Application of Integrals Class 12: notes and important questions

Application of Integrals — revision notes

1. What is in the syllabus

Area under simple curves only: straight lines, circles, parabolas and ellipses in standard form (centre or vertex at the origin), together with regions cut off by straight lines. Area between two curves (e.g. two parabolas, circle and parabola) has been deleted.

2. Key results

• Vertical strips: area between \(y = f(x) \ge 0\), the \(x\)-axis and \(x = a\), \(x = b\) is \(\displaystyle\int_a^b y\,dx\).
• Horizontal strips: area between \(x = g(y) \ge 0\), the \(y\)-axis and \(y = c\), \(y = d\) is \(\displaystyle\int_c^d x\,dy\).
• Region between a curve and a line: \(\displaystyle\int (\text{upper} - \text{lower})\,dx\) or \(\displaystyle\int (\text{right} - \text{left})\,dy\), limits = points of intersection.
• If the curve is below the axis, the integral is negative: area \(= \left|\displaystyle\int_a^b y\,dx\right|\). If the curve crosses the axis, split the interval at the crossing.
• \(\displaystyle\int \sqrt{a^2 - x^2}\,dx = \dfrac x2\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac xa + C\).
• Circle \(x^2 + y^2 = a^2\): area \(\pi a^2\). Ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\): area \(\pi ab\) (from \(4\displaystyle\int_0^a \dfrac ba\sqrt{a^2 - x^2}\,dx\)).
• Parabolic segment: the region between a parabola and a chord perpendicular to its axis is \(\dfrac23\) of the enclosing rectangle.

3. Method (write these steps in the exam)

1. Draw a rough graph and shade the region.
2. Find the intersection points (these are the limits).
3. Choose vertical or horizontal strips so that the region needs as few integrals as possible.
4. Use symmetry (double the first-quadrant part) where valid.
5. Integrate, then write the answer in sq units.

Worked example 1

Area bounded by \(y^2 = 4x\) and \(x = 1\): by symmetry, \(2\displaystyle\int_0^1 2\sqrt x\,dx = \dfrac83\) sq units.

Worked example 2

Area between \(y^2 = 3x\) and \(y = x\): they meet at \(x = 0, 3\). Area \(= \displaystyle\int_0^3 (\sqrt{3x} - x)\,dx = 6 - \dfrac92 = \dfrac32\) sq units.

Worked example 3

Area of the circle \(x^2 + y^2 = 9\) between \(x = 0\) and \(x = \dfrac32\) (upper half): \(\left[\dfrac x2\sqrt{9 - x^2} + \dfrac92\sin^{-1}\dfrac x3\right]_0^{3/2} = \dfrac{9\sqrt3}{8} + \dfrac{3\pi}{4}\).

Common errors

• Reporting a negative or zero “area” because part of the curve is below the axis.
• Forgetting to double when using symmetry — or doubling a region that is not symmetric.
• Wrong limits: using the intercepts of the line instead of the intersection points with the curve.
• Integrating \(y\) with respect to \(y\) (mixing up vertical and horizontal strips).
• Dropping the \(\dfrac{a^2}{2}\sin^{-1}\dfrac xa\) term or evaluating \(\sin^{-1}\) in degrees.

Board-exam tips

• A neat, labelled rough graph with the region shaded usually earns a mark in the 5-mark question.
• Always write the final answer with “sq units” (or the actual units in a context question).
• Check your answer against a simple bound (e.g. the enclosing rectangle or the whole circle).