medium difficulty10 min

Area Under a Curve Using Definite Integral

The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is found by adding thin vertical strips from x = a to x = b. If f(x) is non-negative on [a, b], the area is ∫_a^b f(x) dx.

In one line: Area under one curve means strip height is curve value minus x-axis value: y - 0.

Practice This Concept

Understand it

Use this result when the required region lies between a single curve and the x-axis over a fixed interval. The integrand represents the height of a vertical strip, and dx represents its small width. If the curve goes below the x-axis, ∫_a^b f(x) dx gives signed area, so actual area needs separate intervals or absolute value handling.

Example

The area bounded by y = x², the x-axis, x = 0 and x = 2 is ∫₀² x² dx = [x³/3]₀² = 8/3 square units.

Formula and rule

Key rule: Area = ∫_a^b y dx = ∫_a^b f(x) dx when the region is bounded by y = f(x), x-axis, x = a, x = b, and f(x) ≥ 0 on [a, b].

If f(x) ≤ 0, actual area = -∫_a^b f(x) dx.

If f(x) changes sign, split the interval at zeros of f(x).

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Find the area bounded by y = 4 - x² and the x-axis.

1Intersections with x-axis: 4 - x² = 0 gives x = -2, 2.
2Since 4 - x² ≥ 0 on [-2, 2], area = ∫_-2² (4 - x²) dx.
3Using symmetry, area = 2∫₀² (4 - x²) dx = 2[4x - x³/3]₀² = 2(8 - 8/3) = 2(16/3) = 32/3 square units.

Method you can reuse

1Identify the curve y = f(x).
2Identify the lower and upper x-limits.
3Check whether the curve is above the x-axis on the interval.
4Write area as ∫_a^b f(x) dx or split if the sign changes.
5Integrate, substitute limits, simplify, and write square units.

Sketch it in your answer

Draw x-axis and y-axis, sketch y = f(x), shade the region between the curve and x-axis from x = a to x = b, and label vertical boundaries x = a and x = b.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Find the area bounded by y = 3x, the x-axis, x = 0 and x = 2.

Think first, then show answer

Area = ∫₀² 3x dx = [3x²/2]₀² = 6 square units.

Common wrong answer

Using ∫_-2² x² - 4 dx for y = 4 - x² gives a negative value, because the integrand has been reversed relative to the shaded region.

Don't mix this up

Students often write ∫_a^b y dx without checking whether y is above or below the x-axis throughout the interval.

Exam tip

Before integrating, mark the limits on the x-axis and check whether the curve stays on one side of the x-axis. This avoids losing marks due to signed-area confusion.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Usually asked as a short or long answer question requiring limits from given boundaries, direct integration, and final area in square units. It may also appear with a curve and two vertical lines, where no intersection solving is needed.

1
Enough for 1 mark
The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is found by adding thin vertical strips from x = a to x = b. If f(x) is non-negative on [a, b], the area is ∫_a^b f(x) dx.
2
Add for 2 marks
Example: The area bounded by y = x², the x-axis, x = 0 and x = 2 is ∫₀² x² dx = [x³/3]₀² = 8/3 square units.
3
Add for 3 marks
Use this result when the required region lies between a single curve and the x-axis over a fixed interval.

MCQ Quiz

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