Application of Integrals (Area under the curve) | ISC Class 12 Maths Notes

This note covers elementary area strips, definite integrals, signed integrals and geometric area, choice of integration variable, areas involving lines, polynomials, circles, parabolas and ellipses, areas between curves, modulus functions, exponential functions and logarithmic functions.

How does integration measure an area under a curve?

A bounded region is a part of the plane enclosed by specified boundaries. These boundaries may include a curve, straight lines and coordinate axes. The horizontal coordinate is x, the vertical coordinate is y, and their axes meet at the origin, the point (0, 0).

Write a curve as y = f(x), where f is a function assigning a y-value to each permitted x-value. The graph consists of the corresponding points (x, y). The x-axis has equation y = 0; the y-axis has equation x = 0.

What is an elementary strip?

Take a region above the x-axis between the vertical lines x = a and x = b, where a and b are fixed numbers with a < b. An elementary strip is a very thin part of this region used to build its area.

A vertical strip has height y and infinitesimal width dx. The symbol dA denotes its elementary area, while A denotes the total area. Thus dA = y dx. Adding the strips by integration gives the area of the whole region.

Result: Area above the x-axis

For a continuous, non-negative function f on [a, b],

$$A=\int_a^b f(x)\,dx.$$

The closed interval [a, b] contains both endpoints and all real numbers between them. Here continuous means that the graph has no break on the interval. The integral sign ∫ represents integration; dx identifies x as the variable of integration.

The shaded region lies above the x-axis, below y = f(x), and between x = a and x = b. A narrow vertical strip is labelled with height y and width dx.

See Fig. 8.1 in your NCERT textbook

Each boundary matters. The curve supplies the strip's height, while the two vertical lines specify where the addition begins and ends. The answer is expressed in square units, meaning the area units corresponding to the units used on the coordinate axes.

How is a definite integral evaluated?

A definite integral has specified lower and upper limits and produces a value. In the notation ∫ from a to b, a is the lower limit and b is the upper limit. The function being integrated is the integrand.

An antiderivative of f is a function F whose derivative is f. A derivative measures the rate of change of a function. The notation F′(x) = f(x) states this relationship, with the prime indicating differentiation with respect to x.

Theorem: Evaluation by an antiderivative

If f is continuous on [a, b] and F is an antiderivative of f, the second fundamental theorem of integral calculus gives

$$\int_a^b f(x)\,dx=[F(x)]_a^b=F(b)-F(a).$$

The bracket notation means upper-limit value minus lower-limit value. Use the same antiderivative at both endpoints. An indefinite integral gives the family of antiderivatives F(x) + C, where C is an arbitrary constant. That constant cancels from the endpoint subtraction and is unnecessary in the final definite value.

What steps keep evaluation clear?

1. Identify the region and decide whether the integrand represents a positive strip length throughout the interval.
2. Find an antiderivative and check its derivative against the integrand.
3. Substitute the upper limit into the complete antiderivative, then substitute the lower limit.
4. Subtract the second value from the first and interpret the result using the region's position.

For example, an antiderivative of x² is x³/3. This is useful for the area below the parabola y = x², a curved graph with its lowest point at the origin. The definite integral still needs boundaries before it gives a particular area.

Note: Finding an antiderivative completes the integration step, but identifying the correct region comes first. A correct antiderivative cannot repair limits or a strip length chosen for a different region.

Why can a definite integral differ from geometric area?

A signed integral combines contributions with their signs. A curve above the x-axis contributes positively; a curve below it contributes negatively when integration proceeds from the smaller x-value to the larger one. Geometric area counts the size of each region positively.

The absolute value or modulus of a real number is its non-negative magnitude, written using vertical bars. For a curve entirely below the x-axis on [a, b], negate its definite integral, or take the absolute value of that integral, to obtain its area.

Result: Add positive areas after splitting

If the curve crosses the x-axis, separate the interval at the relevant zeros, meaning the x-values where f(x) = 0. Find each piece's positive area before adding. Equivalently, for continuous f,

$$A=\int_a^b |f(x)|\,dx.$$

Taking the absolute value of one integral over the whole interval can leave cancellation between positive and negative pieces. It does not generally give the total geometric area.

Worked example 1. Find the area bounded by y = 3x + 2, the x-axis, x = −1 and x = 1.

The line crosses the x-axis when 3x + 2 = 0, giving x = −2/3. It lies below the axis from −1 to −2/3 and above it from −2/3 to 1.

$$A=-\int_{-1}^{-2/3}(3x+2)\,dx+\int_{-2/3}^{1}(3x+2)\,dx.$$

Use the antiderivative 3x²/2 + 2x. The first positive area is 1/6 and the second is 25/6.

Answer: A = 1/6 + 25/6 = 13/3 square units.

The line y = 3x + 2 forms one shaded triangular region below the axis beside x = −1 and another above it ending at x = 1.

See Fig. 8.9 in your NCERT textbook

The two pieces share a boundary point, but their areas are added. The negative sign before the first integral converts its negative signed contribution into the positive size of that piece.

When should horizontal strips replace vertical strips?

A horizontal strip is useful when the curve is conveniently written as x = g(y), where g assigns an x-value to each permitted y-value. Its infinitesimal thickness is dy, so integration proceeds with respect to y.

Let c and d be the lower and upper y-values, with c < d. If the curve lies to the right of the y-axis, the strip extends from x = 0 to x = g(y). Its length is g(y), and its elementary area is g(y) dy.

Result: Area beside the y-axis

For continuous g with g(y) ≥ 0 on [c, d],

$$A=\int_c^d g(y)\,dy.$$

The limits now describe vertical position. They must be y-values. The strip's length is an x-distance, but its thickness and integration variable are dy. This distinction prevents the common mistake of pairing a horizontal strip with x-limits.

The shaded region lies between the y-axis and x = g(y), with horizontal boundaries y = c and y = d. A strip is labelled with length x and thickness dy.

See Fig. 8.2 in your NCERT textbook

Worked example 2. Find the area bounded by the parabola y² = 4x, the y-axis and the line y = 3.

The parabola meets the y-axis at y = 0. Write it as x = y²/4. A horizontal strip extends from x = 0 to x = y²/4, while y runs from 0 to 3.

$$A=\int_0^3\frac{y^2}{4}\,dy=\left[\frac{y^3}{12}\right]_0^3.$$

Answer: A = 27/12 = 9/4 square units.

The specified line y = 3 selects the upper enclosed region. Including the corresponding region below the x-axis would change the question. Choosing horizontal strips makes both the requested boundaries and the simple polynomial integral visible.

How are areas under polynomial curves calculated?

A polynomial is a sum of constant multiples of non-negative whole-number powers of a variable. The area method uses the same strip construction as before. First establish the sign of the polynomial on the given interval, then integrate the appropriate strip height.

What does the power rule contribute?

For a non-negative whole number n, an antiderivative of xⁿ is xⁿ⁺¹/(n + 1). Here n denotes the exponent. Polynomial terms can be integrated separately, with their coefficients, the constant multipliers of the powers, retained. The limits are applied after an antiderivative has been found.

Worked example 3. Find the area bounded by y = x², x = 1, x = 2 and the x-axis.

The curve is above the axis on this interval, so its height is x².

$$A=\int_1^2 x^2\,dx=\left[\frac{x^3}{3}\right]_1^2=\frac{8-1}{3}.$$

Answer: A = 7/3 square units.

Worked example 4. Find the area bounded by y = x⁴, x = 1, x = 5 and the x-axis.

The height is non-negative throughout the interval. Integrating and applying the endpoints gives

$$A=\int_1^5 x^4\,dx=\left[\frac{x^5}{5}\right]_1^5=\frac{3125-1}{5}.$$

Answer: A = 3124/5 square units.

These examples use the stated vertical boundaries. The fact that each curve also reaches the origin does not make zero a limit for these particular regions. Limits belong to the requested area, rather than automatically to the intercepts of a curve.

If a polynomial changes sign within the interval, the power rule is unchanged, but the area setup needs separate pieces. Solving for its zeros and checking the signs therefore precedes the integration step.

How does symmetry help find the area of a circle?

A circle consists of points at a fixed distance, its radius, from its centre. In x² + y² = a², a is now a positive radius, rather than an integration endpoint chosen independently. The centre is the origin.

The circle is symmetric about both coordinate axes: reflecting it across either axis leaves it unchanged. The axes divide the plane into four quadrants. In the first-quadrant region, including its axis boundaries, both coordinates are non-negative. The required branch, meaning the portion described by one choice of y, is y = √(a² − x²), where √ denotes the non-negative square root.

Why choose the positive square root?

The negative branch describes the lower semicircle, which is the lower half of the circle. For a first-quadrant strip, the positive branch gives its height above the x-axis.

Worked example 5. Use integration to find the area enclosed by x² + y² = a², where a > 0.

Four equal quadrant areas make the whole circle. Hence

$$A=4\int_0^a\sqrt{a^2-x^2}\,dx.$$

Use the antiderivative (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a). Here sin⁻¹ means inverse sine, with angles in radians, the angle measure defined by arc length divided by radius.

At x = a the square-root term is zero and sin⁻¹(1) = π/2; at x = 0 both terms are zero. The constant π is the ratio of a circle's circumference to its diameter.

Answer: A = 4 × (πa²/4) = πa² square units.

Horizontal strips produce the same answer using x = √(a² − y²), with y running from 0 to a. In either method, the factor four is justified by equal quadrant regions, not by the appearance of a square root.

How is the area of an ellipse obtained?

An ellipse in the standard form x²/a² + y²/b² = 1 is a closed curve centred at the origin. Here a and b are positive semi-axis lengths, the distances from the centre to the curve along the x-axis and y-axis respectively.

Its intercepts, the points where it meets the axes, are (a, 0), (−a, 0), (0, b) and (0, −b). Reflection in either axis preserves the equation, so the four quadrant regions have equal areas.

How is the circle integral reused?

Worked example 6. Find the area enclosed by x²/a² + y²/b² = 1, where a > 0 and b > 0.

In the first quadrant, y = (b/a)√(a² − x²), and x runs from 0 to a. Therefore

$$A=\frac{4b}{a}\int_0^a\sqrt{a^2-x^2}\,dx.$$

The integral is πa²/4, as obtained in the circle calculation.

Answer: A = (4b/a)(πa²/4) = πab square units.

The ellipse is labelled at its four axis intercepts. Its first-quadrant region is shaded, and a vertical strip has height y and width dx.

See Fig. 8.7 in your NCERT textbook

Worked example 7. Find the area bounded by x²/16 + y²/9 = 1.

Comparing with standard form gives a² = 16 and b² = 9. Thus a = 4 and b = 3. The denominators are squares of semi-axis lengths, not the lengths themselves.

Answer: A = πab = π × 4 × 3 = 12π square units.

Using horizontal strips instead gives x = (a/b)√(b² − y²), integrated from y = 0 to y = b and multiplied by four. Notice that the upper limit changes with the variable: a belongs to the x-direction, while b belongs to the y-direction.

How is the area enclosed between two curves found?

Let y = f(x) and y = g(x) be two continuous curves on [a, b]. Here f and g describe their heights, and a and b are the left and right limits of the chosen region. The intersection points are points satisfying both equations.

If f(x) ≥ g(x) throughout the interval, a vertical strip begins on the lower curve and ends on the upper curve. Its height is f(x) − g(x), even if both curves lie below the x-axis.

Result: Upper curve minus lower curve

$$A=\int_a^b [f(x)-g(x)]\,dx.$$

If the order changes, split at the relevant intersections and use the positive difference on each interval. Equivalently, integrate |f(x) − g(x)|. An intersection is found by solving f(x) = g(x), but the stated boundaries still decide which intersections belong to the requested region.

How does an absolute-difference integral work?

Worked example 8. Evaluate the definite integral of |x³ − x| from x = −1 to x = 2.

The expression x³ − x factors as x(x − 1)(x + 1), so its zeros are −1, 0 and 1. Its sign is non-negative on [−1, 0], non-positive on [0, 1], and non-negative on [1, 2].

$$\int_{-1}^{2}|x^3-x|\,dx=\int_{-1}^{0}(x^3-x)\,dx+\int_0^1(x-x^3)\,dx+\int_1^2(x^3-x)\,dx.$$

Using x⁴/4 − x²/2 for x³ − x gives positive contributions 1/4, 1/4 and 9/4.

Answer: The integral is 11/4.

This integral also measures the area between y = x³ and y = x over the stated interval, with x = 2 closing its right-hand edge. Their vertical separation is |x³ − x|. The change in upper curve explains the splitting.

For horizontal strips, write the right boundary as x = r(y) and the left boundary as x = l(y), where r and l are their respective x-coordinate functions. Integrate r(y) − l(y) between the lower and upper y-limits, splitting wherever their order changes.

How are modulus functions handled in area problems?

A modulus function uses absolute value. For a real expression u, |u| = u when u ≥ 0, and |u| = −u when u < 0. Here u stands for the expression inside the modulus signs.

To integrate such a function, determine where its inside expression vanishes. On each interval between those values, replace the modulus by the appropriate formula. This is a piecewise description, meaning that different formulae apply on different parts of the domain, the permitted set of inputs.

Where must the formula change?

Worked example 9. Evaluate the integral of |x + 3| from x = −6 to x = 0.

The expression x + 3 vanishes at x = −3. Thus |x + 3| = −x − 3 on [−6, −3] and |x + 3| = x + 3 on [−3, 0].

$$\int_{-6}^{0}|x+3|\,dx=\int_{-6}^{-3}(-x-3)\,dx+\int_{-3}^{0}(x+3)\,dx.$$

The two integrals are 9/2 and 9/2. Answer: The integral is 9, also the area in square units below this graph and above the x-axis on the given interval.

Does a modulus inside a product make the product positive?

Worked example 10. Find the area bounded by y = x|x|, the x-axis, x = −1 and x = 1.

For x < 0, x|x| = −x²; for x ≥ 0, x|x| = x². The graph is below the axis on the negative interval, so negate that part's signed integral.

$$A=-\int_{-1}^{0}(-x^2)\,dx+\int_0^1x^2\,dx=\frac13+\frac13.$$

Answer: A = 2/3 square units.

The outside factor x retains its sign. Consequently x|x| can be negative although |x| cannot. Distinguish removing the modulus from converting a signed integral to area; these are separate decisions in this example.

How do exponential and logarithmic curves fit the area method?

An exponential function has its variable in the exponent. For y = eˣ, e denotes the base of natural logarithms. This function is positive for every real x, and its antiderivative is eˣ.

What are the boundaries for an exponential area?

For finite real numbers a < b, the region between y = eˣ, the x-axis, x = a and x = b has area

$$A=\int_a^b e^x\,dx=e^b-e^a.$$

The positivity of eˣ makes the signed integral equal to the geometric area on this interval. If a second curve forms the other boundary, use the difference between the two heights. The antiderivative alone does not identify that difference.

What must be checked for a logarithmic area?

The natural logarithm, written log x here, is the inverse of eˣ: if y = log x, then eʸ = x. Its real domain is x > 0. It vanishes at x = 1, is negative for 0 < x < 1, and positive for x > 1.

An antiderivative of log x is x log x − x. For 1 ≤ a < b, the area between the logarithmic curve and the x-axis is

$$A=\int_a^b\log x\,dx=[x\log x-x]_a^b.$$

For 0 < a < b ≤ 1, negate this integral. For 0 < a < 1 < b, split it at x = 1 and add the positive areas. These conditions keep both the domain and the sign explicit.

Integration by parts is the rule that integrates a product by transferring differentiation from one factor to another. To integrate log x, regard it as log x multiplied by 1. Integrating 1 gives x, while differentiating log x gives 1/x.

The rule then gives x log x minus the integral of x × (1/x), producing x log x − x. A finite closed interval used with the stated evaluation theorem must remain within the logarithm's domain.

How can a complete area solution be planned and checked?

A complete solution connects the boundaries to the integral and the integral to a positive area. Start with a rough sketch showing the relevant curves and lines. Its purpose is to identify the enclosed region, intercepts and changes of boundary.

Which decisions come before calculation?

1. List the given boundaries, including axes and specified vertical or horizontal lines.
2. Find intersections needed for limits or changes in the order of the curves.
3. Choose vertical or horizontal strips and express both ends of a strip in the same variable.
4. Split where the strip formula changes, then integrate and add the positive contributions.

A symmetry check can shorten the work, but the region itself must share the symmetry being used. A complete circle has four equal quadrant regions; a selected region bounded by an additional line need not have those same four equal pieces.

Finally, verify the antiderivative, the order of endpoint subtraction and any symmetry multiplier. Retain exact fractions and multiples of π when they arise. State the answer in square units, and ensure it represents all the requested pieces.

Glossary

• Bounded region — A part of the coordinate plane enclosed by the specified curves, lines or coordinate axes.
• Elementary area — The area of a very thin strip used in constructing an integral for a region.
• Definite integral — An integral with specified limits that evaluates the accumulated signed contribution over an interval.
• Integrand — The function being integrated, written after the integral sign and before the differential.
• Antiderivative — A function whose derivative equals the function that is to be integrated.
• Limits of integration — The endpoint values between which integration is performed in the chosen variable.
• Geometric area — The non-negative size of a region, obtained by adding the positive areas of its pieces.
• Absolute value — The non-negative magnitude of a real number, represented by enclosing it in vertical bars.
• Intersection point — A point whose coordinates satisfy the equations of both curves being considered.
• Symmetry about an axis — The property that reflection across the stated axis leaves the figure unchanged.
• Semi-axis length — The distance from an ellipse's centre to an intercept along one of its axes.
• Piecewise description — A description using different formulae on different specified parts of a function's domain.

Common errors and misconceptions

• Misconception: A definite integral is the total geometric area in every case. Correct: Parts below the x-axis contribute negatively; split into pieces and add their positive areas.
• Misconception: Taking the modulus after adding signed contributions removes every area error. Correct: Cancellation may already have occurred. Use a positive strip length on each interval before adding.
• Misconception: Horizontal strips use the same numerical limits as vertical strips. Correct: Horizontal strips require y-limits; vertical strips require x-limits, obtained from the appropriate boundaries.
• Misconception: Either square-root branch works for a quadrant calculation. Correct: Choose the branch belonging to the actual region, such as the positive y-branch in the first quadrant.
• Misconception: The denominators in an ellipse's standard equation are semi-axis lengths. Correct: They are the squares of those lengths, so take positive square roots before using the area formula.
• Misconception: A factor four can be used for any region involving a circle. Correct: It requires four equal pieces of the requested region, as in the complete circle.
• Misconception: The expression x|x| is non-negative because it contains modulus signs. Correct: The outside factor x is negative when x < 0, giving x|x| = −x² there.
• Misconception: The logarithmic area formula applies across zero. Correct: The real function log x requires x > 0; the stated closed-interval method must respect that domain.

Exam-style questions with model answers

Q1. A continuous curve y = f(x) lies below the x-axis throughout [a, b], where a < b. State an integral for its area bounded by the curve, the x-axis, x = a and x = b, and explain its sign. [2 marks]

1. The area is A = −∫ from a to b of f(x) dx, measured in square units.
2. The integral is negative because f(x) is negative throughout the interval, so its negative gives the positive geometric area.

Q2. Find the area bounded by y = x², the x-axis, x = 1 and x = 2. [3 marks]

1. The curve lies above the x-axis throughout [1, 2], so a vertical strip has height x² and width dx. Therefore A = ∫ from 1 to 2 of x² dx.
2. An antiderivative is x³/3. Applying the upper and lower limits gives A = 2³/3 − 1³/3.
3. Thus A = 8/3 − 1/3 = 7/3 square units, with no sign correction required.

Q3. Find the area bounded by the parabola y² = 4x, the y-axis and the line y = 3. Use horizontal strips. [3 marks]

1. Write the parabola as x = y²/4. Its meeting point with the y-axis has y = 0, so the required y-limits are 0 and 3.
2. A horizontal strip has length y²/4 − 0 and thickness dy. Hence A = ∫ from 0 to 3 of (y²/4) dy.
3. Using the antiderivative y³/12 gives A = 27/12 − 0 = 9/4 square units.

Q4. Find the total area bounded by y = 3x + 2, the x-axis, x = −1 and x = 1. Show how the change of sign is handled. [5 marks]

1. Set y = 0 to find the axis crossing: 3x + 2 = 0 gives x = −2/3, which lies inside the stated interval.
2. The line lies below the axis on [−1, −2/3] and above it on [−2/3, 1]. The two positive areas must be added.
3. An antiderivative is 3x²/2 + 2x. The integral over [−1, −2/3] is −1/6, so the first geometric area is 1/6.
4. Using the same antiderivative over [−2/3, 1] gives a positive integral of 25/6, which is the second geometric area.
5. The total area is therefore A = 1/6 + 25/6 = 13/3 square units. The split prevents cancellation between the two pieces.

Q5. Using integration, derive the area enclosed by x² + y² = a², where a > 0 is the radius. You may use the antiderivative (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) of √(a² − x²), where sin⁻¹ denotes inverse sine in radians. [5 marks]

1. The circle is centred at the origin and is symmetric about both coordinate axes. Its complete area is four times its first-quadrant area.
2. In that quadrant, y = √(a² − x²) and x runs from 0 to a, so A = 4∫ from 0 to a of √(a² − x²) dx.
3. Apply the supplied antiderivative. At x = a, the square-root term vanishes and sin⁻¹(1) = π/2, giving the value πa²/4.
4. At x = 0, both terms vanish, so subtracting the lower-limit value leaves the quadrant area equal to πa²/4.
5. Multiplying by four gives A = πa² square units, where π is the ratio of circumference to diameter.

Q6. Find the area bounded by x²/16 + y²/9 = 1. Identify the semi-axis lengths and show the quadrant integral, using ∫ from 0 to 4 of √(16 − x²) dx = 4π. [4 marks]

1. Comparing with x²/a² + y²/b² = 1 gives a = 4 and b = 3, the positive semi-axis lengths along the coordinate axes.
2. In the first quadrant the curve is y = (3/4)√(16 − x²), with x between 0 and 4.
3. By symmetry, the total area is four times the quadrant integral: A = 4 × (3/4)∫ from 0 to 4 of √(16 − x²) dx.
4. Using the given integral, A = 3 × 4π = 12π square units.

Q7. Find the area bounded by y = x|x|, the x-axis, x = −1 and x = 1, showing the formula on each side of zero. [4 marks]

1. For x < 0, |x| = −x, so y = −x². For x ≥ 0, |x| = x, so y = x².
2. The negative-side region lies below the axis. Its positive area is −∫ from −1 to 0 of (−x²) dx = 1/3.
3. The positive-side region lies above the axis. Its area is ∫ from 0 to 1 of x² dx = 1/3.
4. Adding the positive contributions gives the total area A = 2/3 square units.

Key takeaways

• A vertical strip uses a height expressed in x and thickness dx; its limits are x-values.
• A horizontal strip uses a length expressed in y and thickness dy; its limits are y-values.
• A definite integral counts signed contributions, while geometric area adds the non-negative sizes of the enclosed pieces.
• Between two curves, use upper minus lower for vertical strips, or right minus left for horizontal strips.
• Find crossings and changes of boundary before integrating, and split the interval wherever the required strip formula changes.
• Symmetry gives the complete circle area πa² and ellipse area πab from four equal quadrant areas.
• Remove modulus signs by checking the enclosed expression's sign on each relevant interval before performing integration.
• For logarithmic curves, respect the positive domain and split an area at x = 1 when the interval crosses it.

Test yourself

What does dx identify in a vertical-strip area integral?

It identifies the strip's infinitesimal width and shows that integration proceeds with respect to x.

Why does an arbitrary integration constant disappear from a definite integral?

The same constant occurs in both endpoint values, so subtracting the lower value from the upper cancels it.

Where must the line y = 3x + 2 be split for area over [−1, 1]?

Split at x = −2/3, where the line crosses the x-axis and changes the sign of its height.

For y² = 4x, the y-axis and y = 3, what are the horizontal-strip length and limits?

The strip length is y²/4 and the y-limits are 0 and 3, giving the integral of y²/4 over that interval.

Which branch of x² + y² = a² gives a first-quadrant vertical-strip height, for a > 0?

Use y = √(a² − x²), since the first-quadrant height is non-negative and x runs from 0 to a.

What are the semi-axis lengths for x²/16 + y²/9 = 1?

They are 4 along the x-axis and 3 along the y-axis, obtained from the positive square roots of the denominators.

How is |x + 3| written when x < −3?

It equals −(x + 3), or −x − 3, because the expression inside the modulus is negative.

What does ∫ from a to b of |f(x) − g(x)| dx measure for continuous curves y = f(x) and y = g(x), with a < b?

It measures the area between the curves over that interval, using their non-negative vertical separation even if their order changes.