application of integrals case study

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Case Study on Application Of Integrals Class 12 Maths PDF

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The case study questions on Application Of Integrals are based on the CBSE Class 12 Maths Syllabus, and therefore, referring to the Application Of Integrals case study questions enable students to gain the appropriate knowledge and prepare better for the Class 12 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Application Of Integrals Class 12 Maths with Solutions in PDF

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The case study on Application Of Integrals Class 12 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Application Of Integrals case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Application Of Integrals Case Study Questions on Class 12 Maths?

There are three major reasons why one should solve Application Of Integrals case study questions on Class 12 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 12 Maths students, therefore, it is important to solve Application Of Integrals Case study questions as it will help better prepare for the Class 12 board exam preparation.
• Develop Problem-Solving Skills: Class 12 Maths Application Of Integrals case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 12 students develop their problem-solving skills, which are essential for success in any profession rather than Class 12 board exam preparation.
• Understand Real-Life Applications: Several Application Of Integrals Class 12 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Application Of Integrals as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Application Of Integrals?

Students can choose their own way to answer Case Study on Application Of Integrals Class 12 Maths, however, we believe following these three steps would help a lot in answering Class 12 Maths Application Of Integrals Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Application Of Integrals questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Application Of Integrals Class 12 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 12 Application Of Integrals?

 A few essential things to know to solve Case Study Questions on Class 12 Application Of Integrals are -

• Basic Formulas of Application Of Integrals: One of the most important things to know to solve Case Study Questions on Class 12 Application Of Integrals is to learn about the basic formulas or revise them before solving the case-based questions on Application Of Integrals.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 12 Maths Application Of Integrals case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

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Applications of Integrals

There is a number of methods of calculations among which are functions, differentiation, and integration. Applications of Integrals are applied in various fields like Mathematics, Science, Engineering. Further, for the calculation of areas or irregular shapes in a two-dimensional space, we use majorly integrals formulas.

Here a brief introduction on integrals is given, with applications of integrals to find areas under simple curves, areas bounded by a curve and a line and area between two curves, and also the application of integrals in other mathematical disciplines along with the solved examples.

Definition of Integral

Given the derivative f’ of the function f, a question that arises is, "Can we determine the function f?" Here, the function f is called antiderivative or integral of f’. The process of finding the antiderivative is called integration. On the other hand, the value of the function found by the process of integration is called an Integral.

For example, the derivative of f(x) = x 3 is f’(x) = 3x 2 ; and the antiderivative of g(x) = 3x 2 is f(x) = x 3. Here, the integral of g(x) = 3x 2 is f(x)=x 3

Definition of integral: An integral is a function, of which a given function is the derivative. Integration is basically used to find the areas of the two-dimensional region and for computing volumes of three-dimensional objects. Therefore, finding the integral of a function with respect to the x-axis refers to finding the area of the curve with respect to the x-axis. The integral is also called as anti-derivative as it is the reverse process of differentiation.

In general, there are two types of integrals. Definite integrals are defined for integrals with limits and indefinite integrals do not include any limits. Here, let us explore more about definite, and indefinite integrals.

Types of Integrals

Definite integrals.

These are the integrals that have a pre-existing value of limits; thus making the final value of integral definite. The definite integrals are used to find the area under the curve with respect to one of the coordinate axes, and with the defined limits. Here we aim at finding the area under the curve g(x) with respect to the x-axis and having the limits from b to a.

Indefinite Integrals

These are the integrals that do not have a pre-existing value of limits; thus making the final value of integral indefinite. The indefinite integrals are used to integrate the algebraic expressions, trigonometric functions, logarithmic, and exponential functions. Here g'(x) is the derivative answer, which on integration results in the original function of g(x). The integration does not give back the constant value of the original expression, and hence a constant 'c' is added to the answer of the integral.

Application of Integrals

From the many applications of integrals, some are listed below:

In Mathematics integrals are used to find:

• Center of mass(Centroid) of an area having curved sides
• The average value of a curve
• The area between two curves
• The area under a curve

In Physics integrals are used to find:

• Centre of gravity
• Center of mass
• Mass and moment of inertia of vehicles
• Mass and momentum of satellites
• The velocity and trajectory of a satellite

Application of integrals also includes finding the area enclosed in the eclipse, the area of the region bounded by the curve, or any enclosed area bounded in the x-axis and y-axis. The application of integrations varies depending upon the fields. Graphic designers use it for the creation of three-dimensional models. Physicists use it to determine the center of gravity, etc.

Let us have a look at one of the common applications of integrals i.e., how to find area under the curve.

How to Find Area Under The Curve?

The area under the curve can be calculated through three simple steps. First, we need to know the equation of the curve(y = f(x)), the limits across which the area is to be calculated, and the axis enclosing the area. Secondly, we have to find the integration (antiderivative) of the curve. Finally, we need to apply the upper limit and lower limit to the integral answer and take the difference to obtain the area under the curve.

Area = \(_a\int^b y.dx \)

= \(_a\int^b f(x).dx\)

=\( [g(x)]^b_a\)

=\( g(b) - g(a)\)

Related Topics:

• Integration
• Integral Calculator
• Integration By Parts
• Differentiation and Integration Formula
• Integration Formulas

Important Notes on Applications of Integrals:

• The value of the function found by the process of integration is called an integral.
• In general, there are two types of integrals: Definite Integrals (the value of the integral is definite) Indefinite Integrals (the value of the integral is indefinite)

Solved Examples on Applications of Integrals

Example 1: Can you find the area under the curve using the application of integrals, for the region bounded by the circle x 2 + y 2 = 16 in the first quadrant?

The given equation of the circle is x 2 + y 2 = 16 Simplifying this equation we have \(y=\sqrt {4^2-x^2}\)

Here we find the area of the quadrant of the circle across the limits [0, 4] and then multiply it by 4 to obtain the area of the circle. \(\mathrm{A}=4\int_{0}^{4} y \cdot d x\) \(=4\int_{0}^{4} \sqrt{4^{2}-x^{2}} \cdot d x\) \(=4\left[\frac{x}{2} \sqrt{4^{2}-x^{2}}+\frac{4^{2}}{2} \operatorname{Sin}^{-1} \frac{x}{4}\right]_{0}^{4}\) \(=4[((4 / 2) \times 0+(16 / 2) \operatorname{Sin}-11)-0]\) \(=4(16 / 2)(\pi / 2)\) \(=16 \pi\)

Answer: Therefore the area of the region bounded by the circle in the first quadrant is 16π sq units.

Example 2: Find the area under the curve using the application of integrals, for the region enclosed by the ellipse x 2 /36 + y 2 /25 = 1.

Solution: The given equation of the ellipse is x 2 /36 + y 2 /25 = 1 This can be transformed to obtain \(y=\frac{5}{6} \sqrt{6^{2}-x^{2}}\) \(\begin{aligned} A &=4 \int_{0}^{6} y \cdot d x \\ &=4 \int_{0}^{6} \frac{5}{6} \cdot \sqrt{6^{2}-x^{2}} \cdot d x \\ &=\frac{20}{6}\left[\frac{x}{2} \cdot \sqrt{6^{2}-x^{2}}+\frac{6^{2}}{2} \sin ^{-1} \frac{x}{6}\right]_{0}^{6} \\ & \left.=\frac{20}{6}\left[\left(\frac{6}{2} \times 0\right)+\frac{6^{2}}{2} \cdot \operatorname{Sin}^{-1} 1\right)-0\right] \\ &=\frac{20}{6} \cdot \frac{36}{2} \cdot \frac{\pi}{2} \\ &=30 \pi \end{aligned}\)

Answer: Therefore the area of the ellipse is 30π sq units.

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Practice Questions on Applications of Integrals

Faqs on applications of integrals, what are the applications of integrals in real life.

Integration finds many uses in the fields of Engineering, Physics, Maths, etc. For example in Physics, Integration is very much needed. For example, to calculate the Centre of Mass, Centre of Gravity, and Mass Moment of Inertia of a sports utility vehicle. To calculate the velocity and trajectory of an object, predict the position of planets, and understand electromagnetism.

What Are the Real-Life Applications of Integrals and Differentiation?

Differentiation and integration can help us solve many types of real-world problems. We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.). And the applications of integrals are useful to find the areas of irregular shapes.

What Is the Application of Integrals in Maths?

Integrals are used to evaluate such quantities as area, volume, work, and, in general, any quantity that can be interpreted as the area under a curve .

What Is Integration in Simple Words?

In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale.

How Do You Define Integration?

An integral is a function, of which a given function is the derivative. Integration is basically used to find the areas of the two-dimensional region and computing volumes of three-dimensional objects. Therefore, finding the integral of a function with respect to x means finding the area with respect to the X-axis and the curve. The integral is also called as anti-derivative as it is the reverse process of differentiation.

What Are the Two Types of Integrals?

There are two forms of integrals.

• Indefinite Integrals: It is an integral of a function when there is no limit for integration. It contains an arbitrary constant.
• Definite Integrals: An integral of a function with limits of integration.

Case study application of integral 3 chapter 8 class 12

Case study chapter 8 (application of integral ).

Case study 3: Read the following and answer the question(Case study application of integral 3) A student designs an open air Honeybee nest on the branch of a tree, whose plane figure is parabolic and the branch of tree is given by a straight line.

(i) Point of intersection of the parabola and straight line are

(a) (4, 0) and (-4, 0)           (b) (4, -4) and (4, 0)

(c) (-4, 4) and (4, 4)            (d) (2, 4) and (-2, 4)

(ii) Length of each horizontal strip of the bounded region is given by

(iii) Length of each vertical strip is given by

(c) 4                                           (d) None of these

(c) 64/3                                   (d) 128/3

(v) Area of each vertical strip is given by

Solution: (i) Answer (c)

Given equation of parabola is

x² = 4 y  —–(i)

And equation of straight line y = 4

∴ From (i), we get

x² = 4×4 = 16

∴ Point of intersection are (4, 4) and (-4, 4)

(ii) Answer (c)

x² = 4 y  —(i)

Length of horizontal strip be = 2×2√y = 4√y

(iii) Answer (a)

(iv) Answer (c)

Area of required bounded region

(v) Answer (b)

Area of each(one) vertical strip

Some Other Case study problem

Case study 1: Read the following and answer the question.(Case study application of integral 1)

Nowadays, almost every boat has a triangular sail. By using a triangular sail design it has become possible to travel against the wind using a technique known as tacking. Tacking allows the boat to travel forward with r triangular sail on the walls and three edges(lines) at the triangular sail are given by the equation x = 0, y = 0 and y + 2x – 4 = 0 respectively.

Solution: For solution click here

Case study 2: Read the following and answer the question(Case study application of integral 2)

Case study 4: Read the following and answer the question

A boy design a pizza by cutting it with a knife on a card board. If pizza is circular in shape which is represented by the

Solution: for solution click here

Case study 5:-A farmer has a triangular shaped field. His, son a science student observes the triangular field has three edges and can be drawn on a plain paper with three lines given by its equations.(Case study application of integral 5)

Based on the above information answer the following question:

(i) Find the area of the shaped region in the figure shown below.

(ii) Find the area of the triangle  ΔABC.

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Chapter 6: Applications of Integration

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• Page ID 10828

In this chapter, we use definite integrals to calculate the force exerted on the dam when the reservoir is full and we examine how changing water levels affect that force. Hydrostatic force is only one of the many applications of definite integrals we explore in this chapter. From geometric applications such as surface area and volume, to physical applications such as mass and work, to growth and decay models, definite integrals are a powerful tool to help us understand and model the world around us.

• 6.0: Prelude to Applications of Integration The Hoover Dam is an engineering marvel. When Lake Mead, the reservoir behind the dam, is full, the dam withstands a great deal of force. However, water levels in the lake vary considerably as a result of droughts and varying water demands.
• 6.1E: Exercises for Section 6.1
• 6.2E: Exercises for Volumes of Common Cross-Section and Disk/Washer Method
• 6.3b: Volumes of Revolution: Cylindrical Shells OS
• 6.3E: Exercises for the Shell Method
• 6.4: Arc Length and Surface Area In this section, we address a simple question: Given a curve, what is its length? This is often referred to as arc length.
• 6.4: Arc Length of a Curve and Surface Area The arc length of a curve can be calculated using a definite integral. The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions of y. The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution. The integrals generated by both the arc length and surface area formulas are often difficult to evaluate.
• 6.4E: Exercises for Section 6.4
• 6.5: Using Integration to Determine Work Work is the scientific term used to describe the action of a force which moves an object. The SI unit of force is the Newton (N), and the SI unit of distance is a meter (m). The fundamental unit of work is one Newton--meter, or a joule (J). That is, applying a force of one Newton for one meter performs one joule of work.
• 6.5b: More Physical Applications of Integration In this section, we examine some physical applications of integration. Several physical applications of the definite integral are common in engineering and physics. Definite integrals can be used to determine the mass of an object if its density function is known. Work can also be calculated from integrating a force function, or when counteracting the force of gravity, as in a pumping problem. Definite integrals can also be used to calculate the force exerted on an object submerged in a liquid.
• 6.5E: Exercises on Work
• 6.6E: Exercises for Section 6.6
• 6.7E: Exercises for Section 6.7
• 6.8E: Exercises for Section 6.8
• 6.9E: Exercises for Section 6.9
• Chapter 6 Review Exercises These are homework exercises to accompany OpenStax's "Calculus" Textmap.

Thumbnail: A region between two functions.

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .

Applications of Integration: Area and Volume

One very useful application of Integration is finding the area and volume of “curved” figures, that we couldn’t typically get without using Calculus. Since we already know that can use the integral to get the area between the $ x$- and $ y$-axis and a function, we can also get the volume of this figure by rotating the figure around either one of the axes.

Area Between Curves

We learned how to get the area under a curve here in the Definite Integrals section , and actually, the area under a curve is technically the area between the curve (function) and the $ x$-axis, since the $ x$-axis is at $ y=0$. Generally, we can get the area between any two curves by subtracting the bottom curve from the top curve everywhere where the top curve is higher than the bottom curve. The cool thing about this is it even works if one of the curves is below the $ x$-axis, as long as the higher curve always stays above the lower curve in the integration interval.

Note that we may need to find out where the two curves intersect (and where they intersect the $ x$-axis) to get the limits of integration. And sometimes we have to divide up the integral if the functions cross over each other in the integration interval.

Here is the formal definition of the area between two curves:

For functions $ f$ and $ g$ where $ f\left( x \right)\ge g\left( x \right)$ for all $ x$ in $ [a,b]$, the area of the region bounded by the graphs and the vertical lines $ x=a$ and $ x=b$ is:

$ \displaystyle \text{Area}=\int\limits_{a}^{b}{{\left[ {f\left( x \right)-g\left( x \right)} \right]}}\,dx$

Let’s try some problems:

Notice this next problem, where it’s much easier to find the area with respect to $ y$, since we don’t have to divide up the graph. When we integrate with respect to $ y$, we will have horizontal rectangles  (parallel to the $ x$-axis) instead of vertical rectangles (perpendicular to the $ x$-axis), since we’ll use “$ dy$” instead of “$ dx$”. If we have the functions in terms of $ x$, we need to use Inverse Functions to get them in terms of $ y$.

If you’re having trouble find the area with respect to $ x$ , trying “tilting the graph” to see if it’s possible to integrate with respect to $ y$ .

Here are more problems where we take the area with respect to $ y$.

Again, if you’re having trouble find the area with respect to $ x$ , trying “tilting the graph” to see if it’s possible to integrate with respect to $ y$ . For the entire interval, the top (right) curve must always be “above” (or to the right of) the bottom graph.

Volumes of Solids by Cross Sections

Now that we know how to get areas under and between curves, we can use this method to get the volume of a three-dimensional solid, either with cross sections, or by rotating a curve around a given axis. Think about it; every day engineers are busy at work trying to figure out how much material they’ll need for certain pieces of metal, for example, and they are using calculus to figure this stuff out!

Let’s first talk about getting the volume of solids by cross-sections of certain shapes. When doing these problems, think of the bottom of the solid being flat on your horizontal paper, and the 3-D part of it coming up from the paper. Cross sections might be squares, rectangles, triangles, semi-circles, trapezoids, or other shapes. (We’ll have to use some geometry to get these areas.)

Cross sections can either be perpendicular to the $ x$-axis or $ y$-axis; in our examples, they will be perpendicular to the $ x$-axis, which is what is we are used to.

Since integration is “infinite summation” , we can just integrate over the interval of cross sections to get a volume. Thus, given the cross-sectional area $ A(x)$ in interval $ [a,b]$, and cross sections are perpendicular to the $ x$-axis, the volume of this solid is $ \text{Volume = }\int\limits_{a}^{b}{{A\left( x \right)}}\,dx$

Here are examples of volumes of cross sections between curves. Slices of the volume  are shown to better see how the volume is obtained:

Here’s one more that’s a little tricky:

Volumes of Solids: The Disk Method

Now let’s talk about getting a volume by revolving a function or curve around a given axis to obtain a solid of revolution .

Since we know now how to get the area of a region using integration, we can get the volume of a solid by rotating the area around a line, which results in a right cylinder, or disk . (Remember that the formula for the volume of a cylinder is $ \pi {{r}^{2}}\cdot \text{height}$). The radius is the distance from the axis of revolution to the function, and the “height” of each disk, or slice is “$ dx$” or “$ dy$”, depending on the orientation. Again, since integration is “infinite summation” , we can just integrate over the interval to get a volume. Note that volumes may be different, depending on which axis is used for rotation!

Let’s try some problems:

Volumes of Solids: The Washer Method

The washer method is similar to the disk method, but it covers solids of revolution that have “holes”, where we have inner and outer functions, thus inner and outer radii.

With washers, we have two revolving solids and we basically subtract the area of the inner solid from the area of the outer one. Note that for this to work, the middle function must be completely inside (or touching) the outer function over the integration interval.

Let’s try some problems:

Volumes of Solids: The Shell Method

The shell method for finding volume of a solid of revolution uses integration along an axis perpendicular to the axis of revolution instead of parallel , as we’ve seen with the disk and washer methods. The nice thing about the shell method is that you can integrate around the $ y$-axis and not have to take the inverse of functions. Also, the rotational solid can have a hole in it (or not), so it’s a little more robust. It’s not intuitive though, since it deals with an infinite number of “surface areas” of rectangles in the shapes of cylinders (shells).

Here are the equations for the shell method :

Since I believe the shell method is no longer required the Calculus AP tests (at least for the AB test), I will not be providing examples and pictures of this method. Please let me know if you want it discussed further.

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On to Integration by Parts  — you are ready!

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Unit 8: Applications of integration

About this unit.

Definite integrals are all about the accumulation of quantities. Let's see how they are applied in order to solve various kinds of problems.

Finding the average value of a function on an interval

• Average value over a closed interval (Opens a modal)
• Calculating average value of function over interval (Opens a modal)
• Mean value theorem for integrals (Opens a modal)
• Average value of a function Get 3 of 4 questions to level up!

Connecting position, velocity, and acceleration functions using integrals

• Motion problems with integrals: displacement vs. distance (Opens a modal)
• Analyzing motion problems: position (Opens a modal)
• Analyzing motion problems: total distance traveled (Opens a modal)
• Motion problems (with definite integrals) (Opens a modal)
• Worked example: motion problems (with definite integrals) (Opens a modal)
• Average acceleration over interval (Opens a modal)
• Analyzing motion problems (integral calculus) Get 3 of 4 questions to level up!
• Motion problems (with integrals) Get 3 of 4 questions to level up!

Using accumulation functions and definite integrals in applied contexts

• Area under rate function gives the net change (Opens a modal)
• Interpreting definite integral as net change (Opens a modal)
• Worked examples: interpreting definite integrals in context (Opens a modal)
• Analyzing problems involving definite integrals (Opens a modal)
• Worked example: problem involving definite integral (algebraic) (Opens a modal)
• Interpreting definite integrals in context Get 3 of 4 questions to level up!
• Analyzing problems involving definite integrals Get 3 of 4 questions to level up!
• Problems involving definite integrals (algebraic) Get 3 of 4 questions to level up!

Finding the area between curves expressed as functions of x

• Area between a curve and the x-axis (Opens a modal)
• Area between a curve and the x-axis: negative area (Opens a modal)
• Area between curves (Opens a modal)
• Worked example: area between curves (Opens a modal)
• Composite area between curves (Opens a modal)
• Area between a curve and the x-axis Get 3 of 4 questions to level up!
• Area between two curves given end points Get 3 of 4 questions to level up!
• Area between two curves Get 3 of 4 questions to level up!

Finding the area between curves expressed as functions of y

• Area between a curve and the 𝘺-axis (Opens a modal)
• Horizontal area between curves (Opens a modal)
• Horizontal areas between curves Get 3 of 4 questions to level up!

Finding the area between curves that intersect at more than two points

• No videos or articles available in this lesson
• Area between curves that intersect at more than two points (calculator-active) Get 3 of 4 questions to level up!

Volumes with cross sections: squares and rectangles

• Volume with cross sections: intro (Opens a modal)
• Volume with cross sections: squares and rectangles (no graph) (Opens a modal)
• Volume with cross sections perpendicular to y-axis (Opens a modal)
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Volumes with cross sections: triangles and semicircles

• Volume with cross sections: semicircle (Opens a modal)
• Volume with cross sections: triangle (Opens a modal)
• Volumes with cross sections: triangles and semicircles Get 3 of 4 questions to level up!

Volume with disc method: revolving around x- or y-axis

• Disc method around x-axis (Opens a modal)
• Generalizing disc method around x-axis (Opens a modal)
• Disc method around y-axis (Opens a modal)
• Disc method: revolving around x- or y-axis Get 3 of 4 questions to level up!

Volume with disc method: revolving around other axes

• Disc method rotation around horizontal line (Opens a modal)
• Disc method rotating around vertical line (Opens a modal)
• Calculating integral disc around vertical line (Opens a modal)
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Volume with washer method: revolving around x- or y-axis

• Solid of revolution between two functions (leading up to the washer method) (Opens a modal)
• Generalizing the washer method (Opens a modal)
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Volume with washer method: revolving around other axes

• Washer method rotating around horizontal line (not x-axis), part 1 (Opens a modal)
• Washer method rotating around horizontal line (not x-axis), part 2 (Opens a modal)
• Washer method rotating around vertical line (not y-axis), part 1 (Opens a modal)
• Washer method rotating around vertical line (not y-axis), part 2 (Opens a modal)
• Washer method: revolving around other axes Get 3 of 4 questions to level up!

The arc length of a smooth, planar curve and distance traveled

• Arc length intro (Opens a modal)
• Worked example: arc length (Opens a modal)
• Arc length Get 3 of 4 questions to level up!

Calculator-active practice

• Contextual and analytical applications of integration (calculator-active) Get 3 of 4 questions to level up!

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Class 12th Maths - Application of Integrals Case Study Questions and Answers 2022 - 2023

By QB365 on 08 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12 Maths Subject - Application of Integrals, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Application of integrals case study questions with answer key.

12th Standard CBSE

Final Semester - June 2015

Consider the following equations of curves : x ? - = y and y = x. On the basis of above information, answer the following questions. (i) The point(s) of intersection of both the curves is (are)

(iv) The value of the integral  \(\int_{0}^{1} x^{2} d x\)  

(v) The value of area bounded by the curves x ? - = y and x = y is

Consider the curve x 2 +y 2 = 16 and line y = x in the first quadrant. Based on the above information, answer the following questions. (i) Point of intersection of both the given curves is

(iv) The value of the integral  \(\int_{2 \sqrt{2}}^{4} \sqrt{16-x^{2}} d x\)  is

(v) Area bounded by the two given curves is

(iv) Area of each slice of pizza when child cut the pizza into 4 equal pieces is

(v) Area of whole pizza is

Consider the following equation of curve I' = 4x and straight line x + y = 3. Based on the above information, answer the following questions. (i) The line x + y = 3 cuts the x-axis and y-axis respectively at

(ii) Point(s) of intersection of two given curves is (are)

(v) Value of area bounded by given curves is

(ii) Area of curve explained in the passage from 0 to  \(\frac{\pi}{2}\)  is

(iii) Area of curve discussed in classroom from  \(\frac{\pi}{2} \text { to } \frac{3 \pi}{2}\)  is

(iv) Area of curve discussed in classroom from  \(\frac{3 \pi}{2} \text { to } 2 \pi\)  is

(v) Area of explained curve from 0 to  \(2 \pi\)  is

(ii) Value of  \(\int_{0}^{\pi / 4} \sin x d x\)  is

(iii) Value of  \(\int_{\pi / 4}^{\pi / 2} \cos x d x\)  is

(iv) Value of  \(\int_{0}^{\pi} \sin x d x\)  is

(v) Value of  \(\int_{0}^{\pi / 2} \sin x d x\)  is

(ii) Equation of line BC is

(iii) Area of region ABCD is

(iv) Area of  \(\Delta A D C\)  is

(iv) Area of \(\Delta A B C\)  is

(iv) Value of  \(\int_{1 / 2}^{1} \sqrt{1-x^{2}} d x\)  is

(v) Area of hidden portion of lower circle is

(iv) The value of  \(2 \int_{0}^{3}\left(1-\frac{x}{3}\right) d x\)  is 

(v) Area of the smaller region bounded by the mirror and scratch is

Consider the following equations of curves y = cos x, y = x + 1 and y = 0. On the basis of above information, answer the following questions. (i) The curves y = cos x and y = x + 1 meet at

(ii) y = cos x meet the x-axis at

(iii) Value of the integral  \(\int_{-1}^{0}(x+1) d x\)  is

(iv) Value of the integral  \(\int_{0}^{\pi / 2} \cos x d x\)  is

(v) Area bounded by the given curves is

*****************************************

Application of integrals case study questions with answer key answer keys.

(i) (a) : Here, teacher explained about cosine curve. (ii) (c) : Required area  \(\int_{0}^{\pi / 2} \cos x d x\)   \(=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1-0=1 \text { sq. unit }\)   (iii) (b) : Required area =  \(\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|=\left|[\sin x]_{\pi / 2}^{3 \pi / 2}\right|\)   \(=\left|\sin \frac{3 \pi}{2}-\sin \frac{\pi}{2}\right|=|-1-1|=|-2|\)   = 2 sq. units [Since, area can't be negative] (iv) (a) : Required area =  \(\int_{3 \pi / 2}^{2 \pi} \cos x d x=[\sin x]_{3 \pi / 2}^{2 \pi}\)   \(=\sin 2 \pi-\sin \frac{3 \pi}{2}=0-(-1)=1 \text { sq. unit }\)   (v) (d) : Required area \(=\int_{0}^{\pi / 2} \cos x d x+\left|\int_{\pi / 2}^{3 \pi / 2} \cos x d x\right|+\int_{3 \pi / 2}^{2 \pi} \cos x d x\)   = 1 + 2 + 1 = 4 sq. units.

(i) (c) : For point of intersection, we have sin x = cos x \(\Rightarrow \frac{\sin x}{\cos x}=1 \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}\)   (ii) (a) :  \(\int_{0}^{\pi / 4} \sin x d x=[-\cos x]_{0}^{\pi / 4}=-\cos \frac{\pi}{4}+\cos 0\)   \(=1-\frac{1}{\sqrt{2}}\)   (iii) (b) :  \(\int_{\pi / 4}^{\pi / 2} \cos x d x=[\sin x]_{\pi / 4}^{\pi / 2}=\sin \frac{\pi}{2}-\sin \frac{\pi}{4}\)   \(=1-\frac{1}{\sqrt{2}}\)   (iv) (c) :   \(\int_{0}^{\pi} \sin x d x=[-\cos x]_{0}^{\pi}=[-\cos \pi+\cos 0]=2\)   (v) (b) :   \(\int_{0}^{\pi / 2} \sin x d x=[-\cos x]_{0}^{\pi / 2}=\left[-\cos \frac{\pi}{2}+\cos 0\right]\)  = 0+1=1

(i) (a) : Equation of line AB is \(y-0=\frac{3-0}{1+1}(x+1) \Rightarrow y=\frac{3}{2}(x+1)\)   (ii) (c) : Equation of line BC is  \(y-3=\frac{2-3}{3-1}(x-1)\)   \(\Rightarrow y=-\frac{1}{2} x+\frac{1}{2}+3 \Rightarrow y=\frac{-1}{2} x+\frac{7}{2}\)   (iii) (d) : Area of region ABCD = Area of \(\triangle A B E\) + Area of region BCDE \(=\int_{-1}^{1} \frac{3}{2}(x+1) d x+\int_{1}^{3}\left(\frac{-1}{2} x+\frac{7}{2}\right) d x\)   \(=\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}+\left[\frac{-x^{2}}{4}+\frac{7}{2} x\right]_{1}^{3}\)   \(=\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]+\left[\frac{-9}{4}+\frac{21}{2}+\frac{1}{4}-\frac{7}{2}\right]\)   = 3 + 5 = 8 sq. units (iv) (a) : Equation of line AC is  \(y-0=\frac{2-0}{3+1}(x+1)\)   \(\Rightarrow y=\frac{1}{2}(x+1)\)   \(\therefore \text { Area of } \Delta A D C=\int_{-1}^{3} \frac{1}{2}(x+1) d x=\left[\frac{x^{2}}{4}+\frac{1}{2} x\right]_{-1}^{3}\)   \(=\frac{9}{4}+\frac{3}{2}-\frac{1}{4}+\frac{1}{2}=4 \text { sq. units }\)   (v) (b) : Area of  \(\Delta A B C\) = Area of region ABCD - Area of  \(\Delta A C D=8-4=4 \mathrm{sq} . \text { units }\)

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Application of Integrals

There is a number of methods of calculations among which are functions, differentiation and integration. Application of Integrals is applied in various fields like Mathematics, Science, Engineering etc. For the calculation of areas, we use majorly integrals formulas. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two curves, and also the application of integrals in the mathematical disciplines along with the solved problem.

Integral Definition

An integral is a function, of which a given function is the derivative. Integration is basically used to find the areas of the two-dimensional region and computing volumes of three-dimensional objects. Therefore, finding the integral of a function with respect to x means finding the area to the X-axis from the curve. The integral is also called as anti-derivative as it is the reverse process of differentiation.

Types of Integrals

There are basically two types of integrals, Definite and Indefinite.  Definite Integral is defined as the integral which contains definite limits,i.e., upper limit and lower limit. It is also named as Riemann Integral. It is represented as;

Indefinite Integral is defined as the integral whose upper and lower limits are not defined.

There are many applications of integrals, out of which some are mentioned below:

• To find the centre of mass(Centroid) of an area having curved sides
• To find the area between two curves
• To find the area under a curve
• The average value of a curve

Integrals are used to calculate

• Centre of gravity
• Mass and momentum of inertia of vehicles
• Mass and momentum of satellites
• Mass and momentum of a tower
• The centre of mass
• The velocity of a satellite at the time of placing it in orbit
• The trajectory of a satellite at the time of placing it in orbit
• To calculate Thrust

Video Lesson

Area under the curve.

Important Questions

Definite Integral Problem

Let us discuss here how the application of integrals can be used to solve certain problems based on scenarios to find the areas of the two-dimensional figure.

Example: Find the area enclosed by the circle x 2 +y 2 =r 2 , where r is the radius of the circle .

Solution: Let us draw a circle in the XY plane with a radius as r.

From the graph,

A has coordinates(0,r) on the x-axis and B has coordinates(r,0) on y-axis.

Area of Circle=4*Area of region OBAO

Now, from the equation of circle,

x 2 +y 2 =r 2

y 2 =r 2 -x 2

The region OABO lies in the first quadrant of the x-y plane.

From the differentiation formula,

Hence, is the answer.

With the above example problem, we hope the concept of integrals is understood. In the same way, we can apply integrals to find the area of enclosed in eclipse, the area of the region bounded by the curve or for any enclosed area bounded in the x-axis and y-axis.

The application of integrations in real life is based upon the industry types, where this calculus is used. Like in the field of engineering, engineers use integrals to determine the shape of building constructions or length of power cable required to connect the two substations etc. In Science, it is used to solve many derivations of Physics topics like the centre of gravity etc. In the field of graphical representation, where three-dimensional models are demonstrated.

The application of integrals class 12 syllabus covers to find the area enclosed by the circle and similar kind of question pattern.

For more related topics of Integrals and  NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives download BYJU’S- The Learning App.

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