separately.

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What a learner can do afterwards

  • Estimate an area with rectangles and compare it with the exact value
  • Evaluate a definite integral using square-bracket notation
  • Split an integral at a root so that area below the axis is counted correctly

1 · Read

Before brackets, there were rectangles. Slice the region under the curve into thin strips, treat each strip as a rectangle, and add them up. The height of each strip comes from the function at your chosen sample: left end, right end, or middle. Left samples lag on rising graphs and come out low, right samples overshoot and come out high, and more strips squeeze the error smaller. Try it on y equals x squared from 0 to 2. Two left-endpoint strips give 0 plus 1, which is 1, too low. Two right-endpoint strips give 1 plus 4, which is 5, too high. The exact area is 8 over 3, about 2.67, sitting between them.

Try it together

Brackets turn estimates into exact values. A definite integral is a single number, while an indefinite integral with no limits is a family of functions. Find an antiderivative, write it in square brackets with your limits, then subtract the bottom value from the top one. For the integral from 1 to 3 of 2x dx, the antiderivative is x squared, so you write open bracket x squared close bracket from 1 to 3, and 9 minus 1 gives 8. Your checkpoint works the same way. The area under y equals x squared from 0 to 3 is the integral from 0 to 3 of x squared dx. Its antiderivative is x cubed over 3, so 27 over 3 minus 0 gives 9.

Area below the axis counts negative inside an integral, so split before you trust the total. Find each root where the curve crosses the axis, integrate each stretch on its own, and flip every negative part positive before adding. Take y equals x squared minus 1 from 0 to 2. The signed integral is 2 over 3, but the curve dips below the axis from 0 to 1. That stretch gives an area of 2 over 3, and the stretch from 1 to 2 gives 4 over 3, so the true area is 2.

Rectangles estimate, brackets give the exact value, and roots tell you where to split so below-axis parts count positive.

2 · Watch

Take it off screen

Print a worksheetA4 with an answer key page for grown-ups. No screen, no internet.

Where it sits

Then practise

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The Definite Integral and the Area Under a Curve · Mathematics, ages 17 to 18 · LightMySky