Is it correct to say: Reporting a negative value as the area?

A negative integral means the region is below the x-axis. The area is the absolute value of that integral.

Is it correct to say: Integrating straight across a point where the curve crosses the axis?

∫₀^{2π} sin x dx = 0, but the area is 4. Split at every root in the interval and add the absolute values of the pieces.

Is it correct to say: Subtracting upper − lower the wrong way round?

With dx use (upper − lower); with dy use (right − left). Check by substituting one point between the limits.

Is it correct to say: Guessing the limits instead of solving for the points of intersection?

Solve the two equations simultaneously and use the coordinates you get. Write them down before setting up the integral.

Is it correct to say: Forgetting the symmetry factor, or using it when the region is not symmetric?

Use 4× for a full circle or ellipse and 2× for a parabola cut by a line perpendicular to its axis — only when the required region really is the whole symmetric shape.

Is it correct to say: Using x-limits in an integral with respect to y?

In ∫ x dy the limits are y-values. Convert the bounding lines to y = c and y = d.

📄 Formula sheet (PDF)

Application of Definite Integration

Area under a curve, area bounded by a curve and the y-axis, area between two curves, and standard areas of the circle, ellipse and parabola — Maharashtra HSC Maths Part II Ch 5

📐 13 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Draw a rough sketch first. It shows which curve is on top, where the region starts and ends, and whether symmetry can be used.
→Find the points of intersection algebraically and write them down — they are your limits of integration.
→For a circle or an ellipse, integrate over the first quadrant only and multiply by 4.
→If the boundary is easier to write as x = g(y) (for example a parabola y² = 4ax with horizontal lines), use horizontal strips and integrate with respect to y.
→Area is never negative. Take the absolute value of any part that lies below the x-axis, and end with 'sq. units'.

📐 Formulas(13)

Area Bounded by a Curve and the x-axis★ Board fav

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

Symbol	Meaning
$A$	Area of the region (sq. units)
$y = f (x)$	Equation of the curve
$a, b$	x-coordinates of the bounding vertical lines

Area Bounded by a Curve and the y-axis

A = \int_{c}^{d} x d y = \int_{c}^{d} g (y) d y, g (y) \geq 0 on [c, d]

Symbol	Meaning
$x = g (y)$	Equation of the curve solved for x
$c, d$	y-coordinates of the bounding horizontal lines

Region Below the x-axis

A = \int_{a}^{b} f (x) d x, f (x) \leq 0 on [a, b]

Curve Crossing the x-axis

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

Area Between Two Curves (vertical strips)★ Board fav

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

Symbol	Meaning
$f (x)$	Upper curve
$g (x)$	Lower curve

Area Between Two Curves (horizontal strips)

A = \int_{c}^{d} [ϕ (y) - ψ (y)] d y, ϕ (y) \geq ψ (y) on [c, d]

Symbol	Meaning
$ϕ (y)$	Right-hand curve, x = φ(y)
$ψ (y)$	Left-hand curve, x = ψ(y)

Limits from Points of Intersection

f (x) = g (x) \Rightarrow x = a, x = b

Quarter-Circle Integral

\int_{0}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{4}

Area of a Circle★ Board fav

x^{2} + y^{2} = a^{2} : A = 4 \int_{0}^{a} a^{2} - x^{2} d x = π a^{2}

Area of an Ellipse★ Board fav

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 : A = 4 \int_{0}^{a} \frac{b}{a} a^{2} - x^{2} d x = π ab

Symbol	Meaning
$a, b$	Semi-axes along the x- and y-axes (semi-major and semi-minor when a > b)

Parabola Cut Off by its Latus Rectum

y^{2} = 4 a x, x = a : A = 2 \int_{0}^{a} 2 a x d x = \frac{8 a ^{2}}{3}

Parabola and a Line Through the Vertex

y^{2} = 4 a x, y = m x (m > 0) : A = \int_{0}^{4 a / m^{2}} (2 a x - m x) d x = \frac{8 a ^{2}}{3 m ^{3}}

Area Between y² = 4ax and x² = 4ay

A = \int_{0}^{4 a} (2 a x - \frac{x ^{2}}{4 a}) d x = \frac{16 a ^{2}}{3}

✏️ Solved Examples

1Solved Exampleeasy2 steps

Find the area of the region bounded by the curve y = x², the x-axis and the lines x = 1 and x = 3.

The curve lies above the x-axis on [1, 3], so integrate y with respect to x.

A = \int_{1}^{3} x^{2} d x

2Solved Exampleboard4 steps

Find the area of the ellipse x²/16 + y²/9 = 1.

Here a = 4, b = 3. In the first quadrant, solve for y.

y = \frac{3}{4} 16 - x^{2}

3Solved Exampleboard4 steps

Find the area of the region bounded by the parabola y² = 4x and the line y = x.

Points of intersection: substitute y = x in y² = 4x.

x^{2} = 4 x \Rightarrow x = 0, x = 4

4Solved ExampleHOTS4 steps

Find the area of the region enclosed between the parabolas y² = 4x and x² = 4y.

From x² = 4y, y = x²/4. Substitute in y² = 4x.

\frac{x ^{4}}{16} = 4 x \Rightarrow x (x^{3} - 64) = 0 \Rightarrow x = 0, x = 4

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Reporting a negative value as the area
✓A negative integral means the region is below the x-axis. The area is the absolute value of that integral.
2
Integrating straight across a point where the curve crosses the axis
✓∫₀^{2π} sin x dx = 0, but the area is 4. Split at every root in the interval and add the absolute values of the pieces.
3
Subtracting upper − lower the wrong way round
✓With dx use (upper − lower); with dy use (right − left). Check by substituting one point between the limits.
4
Guessing the limits instead of solving for the points of intersection
✓Solve the two equations simultaneously and use the coordinates you get. Write them down before setting up the integral.
5
Forgetting the symmetry factor, or using it when the region is not symmetric
✓Use 4× for a full circle or ellipse and 2× for a parabola cut by a line perpendicular to its axis — only when the required region really is the whole symmetric shape.
6
Using x-limits in an integral with respect to y
✓In ∫ x dy the limits are y-values. Convert the bounding lines to y = c and y = d.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the area bounded by y = sin x and the x-axis between x = 0 and x = 2π.
Q2
Find the area of the circle x² + y² = 25.
Q3
Find the area of the region bounded by the parabola y² = 8x and its latus rectum.
Q4
Find the area of the region bounded by y = x² and y = x.
Q5
Find the area of the region in the first quadrant bounded by x² = 4y, the y-axis and the lines y = 1 and y = 4.
Q6
Find the area bounded by y = x³, the x-axis and the lines x = −1 and x = 2.

📝 Notes

Application of Definite Integration

This chapter uses the definite integral to find areas of plane regions. Every question comes down to three decisions: which strips to use, what the limits are, and what the height (or length) of a strip is.

A reliable method

Sketch the curves and shade the required region.
Find the points of intersection — these give the limits.
Decide on vertical strips (integrate in $x$ ) or horizontal strips (integrate in $y$ ).
Write the strip length: upper − lower, or right − left.
Integrate, then check the answer is positive and reasonable compared with the sketch.

Vertical or horizontal strips?

Use vertical strips when both boundaries are easy to write as $y = f(x)$ . Switch to horizontal strips when the curves are naturally $x = g(y)$ , such as $y^2 = 4ax$ bounded by horizontal lines, or when vertical strips would need the region split into several pieces.

Symmetry saves time

A circle $x^2 + y^2 = a^2$ and an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are symmetric about both axes, so integrate over the first quadrant and multiply by 4. A parabola $y^2 = 4ax$ is symmetric about the x-axis, so a region cut off by a vertical line is twice the part above the axis.

Standard areas worth remembering

Circle $\pi a^2$ , ellipse $\pi ab$ , parabola and latus rectum $\frac{8a^2}{3}$ , and the two parabolas $y^2 = 4ax$ , $x^2 = 4ay$ enclosing $\frac{16a^2}{3}$ . Use them to check your answer, but always show the integration — the method carries the marks.