Application of Integrals: Chapter 8 Links
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Application of Integrals 8.1 introduces one of the most important applications of definite integrals—finding the area of a region bounded by curves. In this exercise, we will use definite integrals to calculate the area enclosed by a curve, the coordinate axes or a pair of straight lines.
\displaystyle \text{Area}=\int_a^b f(x)\,dx
Before evaluating an integral, it is important to identify the correct limits of integration and determine whether the curve lies above or below the x-axis. Since area is always a positive quantity, the absolute value of the integral is taken whenever required.
In this exercise, we will learn how to find the area enclosed by straight lines, curves and the coordinate axes using definite integrals. A clear understanding of the graphs and the region enclosed (also available in video format) will help you solve these questions accurately for the CBSE board examinations, CUET and other competitive exams.
Key Concepts
Before solving Application of Integrals 8.1, it is important to understand how a definite integral is used to find the area of a region bounded by curves. In most questions, drawing a rough sketch of the graph first makes it much easier to identify the required region, the limits of integration and the correct integral.
1. Area Under a Curve
The area of a region bounded by a curve and the coordinate axes can be found using a definite integral.
\displaystyle \text{Area}=\int_a^b f(x)\,dx
This formula is applicable when the curve lies completely above the x-axis over the interval [a,b].
2. Area Above and Below the x-axis
If the curve lies above the x-axis, the value of the definite integral itself gives the required area.
\displaystyle \text{Area}=\int_a^b f(x)\,dx
If the curve lies below the x-axis, the definite integral may be negative. Since area is always positive, we take its absolute value.
\displaystyle \text{Area}=\left|\int_a^b f(x)\,dx\right|
In other words, we first evaluate the definite integral and, if the answer is negative, simply ignore the negative sign while writing the area.
3. Curve Crossing the x-axis
If a curve crosses the x-axis within the given interval, divide the region at the point(s) where it cuts the x-axis and calculate the area of each part separately.
\displaystyle \text{Area}=\left|\int_a^c f(x)\,dx\right|+\left|\int_c^b f(x)\,dx\right|
This ensures that every part of the required area is counted as a positive quantity.
4. Area with Respect to the x-axis and y-axis
Depending on the given region, the area may be calculated by integrating with respect to x or y.
- Area with respect to the x-axis (vertical strips): \displaystyle \text{Area}=\int_a^b\big(y_{\text{upper}}-y_{\text{lower}}\big)\,dx
- Area with respect to the y-axis (horizontal strips): \displaystyle \text{Area}=\int_c^d\big(x_{\text{right}}-x_{\text{left}}\big)\,dy
Choose the variable of integration that gives the simplest expression and avoids splitting the required region into multiple parts.
5. Steps to Find Area Using Definite Integrals
- Draw a rough sketch of the given graph(s).
- Identify the region whose area is to be calculated.
- Find the points of intersection to determine the limits of integration.
- Form the appropriate definite integral.
- Evaluate the integral and write the final area as a positive quantity.
6. Standard Curves Used in Area Problems
The following standard curves frequently appear in questions based on the application of integrals. A quick sketch of these graphs helps in identifying the required region and setting up the correct definite integral.
| Standard Curves and Their Important Features |
|---|
| Coordinate Axes x-axis : y=0 y-axis : x=0 |
| Straight Lines General form : Ax+By+C=0 Slope-intercept form : y=mx+c Intercept form : \displaystyle \frac{x}{a}+\frac{y}{b}=1 Horizontal line (parallel to x-axis) : y=c Vertical line (parallel to y-axis) : x=c Special line : y=x (Makes an angle of 45^\circ with the positive x-axis.) The intercept form cuts the x-axis at (a,0) and the y-axis at (0,b), making it easy to sketch the graph. |
| Circle Standard equation : x^2+y^2=r^2 General equation : x^2+y^2+2gx+2fy+c=0 Centre : (0,0) (Standard Form) Radius : r Area : \pi r^2 Upper semicircle : y=\sqrt{r^2-x^2} Lower semicircle : y=-\sqrt{r^2-x^2} |
| Ellipse Horizontal major axis : \displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a>b Vertical major axis : \displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1,\quad a>b Centre : (0,0) Major axis : 2a Minor axis : 2b Vertices : (\pm a,0) or (0,\pm a) Area : \pi ab |
| Parabolas y^2=4ax Opens right • Vertex : (0,0) Focus : (a,0) Directrix : x=-a Length of Latus Rectum : 4a y^2=-4ax Opens left • Vertex : (0,0) Focus : (-a,0) Directrix : x=a Length of Latus Rectum : 4a x^2=4ay Opens upward • Vertex : (0,0) Focus : (0,a) Directrix : y=-a Length of Latus Rectum : 4a x^2=-4ay Opens downward • Vertex : (0,0) Focus : (0,-a) Directrix : y=a Length of Latus Rectum : 4a |
| Other Common Curves Modulus graph : y=|x| Sine curve : y=\sin x Cosine curve : y=\cos x |
| Choosing the Variable of Integration Integrate with respect to x when the region is more conveniently divided into vertical strips. \displaystyle \text{Area}=\int_a^b\left(y_{\text{upper}}-y_{\text{lower}}\right)\,dx Integrate with respect to y when the region is more conveniently divided into horizontal strips. \displaystyle \text{Area}=\int_c^d\left(x_{\text{right}}-x_{\text{left}}\right)\,dy |
Tip: Before forming the definite integral, always draw a rough sketch of the given graph(s). A simple sketch helps identify the required region, choose the correct limits of integration and avoid sign errors.
NCERT Solved example: Area of Circle Using Integration
Example: Find the Area Enclosed by the Circle x^2+y^2=a^2
Solution
Method 1: Using Vertical Strips (Integrating with respect to x)
Think First… Before forming the integral, identify the given curve and draw a rough sketch. A simple graph helps determine the required region, the limits of integration and whether to integrate with respect to x or y.
Step 1: Identify the Curve
The given equation
x^2+y^2=a^2
represents a circle with
- Centre : (0,0)
- Radius : a
Step 2: Points of Intersection
The circle cuts the coordinate axes at the following points:
- x-axis at (a,0) and (-a,0)
- y-axis at (0,a) and (0,-a)
Step 3: Rough Sketch
Draw a rough sketch of the circle and shade the region in the first quadrant bounded by the curve, the x-axis and the y-axis.
Since the circle is symmetrical about both the x-axis and the y-axis, it is sufficient to find the area of the shaded region in the first quadrant and then multiply the result by 4.
Step 4: Choose the Variable of Integration
Think First… We shall first integrate with respect to x. Therefore, we consider vertical strips of small thickness dx.
From the equation of the circle,
x^2+y^2=a^2
we get
y^2=a^2-x^2
Taking square roots,
y=\pm\sqrt{a^2-x^2}
Since the shaded region lies entirely in the first quadrant, y is positive.
Therefore,
y=\sqrt{a^2-x^2}
Step 5: Form the Required Integral
Required Area
\displaystyle =4\times\text{Area of the shaded region AOBA}
\displaystyle =4\int_0^a y\,dx
Substituting y=\sqrt{a^2-x^2}, we get
\displaystyle =4\int_0^a\sqrt{a^2-x^2}\,dx
Using the standard result,
\displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C
Step 6: Integrate within Limits to get the Area
Therefore,
Redq. Area \displaystyle =4\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)\right]_0^a
\displaystyle =4\left[\left(\frac{a}{2}\times0+\frac{a^2}{2}\times\frac{\pi}{2}\right)-\left(0+\frac{a^2}{2}\times0\right)\right]
\displaystyle =4\left(\frac{a^2\pi}{4}\right)=\pi a^2
Hence, the required area enclosed by the circle is
\boxed{\displaystyle \text{Area}=\pi a^2}
Verification
The area of a circle of radius a is
\displaystyle \pi r^2=\pi a^2
This agrees with the result obtained using definite integrals.
✅️ Hence, our answer is verified.
Method 2: Using Horizontal Strips (Integrating with respect to y)
Think First… This time, we integrate with respect to y. Hence, we consider horizontal strips of small thickness dy. The length of each strip is equal to the x-coordinate of the circle.
Step 1: Express x in terms of y
From the equation of the circle,
x^2+y^2=a^2
we get
x^2=a^2-y^2
Taking square roots,
x=\pm\sqrt{a^2-y^2}
Since the shaded region lies in the first quadrant, x is positive.
Therefore,
x=\sqrt{a^2-y^2}
Step 2: Form the Required Integral
Required Area
\displaystyle =4\times\text{Area of the shaded region AOBA}
\displaystyle =4\int_0^a x\,dy
Substituting x=\sqrt{a^2-y^2}, we get
\displaystyle =4\int_0^a\sqrt{a^2-y^2}\,dy
Using the standard result,
\displaystyle \int\sqrt{a^2-y^2}\,dy=\frac{y}{2}\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{y}{a}\right)+C
Therefore,
Reqd. Area \displaystyle =4\left[\frac{y}{2}\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{y}{a}\right)\right]_0^a
\displaystyle =4\left[\left(\frac{a}{2}\times0+\frac{a^2}{2}\times\frac{\pi}{2}\right)-\left(0+\frac{a^2}{2}\times0\right)\right]
\displaystyle =4\left(\frac{a^2\pi}{4}\right)=\pi a^2
Hence, the required area enclosed by the circle is
\boxed{\displaystyle \text{Area}=\pi a^2}
Key Takeaway The area enclosed by a region remains the same whether we integrate with respect to x (using vertical strips) or with respect to y (using horizontal strips). In practice, choose the method that gives the simpler integral and requires fewer calculations.
Maths Better Tip (Pre-Rough Sketch): Before writing the solution, first draw a small rough sketch in the rough work area. This helps you understand the region enclosed, identify the boundary curves, choose the correct limits of integration and decide whether to integrate with respect to x or y. Once the region is clear, draw the final rough sketch neatly as part of your answer.
Let us now solve all the NCERT questions step by step in the only Exercise of Application of Integrals 8.1.
Question 1: Application of Integrals 8.1
1. Find the area of the region bounded by the ellipse \displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1
Solution
The given equation represents an ellipse with centre at (0,0), semi-major axis a=4 and semi-minor axis b=3.
Since the ellipse is symmetrical about both the x-axis and the y-axis, we find the area of the region in the first quadrant and multiply it by 4.
From \displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1
We get \displaystyle y^2=9\left(1-\frac{x^2}{16}\right)=\frac{9}{16}(16-x^2)
Therefore, \displaystyle y=\frac34\sqrt{16-x^2}
Hence, as shown in the graph, the shaded region will give:
The Required Area \displaystyle =4\int_0^4y\,dx
\displaystyle =4\int_0^4\frac34\sqrt{16-x^2}\,dx
\displaystyle =3\int_0^4\sqrt{16-x^2}\,dx
Using the standard result,
\displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C
we get
Area \displaystyle =3\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)\right]_0^4
\displaystyle =3\left[(0+8\times\frac{\pi}{2})-(0+0)\right]
\displaystyle =12\pi
Therefore,
\boxed{\displaystyle \text{Reqd. Area}=12\pi\text{ square units}}
Verification: The area of an ellipse is \pi ab. Here, a=4 and b=3. Therefore,
\displaystyle \pi ab=\pi(4)(3)=12\pi
✅️ Hence, our answer is verified.
You may already be following Maths Better for the NCERT Solutions of the Class 12 Maths chapters covered so far, including
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
- Application of Derivatives
- Integrals
Having completed Integrals, you are now learning how to apply those concepts to find the area of regions bounded by curves. Continue practising these Application of Integrals questions carefully, and now let’s proceed to the next question.
Question 2: Application of Integrals 8.1
2. Find the area of the region bounded by the ellipse \displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1
Solution
The given equation represents an ellipse with centre at (0,0), semi-major axis a=3 and semi-minor axis b=2.
Since the ellipse is symmetrical about both the x-axis and the y-axis, we find the area of the region in the first quadrant and multiply it by 4.
From \displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1
We get \displaystyle y^2=9\left(1-\frac{x^2}{4}\right)=\frac94(4-x^2)
Therefore, \displaystyle y=\frac32\sqrt{4-x^2}
Hence, as shown in the graph, the shaded region will give:
The Required Area \displaystyle =4\int_0^2y\,dx
\displaystyle =4\int_0^2\frac32\sqrt{4-x^2}\,dx
\displaystyle =6\int_0^2\sqrt{4-x^2}\,dx
Using the standard result,
Area \displaystyle =6\left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)\right]_0^2
\displaystyle =6\left[(0+2\times\frac{\pi}{2})-(0+0)\right]
\displaystyle =6\pi
Therefore,
\boxed{\displaystyle \text{Area}=6\pi\text{ square units}}
Verification: The area of an ellipse is \pi ab. Here, a=3 and b=2. Therefore,
\displaystyle \pi ab=\pi(3)(2)=6\pi
✅️ Hence, our answer is verified.
Maths Better Tip: Whenever the required region is bounded by an ellipse, you can quickly verify your answer using the standard area formula \pi ab, where a and b are the lengths of the semi-major and semi-minor axes respectively. This provides a quick and reliable check of your final answer.
Question 3: Application of Integrals – MCQ
3. Choose the correct answer.
Area lying in the first quadrant and bounded by the circle x^2+y^2=4 and the lines x=0 and x=2 is
- (A) \pi
- (B) \displaystyle \frac{\pi}{2}
- (C) \displaystyle \frac{\pi}{3}
- (D) \displaystyle \frac{\pi}{4}
Solution
Think First… The required region lies entirely in the first quadrant. Since the limits are given as x=0 and x=2, it is convenient to integrate with respect to x using vertical strips.
From the equation of the circle,
x^2+y^2=4
We get \displaystyle y=\sqrt{4-x^2}
Therefore, the required area is
\displaystyle \int_0^2\sqrt{4-x^2}\,dx
Using the standard result,
Area \displaystyle =\left[\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)\right]_0^2
\displaystyle =(0+\pi)-0=\pi
✅️ Hence, the correct answer is (A).
Maths Better MCQ Shortcut: Observe the geometry of the figure before evaluating the integral. Here, the required region is simply a \dfrac14 of a circle of radius 2. Therefore, the area can be found directly as \displaystyle \frac14\pi(2)^2=\pi. Such observations can save valuable time in MCQs and competitive exams.
Question 4: Application of Integrals 8.1 – MCQ
4. Choose the correct answer.
Area of the region bounded by the curve y^2=4x, the y-axis and the line y=3 is
- (A) 2
- (B) \displaystyle \frac94
- (C) \displaystyle \frac93
- (D) \displaystyle \frac92
Solution
Think First… The curve is given in the form x=f(y). Hence, it is convenient to integrate with respect to y using horizontal strips.
From y^2=4x
We get \displaystyle x=\frac{y^2}{4}
The required region, as shown in the graph, is bounded by the y-axis, i.e., x=0, and the line y=3. Therefore, the limits of integration are 0 to 3.
Hence, the required area is, say
A \displaystyle =\int_0^3\left(\frac{y^2}{4}\right)dy
\displaystyle =\frac14\int_0^3y^2\,dy
\displaystyle =\frac14\left[\frac{y^3}{3}\right]_0^3
Substituting the limits, we get
Area \displaystyle =\frac14\left(\frac{27}{3}\right)=\frac94
✅️ Hence, the correct answer is (B).
Maths Better MCQ Shortcut: The equation is already given in the form x=f(y). Hence, integrating with respect to y using horizontal strips gives the simplest solution. Whenever possible, choose the variable of integration that makes the integral shorter and avoids unnecessary calculations.
Common Mistakes to Avoid
- Skipping the pre-rough sketch: Before starting the solution, draw a small rough sketch in the rough work area. It helps identify the required region, the boundary curves and the correct limits of integration.
- Not drawing the graph in the answer: A neat rough sketch as part of the solution makes the enclosed region, limits and method of integration much easier to understand and explain.
- Choosing the wrong variable of integration: Before forming the integral, decide whether integrating with respect to x or y gives the simpler solution. Use vertical strips for dx and horizontal strips for dy.
- Using incorrect limits of integration: Always find the points of intersection carefully from the graph or the given equations before writing the limits.
- Ignoring symmetry: If the figure is symmetrical about the x-axis, y-axis or both, calculate the area of a convenient portion and multiply accordingly to reduce the calculations.
- Ignoring the geometry of the figure: Before writing the integral, identify the boundary curves, the enclosed region and the points of intersection. Spending a minute understanding the figure often saves several minutes of unnecessary calculations.
- Using the wrong boundary curve: While integrating, always identify the upper and lower curves (or the right and left curves) correctly before forming the integral.
- Writing the wrong integrand: For integration with respect to x, use y_{\text{upper}}-y_{\text{lower}}. For integration with respect to y, use x_{\text{right}}-x_{\text{left}}.
- Not expressing the curve correctly: Before integrating, rewrite the equation in the required form, such as y=f(x) or x=f(y), depending on the chosen variable of integration.
- Ignoring the sign of the integral: If the definite integral evaluates to a negative value, remember that the area of a region is always positive. Hence, write the final answer as a positive quantity.
- Not verifying the final answer: Whenever possible, verify your result using a standard area formula, such as \pi r^2 for a circle or \pi ab for an ellipse. This is a quick way to confirm your calculations.
Continue Learning
Congratulations on completing Application of Integrals 8.1! In this exercise, you learned how to use definite integrals to find the area of regions bounded by curves and coordinate axes. More importantly, you practised identifying the required region, choosing the appropriate variable of integration and forming the correct definite integral before carrying out the calculations.
Before moving to the Miscellaneous Exercise, make sure that you revise:
- Drawing a pre-rough sketch and a neat rough sketch to identify the enclosed region correctly
- Finding the area of a region using definite integrals
- Choosing whether to integrate with respect to x or y based on the given figure
- Using vertical strips for dx and horizontal strips for dy
- Finding the correct limits of integration from the graph or points of intersection
- Making use of symmetry to simplify calculations whenever applicable
- Writing the integrand correctly as y_{\text{upper}}-y_{\text{lower}} or x_{\text{right}}-x_{\text{left}}
- Expressing the given curve in the required form, such as y=f(x) or x=f(y), before integrating
- Writing the final area as a positive quantity whenever required
- Verifying your answer using standard area formulas, such as \pi r^2 for a circle or \pi ab for an ellipse, whenever possible
Explore More
Finding the area using definite integrals becomes much easier once you develop the habit of understanding the geometry of the given figure before starting the calculations. Always begin with a pre-rough sketch, identify the boundary curves, determine the points of intersection and decide whether to integrate with respect to x or y. Form the definite integral carefully, choose the correct limits of integration and remember that the final area is always a positive quantity. Whenever possible, verify your answer using a standard area formula, such as \pi r^2 for a circle or \pi ab for an ellipse. With regular practice, you will learn to visualise the required region quickly and solve application of integrals problems with greater speed, accuracy and confidence.
All the best and keep learning 👍
