xSkill: notes, lectures and practice for CBSE and CSE

1.1 What this chapter asks A definite integral measures a signed area. When the curve lies above the x-axis, the integral is the ordinary area between the curve, the axis and the two vertical lines. When part of the curve lies below the axis, that part contributes a negative value, so the geometric area is the integral of the absolute value, which is found by splitting the interval at the points where the curve crosses the axis. Every question in the chapter begins with a rough sketch: the sketch shows which curve is above, where they meet, and whether horizontal strips are easier than vertical ones. For two curves the area between them is the integral of the upper curve minus the lower curve, between the x-values of their points of intersection. If the curves are easier to describe as x in terms of y, integrate with respect to y and take the right curve minus the left curve.

A = \int_{a}^{b} f (x) d x, f (x) \geq 0

Area under a curve above the x-axis

A = \int_{a}^{b} [f (x) - g (x)] d x, f (x) \geq g (x)

Area between two curves using vertical strips

A = \int_{c}^{d} [x_{right} (y) - x_{left} (y)] d y

Area between two curves using horizontal strips

A = \int_{a}^{b} ∣ f (x) ∣ d x = \int_{a}^{c} f (x) d x - \int_{c}^{b} f (x) d x

Signed area converted to geometric area when the curve crosses the axis at c

circle x^{2} + y^{2} = r^{2} : A = π r^{2}, ellipse \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 : A = π ab

Standard areas recovered from the integral

parabola y^{2} = 4 a x from x = 0 to x = a : A = \frac{8 a ^{2}}{3}

Area between the parabola and its latus rectum

1.2 The method in four steps First sketch both curves on the same axes. Second find the points of intersection by solving the equations together. Third decide which curve is upper on each piece, splitting at any crossing point. Fourth set up the integral and evaluate it. A question that gives the area and asks for a parameter is solved by the same steps and then by equating the computed expression to the given value. The common choice between vertical and horizontal strips is decided by the equations: when the curve is given as x equal to a function of y, horizontal strips avoid solving for y. A region bounded by a line and a parabola can often be computed either way, and the easier direction is the one whose limits are constants.

Exam

Draw the sketch even when the question does not ask for it. The sketch supplies the limits, the upper and lower curves and the split points, and examiners award method marks for showing those three things.

1.3 A note on units and notation Every area in this chapter is quoted in square units. The integrand has the unit of y and dx has the unit of x, so the product has the unit of area. Write the limits before the integrand in every line so that the direction of integration is visible, and never mix the x-limits with the y-limits inside the same integral.