NCERT Solutions for Maths Class 12 Exercise 8.1 Application of Integrals - FREE PDF Download
The NCERT Solutions for Maths Class 12 Exercise 8.1 in Chapter 8 - Application of Integrals, provided by Vedantu, to help students understand the practical applications of integrals. This exercise focuses on finding the area under curves and between curves, a crucial concept in calculus. By mastering these problems, students can develop a strong foundation in integral calculus, which is essential for advanced mathematical studies and applications.
- 4.1Exercise 8.1
It is important to focus on understanding the step-by-step methods used to solve these integral problems. Vedantu's ex 8.1 class 12 expert solutions simplify complex calculations, making it easier for students to grasp the concepts. Practice regularly to ensure a thorough understanding of the application of integrals in real-life scenarios and mathematical problems.
Glance on NCERT Solutions Maths Chapter 8 Exercise 8.1 Class 12 | Vedantu
Exercise 8.1 Class 12 explains the application of integrals in finding areas under curves.
An integral is a mathematical tool used to calculate the area under a curve by summing infinitesimal slices.
The exercise focuses on definite integrals, which compute the exact area between a curve and the x-axis within specified limits.
The area between two curves is found by integrating the difference between their functions over a given interval.
Problems include finding the areas of regions bounded by different curves and lines, such as parabolas, circles, and straight lines.
The exercise explores integrating complex functions to determine the area enclosed by intersecting curves.
There are 4 fully solved questions in Chapter 8 Ex 8.1 Class 12 Application of Integrals
Formulas Used in Class 12 Chapter 8 Exercise 8.1
Area under a Curve: $\int_{a}^{b}f\left ( x \right )dx$
Area between Two Curves: $\int_{a}^{b}[f\left ( x \right )-g\left ( x \right )]dx$
Area bounded by the y-axis: $\int_{c}^{d}[f\left ( y \right )-g\left ( y \right )]dy$
Exercise 8.1
1. Find the area of the region bounded by the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\].
Ans: Rewrite the given equation of ellipse
\[\frac{{{x}^{2}}}{{{4}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]
The intersection point of the ellipse and the x-axis are at \[x=\pm 4\].
The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].
The given equation of the ellipse can be represented as
If we change \[x\] to \[-x\] or \[y\] to \[y\] the equation remains the same.
So, the ellipse is symmetrical about the x-axis and the y-axis.
Area bounded by ellipse \[=4\times\] Area of OABO
Area of OABO \[=\int\limits_{0}^{4}{ydx}\]
\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\]
\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{16}\]
\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{16} \right)\]
\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{16}}\]
Area of OABO \[=\int\limits_{0}^{4}{3\sqrt{1-\frac{{{x}^{2}}}{16}}dx}\]
Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}dx}\]
Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{{{4}^{2}}-{{x}^{2}}}dx}\]
Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation
Area of OABO \[=\frac{3}{4}\left[ \frac{x}{2}\sqrt{16-{{x}^{2}}}+\frac{16}{2}{{\sin }^{-1}}\frac{x}{4} \right]_{0}^{4}\]
Area of OABO \[=\frac{3}{4}\left[ 0+8{{\sin }^{-1}}(1)-0-8{{\sin }^{-1}}(0) \right]_{0}^{4}\]
Area of OABO \[=\frac{3}{4}\left( \frac{8\pi }{2} \right)\]
Area of OABO \[=3\pi \]
Therefore, the area bounded by ellipse \[=\text{4}\times \text{3}\pi\]
The area bounded by ellipse \[=\text{12}\pi \text{ sq}\text{. units}\]
2. Find the area of the region bounded by the ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\].
Ans: Rewrite the given equation of the ellipse
\[\frac{{{x}^{2}}}{{{2}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]
The intersection point of the ellipse and the x-axis are at \[x=\pm 2\].
The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].
The given equation of the ellipse can be represented as a vertical ellipse.
It can be observed that the ellipse is symmetrical about the x-axis and y-axis.
Since, if we change \[x\] to \[-x\] or \[y\] to \[-y\] the equation remains the same.
Area bounded by ellipse \[=4\times\]Area of OABO
Area of OABO \[=\int\limits_{0}^{2}{ydx}\]…………………….(A)
\[\Rightarrow \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\]
\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{4}\]
\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{4} \right)\]
\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{4}}\text{ }\]
\[\Rightarrow y=\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }\]…………………..(1)
Area of OABO \[=\int\limits_{0}^{2}{\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }dx}\] using equation (1) in (A)
Area of OABO \[=\frac{3}{2}\int\limits_{0}^{2}{\sqrt{{{2}^{2}}-{{x}^{2}}}dx}\]
Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation
Area of OABO \[=\frac{3}{2}\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]
Area of OABO \[=\frac{3}{2}\left[ \frac{2\pi }{2} \right]\]
Area of OABO \[=\frac{3\pi }{2}\]
Area bound by the ellipse \[=\text{4}\times \frac{\text{3}\pi }{2}\]
Area bound by the ellipse \[=\text{6}\pi \text{ sq}\text{. units}\]
Choose the correct answer in the following Exercises 3 and 4.
3. Area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is
\[\pi \]
\[\frac{\pi }{2}\]
\[\frac{\pi }{3}\]
\[\frac{\pi }{4}\]
Ans:
Area of \[OAB=\int_{0}^{2}{ydx}\]
Since, \[{{\text{x}}^2} + {y^2} = 4\] $ [\text {Equation of circle}]$
\[{{y}^{2}}=4-{{\text{x}}^{2}}\]
\[y=\sqrt{4-{{\text{x}}^{2}}}\]
So,
Area of \[OAB=\int_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\]
Area of \[OAB=\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]
Area of \[OAB=\left[ \frac{2}{2}\sqrt{4-{{2}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{2}{2}-\left( \frac{0}{2}\sqrt{4-{{0}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{0}{2} \right) \right]\]
Area of \[OAB=\left[ \sqrt{4-4}+2{{\sin }^{-1}}1-0 \right]\]
Area of \[OAB=\left[ 2\times \frac{\pi }{2} \right]\]
Area of \[OAB=\pi \] square units
Hence, the area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is \[\pi \] square units.
Hence, option (A) is the correct answer.
4. Area of the region bounded by the curve \[{{y}^{2}}=4x\], y-axis and the line \[y=3\] is
2
\[\frac{9}{4}\]
\[\frac{9}{3}\]
\[\frac{9}{2}\]
Ans:
The area bounded by the curve \[{{y}^{2}}=4x\], y-axis and the line \[y=3\] is the area of the region \[OAB\].
Area of \[OAB=\int_{0}^{3}{xdy}\]
Area of \[OAB=\int_{0}^{3}{\frac{{{y}^{2}}}{4}dy}\]
Area of \[OAB=\frac{1}{4}\int_{0}^{3}{{{y}^{2}}dy}\]
Area of \[OAB=\frac{1}{4}\left[ \frac{{{y}^{3}}}{3} \right]_{0}^{3}\]
Area of \[OAB=\frac{1}{4}\left[ \frac{{{3}^{3}}}{3}-0 \right]\]
Area of \[OAB=\frac{1}{4}\left[ \frac{27}{3} \right]\]
Area of \[OAB=\frac{1}{4}\left[ 9 \right]\]
Area of \[OAB=\frac{9}{4}\]square units
Hence, option (B) is the correct answer.
Conclusion
NCERT Solutions for Class 12 Chapter 8 Exercise 8.1, provided by Vedantu, offer a comprehensive understanding of the application of integrals. This exercise focuses on calculating areas under curves and between curves, which is a fundamental aspect of integral calculus. It's important to concentrate on class 12 ex 8.1 setting up integrals correctly with the appropriate limits and understanding the difference between definite integrals and the area between curves. Regular practice of these problems will enhance your problem-solving skills and solidify your grasp of integral applications, which are crucial for advanced mathematics and various real-life applications.
Class 12 Maths Chapter 8: Exercises Breakdown
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NCERT Solutions for Class 12 Maths | Chapter-wise List
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FAQs on NCERT Solutions for Class 12 Maths Chapter 8 - Application of Integrals Exercise 8.1
1. What is the main focus of Class 12 Ex 8.1?
Exercise 8.1 focuses on applying integrals to find areas under curves and between curves. This is a fundamental concept in integral calculus, essential for understanding advanced mathematical applications.
2. What is a definite integral in Class 12 Ex 8.1?
A definite integral calculates the area under a curve between two specified points on the x-axis. It is represented as $\int_{a}^{b}f\left ( x \right )dx$, where a and b are the limits.
3. How do you find the area between two curves in Ex 8.1 Class 12 Maths NCERT Solutions?
In Ex 8.1 Class 12 Maths NCERT Solutions the area between two curves is found by integrating the difference of their functions over a given interval. This is represented as $\int_{a}^{b}[f\left ( x \right )-g\left ( x \right )]dx$ are the functions, and a and b are the limits.
4. Why is setting correct limits of integration important in ex 8.1 class 12 maths ncert solutions?
Setting correct limits of integration ensures accurate calculations of the area under or between curves. The limits define the interval over which the integral is evaluated. Incorrect limits can lead to inaccurate results and misunderstandings of the problem. This is especially important in applications where precision is crucial. Correct limits help in obtaining the true measure of the area or volume being calculated.
5. What types of problems are included in this Class 12 Maths Ex 8.1?
Problems in this exercise involve finding areas of regions bounded by various curves and lines. These include parabolas, circles, and straight lines, which are common shapes in geometry. The exercise helps students understand the application of integrals to calculate areas between these shapes. By solving these problems, students gain practical experience in setting up and evaluating integrals. This understanding is essential for more advanced studies in mathematics.
6. What is the significance of the Pythagorean theorem in Class 12 Maths Ex 8.1?
While the Pythagorean theorem itself is not directly used, understanding geometric properties is crucial. It aids in visualizing and solving problems related to areas under curves. This geometric understanding helps in accurately setting up and solving integral problems.
7. How do integrals help in real-life applications in Class 12 Maths Ex 8.1?
Integrals are used in various fields to calculate areas, volumes, and other quantities. They are essential in disciplines like physics, engineering, and economics. In class 12 maths ex 8.1 Integrals help model real-life scenarios and solve practical problems effectively.
8. What should students focus on while solving these problems in Class 12 Maths Chapter 8 Exercise 8.1 Solutions?
Students should focus on setting up integrals correctly with appropriate limits in class 12 maths ex 8.1. Understanding the geometric interpretation of the problem is crucial for accurate solutions. Clear comprehension of the problem ensures precise and effective integral calculations.
9. Are there any shortcuts to solve these problems in Class 12 Maths Chapter 8 Exercise 8.1 Solutions?
There are no significant shortcuts, but practising similar problems can improve speed and accuracy. Familiarity with different types of problems and methods helps. Regular practice and understanding of integral concepts are key to efficiency in class 12 maths ex 8.1.
10. How can Vedantu's solutions help in mastering this topic in Class 12 Maths Chapter 8 Exercise 8.1 Solutions?
Vedantu's solutions provide step-by-step explanations for each problem. These detailed solutions help students understand the application of integrals thoroughly. They enhance problem-solving skills and ensure students can tackle integral problems effectively.
