This chapter applies definite integrals to find areas: the area under a curve \( y = f(x) \) between two vertical lines, the area under a curve \( x = g(y) \) between two horizontal lines, and the areas of standard symmetrical figures such as circles and ellipses. The sheet also covers the absolute-value rule you need when a curve dips below the x-axis.
Every formula below is grouped by topic, with the meaning of each symbol, when to use it, and worked examples with original numbers. For the derivations and explanations, see the Class 12 maths notes on Application of Integrals.
Formulas at a Glance
| Purpose (what you are finding) | Formula |
|---|---|
| Area under a curve with vertical strips | \( A = \int_a^b f(x)\,dx \) |
| Area under a curve with horizontal strips | \( A = \int_c^d g(y)\,dy \) |
| Area when the curve lies below the x-axis (numerical value only) | \( A = \left| \int_a^b f(x)\,dx \right| \) |
| Area of a circle of radius \( a \) (by symmetry of four quadrants) | \( A = \pi a^2 \) |
| Area of an ellipse \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \) (by symmetry) | \( A = \pi ab \) |
All Formulas, Grouped by Topic
These groupings follow the textbook order: area under simple curves first, then areas of standard symmetrical figures (NCERT Class 12 Mathematics Part II, Chapter 8).
Area under Simple Curves
The area of the region bounded by the curve \( y = f(x) \), the x-axis and the lines \( x = a \) and \( x = b \), where \( b \gt a \), is given by \[ \text{Area} = \int_a^b y\,dx = \int_a^b f(x)\,dx \]
The area of the region bounded by the curve \( x = g(y) \), the y-axis and the lines \( y = c \) and \( y = d \), where \( d \gt c \), is given by \[ \text{Area} = \int_c^d x\,dy = \int_c^d g(y)\,dy \]
These two are the defining formulas of the chapter (NCERT Summary, p. 299). They are the same idea — adding thin strips — with the roles of x and y swapped.
Curve Below the x-axis
If the curve lies below the x-axis, then \( f(x) \lt 0 \) on the whole interval and the integral comes out negative. Area is a numerical value, so take its absolute value:
\[ A = \left| \int_a^b f(x)\,dx \right| \]
If part of the curve is above and part is below the x-axis, split the integral at the points where the curve meets the x-axis, take the absolute value of the negative part, and add (NCERT §8.2).
Areas of Circles and Ellipses by Symmetry
For the circle \( x^2 + y^2 = a^2 \), which is symmetrical about both axes, the whole area is four times the first-quadrant area:
\[ A = 4\int_0^a y\,dx = 4\int_0^a \sqrt{a^2 – x^2}\,dx = \pi a^2 \]
For the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), again using four times the first-quadrant area:
\[ A = 4\int_0^a \frac{b}{a}\sqrt{a^2 – x^2}\,dx = \pi ab \]
Both results are worked out in the chapter using the integral \( \int \sqrt{a^2 – x^2}\,dx = \frac{x}{2}\sqrt{a^2 – x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \) (NCERT §8.2). The factor 4 appears because the region is split into four congruent quadrants.
What Each Symbol Means
| Symbol | What it means | Unit / nature |
|---|---|---|
| \( A \) | Area of the bounded region | square units (e.g. \( \text{units}^2 \)) |
| \( f(x) \) | Height of the curve (y-coordinate) at x | length |
| \( g(y) \) | Width of the curve (x-coordinate) at y | length |
| \( a, b \) | x-limits of the region; \( b \gt a \) | length |
| \( c, d \) | y-limits of the region; \( d \gt c \) | length |
| \( dx \) | Width of a vertical elementary strip | length |
| \( dy \) | Width of a horizontal elementary strip | length |
| \( a \) (in circle/ellipse) | Radius of the circle; semi-transverse axis (larger denominator) of the ellipse | length |
| \( b \) (in ellipse) | Semi-conjugate axis of the ellipse | length |
When to Use Each Formula
- Use \( \int_a^b f(x)\,dx \) when the region is bounded by the x-axis and the vertical lines \( x = a \), \( x = b \) — take vertical strips.
- Use \( \int_c^d g(y)\,dy \) when the region is bounded by the y-axis and the horizontal lines \( y = c \), \( y = d \) — take horizontal strips. Example: the region bounded by \( x = g(y) \), the y-axis and the line \( y = 3 \).
- Take the absolute value when the curve lies entirely below the x-axis. When the curve crosses the x-axis, split at the crossing points and add the absolute values of each part.
- Use the symmetry factor 4 only for figures symmetrical about both axes (circle, ellipse in standard form). Compute the first-quadrant area and multiply by 4.
Worked Examples
Example 1: Area under a parabola by vertical strips
Problem: Find the area bounded by \( y = x^2 \), the x-axis, and the lines \( x = 1 \) and \( x = 3 \).
Step 1: The region is bounded by the x-axis and vertical lines, so use the vertical-strip formula.
\[ A = \int_1^3 x^2\,dx \]
Step 2: Integrate.
\[ A = \left[ \frac{x^3}{3} \right]_1^3 = \frac{27}{3} – \frac{1}{3} = \frac{26}{3} \]
Final answer: \( \frac{26}{3} \) square units.
Example 2: Curve crossing the x-axis — use absolute values
Problem: Find the area bounded by the line \( y = 2x + 1 \), the x-axis and the ordinates \( x = -1 \) and \( x = 1 \).
Step 1: The line meets the x-axis where \( 2x + 1 = 0 \), so \( x = -\frac{1}{2} \).
The curve is below the x-axis on \( (-1, -\frac{1}{2}) \) and above it on \( (-\frac{1}{2}, 1) \), so split the integral.
\[ A = \left| \int_{-1}^{-1/2} (2x+1)\,dx \right| + \int_{-1/2}^{1} (2x+1)\,dx \]
Step 2: Evaluate each part.
\[ \left[ x^2 + x \right]_{-1}^{-1/2} = \frac{-1}{4}, \qquad \left[ x^2 + x \right]_{-1/2}^{1} = \frac{9}{4} \]
\[ A = \left| \frac{-1}{4} \right| + \frac{9}{4} = \frac{10}{4} = \frac{5}{2} \]
Final answer: \( \frac{5}{2} \) square units.
Example 3: Area of an ellipse using symmetry
Problem: Find the area enclosed by the ellipse \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \).
Step 1: Here \( a = 5 \) and \( b = 4 \).
The ellipse is symmetrical about both axes, so use the result from §8.2.
\[ A = \pi ab = \pi \times 5 \times 4 = 20\pi \]
Final answer: \( 20\pi \) square units.
For practice on the textbook’s own questions, see the NCERT solutions for Application of Integrals.
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Writing a negative value as the area when the curve is below the x-axis | Take the absolute value: \( A = \left| \int_a^b f(x)\,dx \right| \) | Area can never be negative. If your integral is negative, apply the modulus. |
| Integrating straight from \( a \) to \( b \) when the curve crosses the x-axis in between | Find the crossing point, split the integral there, and add the absolute value of the negative part | Sketch the curve; each part of the region must contribute a positive area. |
| Using the vertical-strip formula \( \int_a^b f(x)\,dx \) for a region bounded sideways by the y-axis | For regions bounded by the y-axis and lines \( y = c \), \( y = d \), write the curve as \( x = g(y) \) and use \( \int_c^d g(y)\,dy \) | Check which axis the strips are perpendicular to: vertical strips → integrate in x; horizontal strips → integrate in y. |
| Using \( \pi ab \) without identifying \( a \) and \( b \) from the standard form | \( a \) and \( b \) are the denominators under \( x^2 \) and \( y^2 \) in \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \) | For \( \frac{x^2}{25}+\frac{y^2}{16}=1 \), the area is \( 20\pi \), not \( 9\pi \). |
Frequently Asked Questions
Why do we take the absolute value of the integral for area?
A definite integral gives a signed value: below the x-axis the integrand is negative, so the integral is negative. Since area is a positive quantity, we take \( \left| \int_a^b f(x)\,dx \right| \) when the curve lies entirely below the x-axis.
When should I use the horizontal strip formula instead of the vertical one?
Use horizontal strips when the region is bounded by the y-axis and horizontal lines \( y = c \) and \( y = d \). You express the curve as \( x = g(y) \) and integrate \( \int_c^d g(y)\,dy \). If the region is bounded by the x-axis and vertical lines \( x = a \), \( x = b \), use the vertical strip formula instead.
Do I always multiply by 4 for circle and ellipse areas?
Only when the figure is symmetrical about both axes, as in the standard forms \( x^2 + y^2 = a^2 \) and \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \). Then the first-quadrant area is one quarter of the total, so the total is four times it.
Reference: NCERT Class 12 Mathematics textbook (Mathematics Part II), Chapter 8, Application of Integrals. The official PDF is available at ncert.nic.in.
Explore Class 12 Maths Formulas
More for this chapter:
- Application of Integrals Notes
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