Anti-differentiation, the constant everyone forgets, and the expression you have to write before you touch the calculator.
Integration undoes differentiation. The surprise is what it measures: slide the right-hand edge and watch the area under the curve build up.
∫12.5 (3x² + 2) dx = 20.1
Write the expression first. The number is what the calculator is for; the expression is what the marks are for.
To differentiate you multiplied by the power and knocked one off it. To integrate you do the opposite, in the opposite order.
Add one to the power, then divide by the new power. So ∫axn dx = an + 1xn+1 + c, for any whole number n except −1.
Why n cannot be −1. Adding one to −1 gives 0, and you would be dividing by zero. Integrating 1⁄x needs a different function altogether, and that is Higher Level. If you meet it in a Standard Level question, re-read the question.
A curve has dydx = 6x² + 2x, and passes through (1, 10). Find y.
Without the boundary condition there are infinitely many answers, all the same shape, stacked at different heights. The point you are given picks one of them. Dropping the "+ c" is not a small slip: it throws away the whole family.
Find the area enclosed by y = 3x² + 2, the x-axis, and the lines x = 1 and x = 4.
Top limit minus bottom limit, and bracket it. The commonest arithmetic error here is losing a sign by not bracketing the second substitution. Writing (64 + 8) − (1 + 2) keeps you safe; writing 64 + 8 − 1 + 2 does not.
The area formula applies where f(x) is above the x-axis. If the curve dips below, the integral counts that part as negative and the answer is no longer the area. At this level a question will normally keep the curve above the axis; if it does not, sketch it first.
1. Find the area under y = 3x² + 2 between x = 1 and x = 4.
2. Integrate 4x³ − 6x + 5.
3. A curve has dydx = 6x² + 2x and passes through (1, 10). Find y when x = 2.
Write the integral expression, with its limits, before you evaluate anything. It is usually an explicit mark, and it is available even when the arithmetic afterwards goes wrong.
The "+ c" on an indefinite integral is a mark. The square brackets with limits on a definite one are the equivalent.
Definite integrals can be done on the calculator, and should be. Spending the time on by-hand evaluation and then losing the setup mark is the wrong trade.
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