Application of Integrals: Introduction and Explanation

Application of Integrals covers the basic properties of integrals as well as the fundamental theorem of calculus.

• There are many different calculating methods, including functions, differentiation, and integration.
• Integrals are used in many fields, including mathematics, science, and engineering.
• We mostly use integral formulae to calculate areas.
• A continuous analog of a sum, an integral is used to calculate areas, volumes, and their generalizations.
• One of the two fundamental operations of calculus is integration, which is the act of computing an integral and the other is differentiation.

Also Read: Differentiation and Integration Formula

Key Terms: Integrals, application of integrals, calculus, area between two curves, simple curves, Integration

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Define Integrals 

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We all have learned formulae to calculate areas of many different geometrical figures such as triangles, rectangles, and circles.

• Many of these formulae are fundamental in the applications of mathematics to many real-life problems that we may face in our day-to-day lives.
• The formulae of elementary geometry allow us to calculate areas of many simple figures.
• However, these same formulae are inadequate for calculating the areas enclosed by curves.
• To solve the areas enclosed by curves, we need some concepts of Integral Calculus. 

The inverse process of differentiation is the integration where we are supposed to find the function whose differential is given.

• The area bounded by the curve is y = f (x), the coordinates x = a, x = b, and the x-axis, which helps us calculate a definite integral as the limit of a sum.
• we shall study a specific application of integrals to find the area under simple curves, the area between lines and arcs of circles, parabolas, and ellipses (standard forms only). 

Integration helps us in finding the areas of the two-dimensional region and also helps in calculating the 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.

Also Read:

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Application of Integral(Explanation)

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The geometric meaning of definite integral is that it is the area under the curve.

Now suppose that we have a curve that is

 

y=f(x) to f(x)dx

The value is f(x) x-axis, x=a, x=b, area of the bounded region or the shaded region is (A)

Where  

A= f(x)dx (A is the shaded region)

Now suppose we have a curve with the equation 

x = f(y) tof(y)dy

Therefore, the value of which is f(y), y-axis, y=c, y=d,

A =f(y)dy

Now suppose we want to find the curve area of the y=f(x)=x³-x, x= -1, x= 1, and the x-axis

We will get A = 

x3-x dx we will get the final answer as zero

Because the definite integral algebraic sum of all the areas bounded by the x/y axis.

Now the graph for the equation y=f(x)=x³-x, will be as follows,

The total area for A1 and A2 is

A1 =

(x3-x)dx  = |x⁴/4-x²/2| where the limit will be from -1 to 0 which will be equal to 1/4.

A2 = 

(x3-x)dx = |x⁴/4-x²/2| where the limit will be from 0 to 1 which will be equal to -1/4.

Now we see that the value of A1 is positive while the value of A2 is negative but we will consider the magnitude of both the equations then we will have.

|-1/4| + |1/4| = 1/2 (1/2 is the total area)

From the above example we can say that we can generalize the formula and the formula for such equation will be

AREA= |fx|dx

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Area Between Two Curves

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Area between two curves

A1 = f(x)- g(x)dx

And A2 =

g(x)- f(x)dx

From the above diagram we conclude two things:

First-

A1 + A2 =

|f(x)- g(x)|dx

Second- The area will always be equal to:

∫ (UPPER CURVE-LOWER CURVE) dx

                         OR 

∫ (RIGHT CURVE-LEFT CURVE)dy

Now we will discuss cases for two curve they are as follows:

All the above mentioned graphs are related to the following equation

A = f(x)- g(x) dx

All the above mentioned graphs are related to the following equation

A = f(y)- g(y) dy

Now we will try to understand the concepts with the help of an example

Ques. Find the area of the region in the first quadrant enclosed by the x-Axis, the line y=x and the circle c² + y² =32.

Ans: The equation that we get from the question are as follows,

Y = x and X² + Y² =32 or we can also write X² + Y² = (4√2)²

Now when we solve the equations Y = x and X² + Y² =32 for the intersection point we get:

x2 +x2(since we know that Y = x) = 32 which will result in 2x²=32 which will give us x=+4and x=-4 ----------------(a)

in the similar way we will get the roots for the y axis so the roots for the y will be y=+4and y=-4----------------(b)

so the points will be (4,4) and (-4,-4).

From the above diagram we can see that the line y = x will go through the center and will make an angle of 45 degree.

Now in order to find the shaded region we will find the intersection point from step (a) and step (b).now we have to calculate the required area

Therefore the required area will be where req area is required are req area(OABA) = area(OMBO) + area(AMBA)

For the area (OMBO)the upper curve is y=x and lower curve is y=0.

For area (AMBA) the upper curve is circle and the lower curve is y=0

Now we will calculate the area for OMBO which is as follows.

Area(OMBO) =

ydx = xdx = |x²/2|with limit 0 and 4 = 8--------(A)

Now we will calculate the area for (AMBA)

= ∫y dx(with upper limit being 4√2 and the lower limit being 4 )

= ∫(√32-x2)dx.

|1/2x*32-X² + 32/2 sin^-1 x/4√2|(with lower limit being 4 and upper limit being 4√2)

(1/2*4√2*0 + 32/2 sin-¹ 4√2/4√2) – (4/2*32-16+32/2 sin^-1 4/4√2)

Now when we solve this above equation we will get the following results.

= 8π-(8+4π) which will give us the following result that is 4π-8--------(B)

On adding the equations A and B we get the following result 4π 

Hence we can say that the area of OABA is 4π.

Following equations can be made from the following that is

1. One curve equations
2. Two curve equations where we will first find the intersection point and then the area which is ∫ (UPPER CURVE-LOWER CURVE) dx
3. To find the area of the triangle equations -

• In these types of questions either three equations will be given or three points will be given.
• If equations for the side are given then we will find the points with the help of intersection.
• If points are given then we will try to find out the equation for all three sides or we will find all the lines that might help us in solving the question.

The video below explains this:

Area under the curve Detailed Video Explanation:

Read More: Conditional Probability Formula

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Things to Remember

• Application of Integrals covers the basic properties of integrals as well as the fundamental theorem of calculus.
• The inverse process of differentiation is the integration where in, we are supposed to find the function whose differential is given.
• We know that the area bounded by the curve is y = f (x), the coordinates x = a, x = b and x-axis, which helps us calculate a definite integral as the limit of a sum.
• Integration helps us in finding the areas of the two-dimensional region and also helps in calculating the 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.
• Geometric meaning of definite integral is that it is the area under the curve.

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Previous Years Questions 

1. Area of the region bounded by two parabolas y=x2y=x2 and x=y2x=y2 is
2. Area bounded by y=x3,y=8y=x3,y=8 and x=0x=0 is
3. What is the area of the region bounded by the line 3x - 5y = 15, x =1, x = 3 and x-axis in sq unit ?​
4. In trapezoidal rule, the curve y=f(x)y=f(x) between two successive ordinates is approximated as​
5. Given e=2.72,e2=7.39,e3=20.09,e4=54.60e=2.72,e2=7.39,e3=20.09,e4=54.60 the approximate value of 4∫0exdx∫04exdx using Simpson?? rule and taking h=1h=1 is​
6. Find the area bounded by the curve x=2−y−y2x=2−y−y2 and y-axis.​
7. Area of the region bounded by y=|x−1|y=|x−1| and y=1y=1 is
8. Area of loop of the curve r=asin2θr=asin2θ is​
9. The area bounded by the curves y= cos x and y= sin x between the ordinates x=0 and x=3π2 is​
10. The area bounded by the curves y=−x2+3 and y=0 is​
11. The set {(x,y):x+y=1}{(x,y):x+y=1} in the xyxy plane represents​
12. The area bounded by y=x2+3y=x2+3 and y=2x+3y=2x+3 is​
13. Area enclosed by the curve π[4(x−√2)2+y2]=8 is​
14. The area of the region bounded by the curves y=x3,y=1x,x=2 is
15. The area of the region bounded by the lines y=2x+1,y=3x+1y=2x+1,y=3x+1 and x=4 is
16. The area of the region bounded by the curve y=x3, its tangent at (1,1) and x−axis is​
17. Area of the region bounded by y=|x| and y=−|x|+2 is​
18. The area of the region enclosed between parabola y2=xy2=x and the line y=mxy=mx is 148.​
19. The area under the curve y=|cosx−sinx|,0≤x≤π2, and above x-axis is :​​