Application of Integrals ISC Class 12: Area Under and Between Curves

What Application of Integrals is about

Weightage: Application of Integrals is part of the unit Calculus, which carries 35 of the 80 marks in the ISC Mathematics, shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Area under a curve and the x-axis

For y = f(x) ≥ 0 on [a, b], the area bounded by the curve, the x-axis and the ordinates x = a and x = b is $A={\int }_a^{b}f(x)dx$ (think of thin vertical strips of height y and width dx). If f(x) ≤ 0 on the interval, the integral is negative and the area is its absolute value. If f changes sign, find where it crosses the axis and split: $A=\left|{\int }_a^{c}fdx\right|+\left|{\int }_c^{b}fdx\right|$.

Worked example

Find the area bounded by y = x², the x-axis and the lines x = 0 and x = 3.

Try it first, then show the solutionSolution

1. y = x² ≥ 0, so $A={\int }_0^3{x}^{2}dx$
2. $={\left[\frac{x^3}{3}\right]}_0^3=9$

Answer: 9 square units

Regions below the axis and with modulus functions

Area is always positive. For y = x² − 4 between x = −2 and x = 2, the curve lies below the axis, so $A=\left|{\int }_{-2}^2(x^2-4)dx\right|$. For a modulus function, split at the corner: y = |x − 1| equals 1 − x for x < 1 and x − 1 for x ≥ 1. A sketch shows which pieces to use.

Worked example

Find the area bounded by y = |x − 1|, the x-axis and the lines x = 0 and x = 3.

Try it first, then show the solutionSolution

1. ${\int }_0^1(1-x)dx=\frac{1}{2}$
2. ${\int }_1^3(x-1)dx={\left[\frac{(x-1{)}^2}{2}\right]}_1^3=2$
3. Total = ½ + 2

Answer: $\frac{5}{2}$ square units

Area of a circle and an ellipse

Both are symmetric about both axes, so find the first-quadrant area and multiply by 4. Circle $x^2+y^2=a^2$: $A=4{\int }_0^a\sqrt{a^2-x^2}dx=4{\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right]}_0^a=4\cdot \frac{\pi a^2}{4}=\pi a^2$. Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$: $y=\frac{b}{a}\sqrt{a^2-x^2}$, so $A=\frac{4b}{a}\cdot \frac{\pi a^2}{4}=\pi ab$.

Worked example

Using integration, find the area of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$.

Try it first, then show the solutionSolution

1. First quadrant: $y=\frac{3}{4}\sqrt{16-x^2}$, x from 0 to 4.
2. $A=4\cdot \frac{3}{4}{\int }_0^4\sqrt{16-x^2}dx=3\times \frac{16\pi }{4}=12\pi$

Answer: 12π square units (= πab with a = 4, b = 3)

Area between two curves

If $f(x)\ge g(x)$ for a ≤ x ≤ b, the area between them is ${\int }_a^b[f(x)-g(x)]dx$: integrate (upper − lower). Steps: (1) sketch both curves; (2) solve f(x) = g(x) for the points of intersection, which give the limits; (3) decide which curve is on top in each interval (test a point); (4) integrate and add. Using horizontal strips, area = ${\int }_c^d[x_{right}-x_{left}]dy$.

Worked example

Find the area enclosed between y = x² and y = x.

Try it first, then show the solutionSolution

1. Intersections: x² = x ⇒ x = 0, 1. Between them, x ≥ x² (e.g. at x = ½).
2. $A={\int }_0^1(x-x^2)dx=\frac{1}{2}-\frac{1}{3}$

Answer: $\frac{1}{6}$ square units

Parabolas and lines

A parabola such as $y^2=4ax$ opens along the x-axis; in the first quadrant $y=2\sqrt{ax}$. For a region bounded by a parabola and a line, find their intersections, sketch, and integrate upper minus lower. Use symmetry about the x-axis when the region is symmetric (double the first-quadrant area).

Worked example

Find the area bounded by the parabola $y^2=8x$ and the line x = 2.

Try it first, then show the solutionSolution

1. The region is symmetric about the x-axis; upper half $y=\sqrt{8x}$.
2. $A=2{\int }_0^2\sqrt{8}\sqrt{x}dx=2\sqrt{8}\cdot \frac{2}{3}{x}^{3/2}{|}_0^2=2\cdot 2\sqrt{2}\cdot \frac{2}{3}\cdot 2\sqrt{2}=\frac{32}{3}$

Answer: $\frac{32}{3}$ square units

Exponential and logarithmic curves

${\int }_a^b{e}^{x}dx=e^b-e^a$, and $\int \log xdx=x\log x-x$ (by parts). y = eˣ lies above the x-axis everywhere; y = log x is negative for 0 < x < 1 and positive for x > 1, so split at x = 1 when needed.

Worked example

Find the area bounded by (a) y = eˣ, the axes and x = 1; (b) y = log x, the x-axis and x = e.

Try it first, then show the solutionSolution

1. (a) ${\int }_0^1{e}^{x}dx=e-1\approx 1.72$
2. (b) log x ≥ 0 on [1, e]: ${\int }_1^e\log xdx=[x\log x-x{]}_1^e=(e-e)-(0-1)=1$

Answer: (a) e − 1 (b) 1 square unit

Region bounded by three lines

Find the three vertices by solving the lines in pairs. Sketch the triangle; draw a vertical line through the middle vertex to split it into two parts, each with a single upper line and a single lower line. Integrate (upper − lower) over each part and add. Check with the formula ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Worked example

Using integration, find the area of the triangle with vertices A(1, 0), B(2, 2) and C(3, 1).

Try it first, then show the solutionSolution

1. Lines: AB: y = 2x − 2; BC: y = 4 − x; AC: $y=\frac{x-1}{2}$ (the lower side throughout).
2. ${\int }_1^2\left(2x-2-\frac{x-1}{2}\right)dx={\int }_1^2\frac{3}{2}(x-1)dx=\frac{3}{4}$
3. ${\int }_2^3\left(4-x-\frac{x-1}{2}\right)dx={\int }_2^3\frac{3}{2}(3-x)dx=\frac{3}{4}$; total $\frac{3}{2}$ (check: ½|1(2 − 1) + 2(1 − 0) + 3(0 − 2)| = 3/2).

Answer: $\frac{3}{2}$ square units

Maths in real life: Application of Integrals

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Area Under a Curve
• Area Between Two Curves
• Area of a Region Bounded by Lines

Application of Integrals formula sheet

Area with the x-axis

$A={\int }_a^b|y|dx={\int }_a^b|f(x)|dx$

Split where the curve crosses the axis.

Area with the y-axis

$A={\int }_c^d|x|dy$

Horizontal strips.

Between two curves

$A={\int }_a^b[f(x)-g(x)]dx,f\ge g$

Limits from intersection points.

Circle

$x^2+y^2=a^2:A=4{\int }_0^a\sqrt{a^2-x^2}dx=\pi a^2$

Ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1:A=\frac{4b}{a}{\int }_0^a\sqrt{a^2-x^2}dx=\pi ab$

Key antiderivative

$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$

Used for circles and ellipses.

Parabola

$y^2=4ax\Rightarrow y=\pm 2\sqrt{ax}$

Use symmetry about the x-axis.

Application of Integrals: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Application of Integrals

The question patterns to expect, and the method that works for each.

Tricks to remember Application of Integrals

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Sketch first, integrate second

  No sketch, no limits: always draw the region before writing the integral.

  Use it when: Every area question.

• Top minus bottom

  Vertical strips: upper curve minus lower curve; horizontal strips: right minus left.

  Use it when: Area between curves.

• Area is never negative

  Below the axis? Take the absolute value of that part separately.

  Use it when: Curves crossing the x-axis.

• Quarter times four

  Circle and ellipse: integrate over the first quadrant, then multiply by 4.

  Use it when: Symmetric regions.

• Check triangles with the vertex formula

  Area by integration must match ½|x₁(y₂ − y₃) + …|.

  Use it when: Regions bounded by three lines.

Application of Integrals: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Application of Integrals

The slips examiners see most often, and what to write instead.

• ✗Mistake: No sketch, so wrong limits

  CISCE examiners' comments, 2024 and 2025: candidates did not understand how to find the limits or the bounded region, drew incorrect sketches, or did not sketch at all, so could not find the area.

  Wrong working

  Integrates from 0 to 4 because '4' appears in the equation.

  Right working

  Sketch, solve for the intersection points, then use them as the limits.

  ✓Do this: Always begin with a labelled sketch.

  Usually costs: 2-3 marks.

• ✗Mistake: Not writing the region as upper minus lower

  CISCE examiners' comments, 2024: for the region between two parabolas, candidates did not express the area as the difference of two functions.

  Wrong working

  Adds the two integrals.

  Right working

  ${\int }_a^b[\text{upper}-\text{lower}]dx$ between the intersection points.

  ✓Do this: Test a point between the limits to see which curve is on top.

  Usually costs: 1-2 marks.

• ✗Mistake: Not splitting a region bounded by lines

  CISCE examiners' comments, 2025: for a field bounded by three lines, candidates failed to find the limits, did not split the region into two parts and erred in writing each part as a difference of functions.

  Wrong working

  One integral from the leftmost to the rightmost vertex with one upper line.

  Right working

  Split at the middle vertex where the upper (or lower) line changes, integrate each part.

  ✓Do this: Mark which line is upper and which is lower in each strip of the sketch.

  Usually costs: 2-3 marks.

• ✗Mistake: Counting area below the axis as negative

  A frequent slip: integrating across a point where the curve crosses the x-axis, so positive and negative parts cancel.

  Wrong working

  Area between y = x³ and the x-axis from −1 to 1 is ${\int }_{-1}^1{x}^{3}dx=0$.

  Right working

  $2{\int }_0^1{x}^{3}dx=\frac{1}{2}$.

  ✓Do this: Split at the crossings and add absolute values.

  Usually costs: 1-2 marks.

Practice questions on Application of Integrals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The area bounded by y = sin x and the x-axis from x = 0 to x = π is:
   (a) 0
   (b) 1
   (c) 2
   (d) π

   Hint

   $[-\cos x{]}_0^{\pi }$.

   Answer

   (c) 2

2. Q2 · 1 mark · MCQBoard level

   The area enclosed by the circle $x^2+y^2=9$ is:
   (a) 3π
   (b) 6π
   (c) 9π
   (d) 81π

   Hint

   πa².

   Answer

   (c) 9π

3. Q3 · 1 mark · MCQBoard level

   The area bounded by y = log x, the x-axis and the line x = e is:
   (a) 1
   (b) e
   (c) e − 1
   (d) 1/e

   Hint

   ∫ log x dx = x log x − x.

   Answer

   (a) 1

4. Q4 · 1 mark · Very short answerBoard level

   Write the integral giving the area between y = 9 − x² and the x-axis, and evaluate it.

   Hint

   Roots ±3.

   Answer

   ${\int }_{-3}^3(9-x^2)dx=36$ square units

5. Q5 · 2 marks · Short answerBoard level

   Find the area bounded by y = x², the x-axis and the lines x = 0 and x = 2.

   Hint

   x³/3.

   Answer

   8/3 square units

6. Q6 · 2 marks · Short answerHarder board level

   Find the area enclosed between y = x² − 4 and the x-axis.

   Hint

   The region is below the axis.

   Answer

   $\left|{\int }_{-2}^2(x^2-4)dx\right|=\frac{32}{3}$

7. Q7 · 2 marks · Short answerHarder board level

   Find the area enclosed by the parabola $y^2=4x$ and the line y = 2x.

   Hint

   Intersections x = 0, 1.

   Answer

   ${\int }_0^1(2\sqrt{x}-2x)dx=\frac{4}{3}-1=\frac{1}{3}$

8. Q8 · 2 marks · Short answerHOTS / competency

   Find the area between y = x³ and the x-axis from x = −1 to x = 1. Why is ${\int }_{-1}^1{x}^{3}dx$ not the answer?

   Hint

   Odd function, opposite signs.

   Answer

   Area = $2{\int }_0^1{x}^{3}dx=\frac{1}{2}$. The integral is 0 because the part below the axis counts negative and cancels the part above.

9. Q9 · 3 marks · Short answerHarder board level

   Using integration, find the area of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$.

   Hint

   First quadrant × 4.

   Answer

   12π square units

10. Q10 · 3 marks · Short answerHarder board level

    Find the area bounded by the parabola $y^2=8x$ and the line x = 2.

    Hint

    Symmetric about the x-axis.

    Answer

    $2{\int }_0^2\sqrt{8x}dx=\frac{32}{3}$ square units

11. Q11 · 3 marks · Short answerHarder board level

    Find the area enclosed between y = x and y = x³ in the first quadrant.

    Hint

    x ≥ x³ on [0, 1].

    Answer

    ${\int }_0^1(x-x^3)dx=\frac{1}{4}$

12. Q12 · 5 marks · Long answerHarder board level

    (a) Sketch y = x² and y = x + 2 and find their points of intersection. [2]
    (b) Using integration, find the area enclosed between them. [3]

    Hint

    Line above parabola between −1 and 2.

    Answer

    (a) Upward parabola with vertex O; line with gradient 1 and y-intercept 2; they meet at (−1, 1) and (2, 4). (b) ${\int }_{-1}^2(x+2-x^2)dx={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2=\frac{10}{3}+\frac{7}{6}=\frac{9}{2}$.

13. Q13 · 2 marks · Short answerHarder board level

    Find the area under the line y = 2x + 1 from x = 0 to x = 3 by integration, and check with geometry.

    Hint

    A trapezium.

    Answer

    $[x^2+x{]}_0^3=12$; trapezium with parallel sides 1 and 7, width 3: ½(1 + 7)(3) = 12.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Area under y = f(x) ≥ 0 from a to b?Show answer

   ${\int }_a^{b}f(x)dx$.

2. 2.Area between two curves?Show answer

   ${\int }_a^b[f(x)-g(x)]dx$ where f is the upper curve.

3. 3.Area of the circle x² + y² = a²?Show answer

   πa².

4. 4.Area of the ellipse x²/a² + y²/b² = 1?Show answer

   πab.

5. 5.How do you find the limits?Show answer

   Solve the equations simultaneously for the points of intersection.

6. 6.What if the curve lies below the x-axis?Show answer

   Take the absolute value of that part's integral.

7. 7.Upper half of $y^2=4ax$?Show answer

   $y=2\sqrt{ax}$.

8. 8.$\int \sqrt{a^2-x^2}dx$?Show answer

   $\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you sketch the standard curves quickly?
• ☐Can you find the area under a curve, handling parts below the axis?
• ☐Can you derive the areas of a circle and an ellipse by integration?
• ☐Can you find intersection points and set up upper − lower?
• ☐Can you split a triangle bounded by three lines into two integrals?
• ☐Can you check a triangle's area with the vertex formula?
• ☐Can you handle modulus, exponential and log curves?

Application of Integrals: frequently asked questions