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Application of the Integrals

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Class 12 Maths Chapter 8
Application of the Integrals
Important Questions

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In application of integrals, there are several important questions that are often explored in both academic and practical contexts. These questions highlight the real-world relevance and the diverse uses of integral calculus.specific questions can vary based on your textbook or curriculum.Basic Integration Formulas,Integration by Substitution,For more specific questions, it would be helpful to refer class 12 application of integrals important questionsto your textbook or study materials.additional academic resources can be very helpful.

Introduction

The application of integrals is a vast and essential part of calculus, playing a crucial role in various scientific, engineering, and mathematical fields. Integral calculus is the study of representing functions as the limit of sums, which is fundamentally the process of accumulation. This concept allows for the calculation of areas, volumes, and other quantities that arise from aggregating infinitesimal data points..Throughout this chapter, emphasis is placed on elucidating critical cbse class 12 maths application of integrals.

What are Application of the Integrals?

Applications of integrals are vast and varied, spanning across many fields.
Area under a Curve: One of the fundamental applications of integrals is finding the area under a curve. This is used extensively in physics and engineering to determine distances, areas, and other spatial properties.
Volume Calculation: Integrals are used to calculate the volume of three-dimensional objects, especially those with irregular shapes. This is crucial in fields like manufacturing, architecture, and civil engineering.
Each of these applications involves integrating functions over specific intervals to calculate quantities that are not easily measurable or observable directly.

Class 12 Application of the Integrals Important Questions and Answers

Q1. The area of the region bounded by the curve x² = 4y and the straight line x = 4y – 2 is
Options
(a) \dfrac{3}{8} \text{sq. units} \\[4.5 bp] (b)\dfrac{5}{8} \text{sq. units}\\[4.5 bp] (c)\space \dfrac{7}{8} \text{sq. units}\\[4.5 bp] (d)\space \dfrac{9}{8} \text{sq. units}

Ans. (d)\space \dfrac{9}{8} \text{sq. units} \\ Explanation:
For the curves x^2 \text{= y and x = 4y - 2,} the points of intersection are x = -1 and x = 2.
Hence, the required area, A = -\int\limits_1^2 \left\{\left[\dfrac{\text{(x + 2)}}{4}\right]- \left[\dfrac{x^2}{4} \right]\right\}\text{ dx}\\ Now, integrate the function and apply the limits, we get
\text{A = } \left( \dfrac{1}{4}\right) \left[\left( \dfrac{10}{3}\right)- \left( \dfrac{-7}{6}\right) \right] \\[4.5 bp] \text{A = }\left( \dfrac{1}{4}\right)\left( \dfrac{9}{2}\right)= \left( \dfrac{9}{8}\right) \text{sq. units}

Q2: The area enclosed between the graph of \space y\space = \space x^3 and the lines x = 0, y = 1, y = 8 is
Options
(a) 7
(b) 14
(c) \dfrac{45}{4}\\ (d) None of these

Ans. (c) \dfrac{45}{4}

Explanation:
Given curve,\text{ y = }x^3\text{ or x =} \sqrt[3]{y}.\\ Hence, the required area,
\\ A = \int\limits_1^8 \sqrt[3]{y} \text{ dy} \\[4.5 bp] A = \left[\dfrac{(y^{\frac{3}{4}})}{\left(\dfrac{4}{3}\right)}\right]_1^8 \\ Now, apply the limits, we get
A = \left(\dfrac{3}{4}\right)(16-1)\\[4.5 bp] A = \left(\dfrac{3}{4}\right)(15) = \dfrac{45}{4}.

Q3. The area of the region bounded by the curve y² = x, the y-axis and between y = 2 and y = 4 is?

Explanation:
Given: y^2\text{ = x}\\ Hence, the required area, A = \int\limits_2^4 y^2 \text{ dy}\\[4.5 bp] A = \left[\dfrac{y^3}{3}\right]_2^4\\[4.5 bp] A = \left(\dfrac{4^3}{3}\right) – \left(\dfrac{2^3}{3}\right)\\[4.5 bp] A = \left(\dfrac{64}{3} \right) – \left(\dfrac{8}{3}\right)\\[4.5 bp] A = \dfrac{56}{3}\text{ sq. units.}

Q4. Area bounded by the curve y = sin x and the x-axis between x = 0 and x = 2π is?

Explanation:
Required Area, A = \int\limits_0^{2π} \text{|sin x| dx}\\[4.5 bp] \text{A = } \int\limits_0^π \text{sin x dx +} \int\limits_0^{2π} \text{(-sin x) dx}\\ Now, substitute the limits, we get
A= 4 sq. units.

Q5: The area of the region bounded by the circle x² + y² = 1 ?

Explanation:
Given circle equation is x^2 + y^2 =1, whose centre is (0, 0) and radius is 1.
Therefore, \space y^2 = 1 -\space x^2 \\[4.5 bp] y = \sqrt{(1\space -\space x^2)} \\[4.5 bp] Hence, the required area,\space A = 4\int\limits_0^1\sqrt{(1\space - \space x^2)}\text{ dx} \\[4.5 bp] Now, integrate the function and apply limits, we get
\text{A = 4}\left(\dfrac{1}{2} \right)\left(\dfrac{π}{2}\right) \\[4.5 bp] A = π sq. units

CBSE Class 12 Maths Chapter wise Important Questions

making of global world class 10 important questions

Conclusion

The applications of integrals are diverse and critical across various scientific, engineering, economic, and mathematical fields. Integrals enable the precise quantification of these changes, offering insights into phenomena in physics, engineering, biology, economics, and more. Exploring cbse class 12 maths application of Integrals important questions, becomes essential for a comprehensive grasp of this fundamental aspect of mathematics. To enhance understanding, platforms like Oswal.io provide a wealth of resources, including question-answer sets, comprehensive class 12 questions and answers.

Frequently Asked Questions

Ans: Integrals find applications in various fields, such as physics, engineering, economics, and biology. They are used to calculate areas, volumes, work done, center of mass, and more.
Ans: To find the area between curves, you typically calculate the definite integral of the absolute difference between the functions representing the curves within the specified bounds.
Ans: Integrals play a crucial role in finding volumes of solids obtained by rotating a region bounded by curves around a specific axis. This process involves slicing the solid into infinitesimally thin discs or washers and summing their volumes using integrals.
Ans: In physics, integrals are applied to find work done by a force. For instance, in the context of moving an object against a force, work done is calculated by integrating the product of force and displacement over a specified path.
Ans: Integrals help determine the center of mass by considering the weighted average of the positions of all the constituent particles or components in a system. Integrating mass distributions allows for the calculation of the center of mass along different axes.
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