H: Integration: Evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

The idea

A definite integral has limits and gives a numerical value. To evaluate ∫ from a to b of f(x) dx, find an antiderivative F(x), then compute F(b) − F(a). The constant of integration cancels in the subtraction, so it is not needed for a definite integral. Geometrically, if f(x) ≥ 0 on [a, b], the integral equals the area under y = f(x) between x = a and x = b. If the curve dips below the x-axis, the integral gives a signed area; for total area, split the interval at roots and take absolute values. For the area between y = f(x) and y = g(x) from x = a to x = b, integrate upper minus lower: ∫(f(x) − g(x)) dx. Find intersection points to determine limits if not given. Example: area between y = x and y = x² from x = 0 to 1 is ∫(x − x²) dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6.

What earns the marks

• Evaluates a definite integral correctly by finding an antiderivative and substituting the upper and lower limits, then subtracting: F(b) − F(a).
• Uses square brackets to show the antiderivative evaluated at the limits, e.g. [F(x)] from a to b, before substituting.
• Interprets a definite integral as the area under a curve when the function is non-negative on the interval.
• Handles areas below the x-axis by splitting the interval at roots and taking absolute values to find total area.
• Finds the area between two curves by integrating (upper curve − lower curve) over the appropriate interval.
• Determines intersection points of two curves to find the limits of integration when they are not given.

Common mistakes and how to put them right

• Subtracting the limits in the wrong order, e.g. F(a) − F(b) instead of F(b) − F(a).

  Put it right: Do upper limit minus lower limit.

• Thinking that omitting the constant of integration in a definite integral is an error.

  Put it right: The constant cancels in F(b) − F(a), so it is not needed for a definite integral; it is only required for an indefinite integral.

• Assuming the definite integral always gives the geometric area when the curve crosses the x-axis.

  Put it right: Split the interval at roots and take absolute values for total area.

• Integrating (lower curve − upper curve) when finding area between curves, giving a negative result.

  Put it right: Subtract the lower curve from the upper curve.

• Not finding intersection points when limits are not given, leading to incorrect limits.

  Put it right: Set the curves equal and solve for x to find limits.

Exam tips

• Write the antiderivative in square brackets with the limits clearly shown before substituting; this helps you avoid sign errors.
• If a curve crosses the x-axis in the interval, split the integral at the root(s) and evaluate each part separately, taking absolute values for total area.
• For area between curves, sketch the graphs to identify which is upper and which is lower, and to see intersection points.
• Check whether the question asks for 'area' or 'value of the integral'; if it asks for area, ensure your answer is positive.