Integration is a mathematical operation that allows us to find the accumulated effect of continuously changing quantities, encapsulating the notion of area under a curve, calculating the area of irregular shapes to determining quantities like volume, work, and probabilities, determining arc length, offering a precise method to measure the distance along a curve. Some of the major application of integral are as below.

- Arc Length: Measuring the length of curves.
- Area under curves: Finding the area enclosed by curves and functions.
- Probability: Determining probabilities of continuous distributions.
- Surface Area: Finding the surface area of 2D, 3D objects.
- Volume: Calculating volumes of solids using methods like cross section, disks, washers, and shells.

#### Arc Length

Solution 👉 Click Here

The concept of arc length refers to the distance along a curve: such as a segment of a circle, a parabola, or any arbitrary curve. Finding the length of this curve is essential in various fields like physics, engineering, and computer graphics.

Let \(y=f(x)\) be a space curve, then arc length of f(x) is defined in terms of lengths of approximating polygonal arcs.

For the purpose of defining arc length between [a,b], we define a partition

\(P =[a=x_o < x_1 < x_2 < x_3 < …< x_{k-1} < x_k = b] \) of [a,b]
Then arc length L between [a,b] of f(x) is defined as

\( l=\displaystyle \lim_{k \to \infty} \sum_{x_0}^{x^k}| dl |\)
or

\( l=\displaystyle \int_a^b | dl |\)
Based on the figure above, the arc length between [a,b] is given by

\( l=\displaystyle \int_a^b dl \)

or \( l=\displaystyle \int_a^b \sqrt{dx^2+dy^2} \)

or \( l=\int_a^b \sqrt{ \left [1+(\frac{dy}{dx})^2 \right ] \times (dx)^2 } \)

or \( l=\int_a^b \sqrt{1+\left (\frac{dy}{dx}\right )^2 } dx \)

Similarly, it can be

\(l= \int_a^b \sqrt{1+\left (\frac{dx}{dy}\right )^2} dy\)

#### Example 1

Calculate the arc length of the graph of \(f(x)=x^2\) over the interval [1,3] .

Solution

Given that

\(y=x^2\)

Hence,
Differentiating w.r.to x, we get

\(\frac{dy}{dx}=2x\)

Now, the arc length is

\(l= \int_a^b \sqrt{1+\left (\frac{dx}{dy}\right )^2} dy\)

or
\(l= \int_1^3 \sqrt{1+(2x)^2} dx\)

or
\(l= \int_1^3 \sqrt{1+4x^2} dx\)

#### Example 2

Determine the arc length of \(x=\frac{2}{3}(y-1)^{3/2}\) between \(1 \le y \le 4\)

Here

\(x=\frac{2}{3}(y-1)^{3/2}\)

or \(\frac{dx}{dy}=(y-1)^{1/2}\)

The arc length is

\(l= \int_a^b \sqrt{1+\left (\frac{dx}{dy}\right )^2} dy\)

or \(l= \int_1^4 \sqrt{1+\left ((y-1)^{1/2} \right)^2} dy\)

or \(l= \int_1^4 \sqrt{1+y-1} dy\)

or \(l= \int_1^4 \sqrt{ y} dy\)

or \(l= \frac{2}{3} y^{2/3}|_1^4 \)

or \(l= \frac{14}{3}\)

#### Exercise

Determine the arc length of
- \(f(x)=2x^{3/2}\) between [0,1]
- \(f(x)=\frac{4}{3} x^{3/2}\) between [0,1]
- \(f(x)=3x^2 \) between \([1,2] \)
- \(f(x)=\frac{1}{x} \) between \([1,4] \)
- \(f(x)=\sin x \) between \([0,\pi] \)

#### Area of a curve

Solution 👉 Click Here

The application of integrals to find the area under a curve is a fundamental concept in calculus with wide-ranging practical implications. This concept is rooted in the idea of slicing up a region under a curve into infinitely thin rectangles, summing their areas, and then taking the limit as the width of the rectangles approaches zero. The resulting sum is precisely the area under the curve between two specified points.

Let \(y=f(x) be a continuous function defined on [a,b], then area under f(x) between [a,b] is defined with respect to tagged partitions of [a,b] . A tagged partition of [a, b] is a finite sequence given by

\( a=x_0 \leq x_1 \leq x_2 \leq x_3 \leq ... \leq x_n=b\)
This partitions the interval [a, b] into n sub-intervals\( [x_{i−1}, x_i] \) of length \(\Delta x_i\) indexed by i.

Now, area of f(x) with respect to such a tagged partition is defined as

Area =\( \displaystyle \lim_{n \to \infty } \sum _{i=1}^n f(x_i) \Delta x_i\)
or

Area =\( \displaystyle \int_a^b f(x)dx\)
#### Example 1

Find the area bounded by a curve \( f(x)= -x^2 + 4x + 5\) with x-axis.

The function is

\( f(x)= -x^2 + 4x + 5\)

The function cuts the x-axis at the points where y=0, this, the point of x-intercepts are

\( y=0\)

or \( -x^2 + 4x + 5=0\)

or \( x=-1,x=5\)

It shows that, x-intercept are \(x=-1\) and \(x=5\). Thus the area between x=-1 to x=5 is

Area\( = \int_a^b f(x) dx\)

or Area\( =\int_{-1}^5 f(x) dx\)

or Area\( =\int_{-1}^5 (-x^2 + 4x + 5) dx \)

or Area\( =36 \)

#### Example 2

Compute the area bounded by a curve \(f (x) = x^4-3x^2\) with x-axis.

The function is

\( f(x)= x^4-3x^2\)

The function cuts the x-axis at the points where y=0, this, the point of x-intercepts are

\( y=0\)

or \( x^4-3x^2=0\)

or \( x=-1.73,x=1.73\)

It shows that, x-intercept are x=-1.73,x=0 and x=1.73. Thus the area between x=-1.73,x=0 and x=1.73 is

Area\( = \int_a^b f(x) dx\)

or Area\( =\int_{-1.73}^{1.73} f(x) dx\)

or Area\( =\int_{-1}^5 (x^4-3x^2) dx \)

or Area\( =-4.16 \)

#### Exercise

- Find area enclosed by parabolas y = x
^{2} and y = 2x – x^{2}.
- Show that the area under the arch of the curve y = sin x is 2.
- Find the area of the region enclosed by parabola y = 2 – x
^{2} and the line y = –x.
- Determine the area of the region enclosed by \( x=\frac{1}{2}y^2\) and y=x-1
- Determine the area of the region enclosed by y= sin x, y=cos x, x=\(\pi\)/2 and x-axis
- Determine the area of the region enclosed by y=2x
^{2}+10,y=4x+16,x=-2 and x=5
- Find the area bounded by y = 3x, and the x-axis and the ordinates x = 0 and x = 4.
- Find the area enclosed by y = x
^{2}-4x+3 and the x-axis.
- Find the area bounded by y-axis and the curve x
^{2}=4(y-2) and the line y=11.
- A curve shown is the parabola is y=9 – x
^{2}. Find h so that the shaded area is two-thirds the total area bounded by the curve and the x-axis.
- Find the area of region between following the curve and x–axis
- (i) y = 3x – 5x
^{2}
- (ii) y = x
^{2} – 4x + 3
- (iii) y = x
^{2} – 8x + 15

#### Area between two curves

Solution 👉 Click Here

As we know, the area bounded by acontinuous function f (x), x–axis, x = a and x = b is given by

\(A = \int_a^b f(x) dx\)
If f(x) and g(x) are two continuous functions throughout [a, b] then area of region between curve f (x) and g(x) from a to b is

\(A = \int_a^b [f (x)-g(x)] dx\)
#### Example 1

Determine the area of the region bounded by \( y=2x^2+10\) and y=4x+16

For the intersection points of two curves,

\(2x^2+10=4x+16\)

or
\(2x^2-4x-6=0\)

or
x=-1,x=3

So, two curves intersects at x=-1,x=3

Now, the area between the curves \( y=2x^2+10\) and y=4x+16 is

\(\int_{-1}^3 \left ( f(x)-g(x) \right ) dx\)

or
\(\int_{-1}^3\left ( 4x+16-(2x^2+10) \right ) dx\)

or
\(\int_{-1}^3\left ( -2x^2+4x+6) \right ) dx \)

or
\(\left ( -\frac{2}{3} x^3-2x^2+6x \right )|_{-1}^3 =\frac{64}{3}\)

#### Example 2

Determine the area of the region bounded by x=-y^{2}+10 and x=(y-2)^{2}

For the intersection points of two curves,

-y^{2}+10=(y-2)^{2}

or
-y^{2}+10=y^{2}-4y+4

or
2y^{2}-4y-6=0

or
y=-1,y=3

So, two curves intersect at y=-1,y=3

Now, the area between the curves x=-y^{2}+10 and x=(y-2)^{2} is

\( \int_{-1}^3 \left ( f(x)-g(x) \right ) dy \)

or
\( \int_{-1}^3\left ( -y^2+10-((y-2)^2) \right ) dy\)

or
\( \int_{-1}^3\left ( -2y^2+4y+6) \right ) dy \)

or
\( \left ( -\frac{2}{3} y^3-2y^2+6y \right )|_{-1}^3 =\frac{64}{3}\)

#### Example 3

Compute the area bounded by curves f(x) = 1-x^{2} and g (x) = x^{4}-3x^{2}.

For the intersection points of two curves.

1-x^{2}=x^{4}-3x^{2}

or
\( x = -\sqrt{1 + \sqrt{2}}, x= \sqrt{1 + \sqrt{2}}\)

So, two curves intersect at x = -1.55, x= 1.55

So, the area between the curves f(x) = 1-x^{2} and g (x) = x^{4}-3x^{2} is

\( \int_{-1.55}^{1.55} \left ( f(x)-g(x) \right ) dy\)

or
\( \int_{-1.55}^{1.55} \left ( 1 - x^2 - (x^4 - 3 x^2)) \right ) dy\)

or
\( \int_{-1.55}^{1.55} \left ( 1 + 2 x^2 - x^4) \right ) dy\)

or
\( \left ( x+\frac{2}{3} x^3+\frac{1}{4}x^5 \right )|_{-1.55}^{1.55} =4.48\)

#### Example 4

Find the area between the curves \(y=x^2\) and \(y=\sqrt{x}\)

Given curves \(y=x^2\) and \(y=\sqrt{x}\) intersect at

\(y=y\)
or

\(x^2=\sqrt{x}\)
or

\(x^2-\sqrt{x}=0\)
or

\(\sqrt{x}(x \sqrt{x}-1)=0\)
x=0 and x=1, so these x=0 and x=1 are the limits of integration.

Now, the area between the curves \(y=x^2\) and \(y=\sqrt{x}\) is

\(\int_0^1 \left ( f(x)-g(x) \right ) dx\)
or

\( \int_0^1 \left ( \sqrt{x}-x^2 \right ) dx\)
or

\( \left ( \frac{2}{3} x^{2/3}-\frac{1}{3} x^3 \right )|_0^1 =\frac{1}{3} \)
#### Example 5

Compute the area bounded by curves between coordinate axes

#### Exercise

Find the Area between the curves

- \( f(x)=x^2-2x;g(x)=0\)
- \( f(x)=x^2-5x+6;g(x)=-x^2+x+6\)
- \( f(x)=\sin x;g(x)=0\) between \(-\pi\) to \(\pi\)
- \( f(x)=\sqrt{x}+1;g(x)=\frac{1}{2}x+1\)
- \(y=2x,y=2−2x,y=0\)
- \( f( y)=y(4−y),g( y)=−y\)
- \( f( y)=y^2+2,g( y)=0,y=−1,y=2\)

#### Area of a Surface of Revolution

Solution 👉 Click Here

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. Here, we find the surface area of the surface of revolution created by revolving the graph of \(𝑦=𝑓(𝑥)\) around the 𝑥 -axis or around y-axis.

For this we use the notion of arc length and frustum of a cone.

Using above figure, the surface area of the frustum is given by

\(A=\pi(r_1+r_2) l \)

Let \(y=f(x)\) be a continuous function, then by the Intermediate Value theorem, , there is a point x between \(r_1,r_2\)such that

\(A=\pi(r_1+r_2) l \)

or \(A=2 \pi f(x) l \)

or \(A=2 \pi f(x) \int_a^b \sqrt{1+\left (\frac{dy}{dx}\right )^2 } dx \)

As with arc length, we can conduct a similar development for functions of 𝑦 to get a formula for the
surface area of surfaces of revolution about the 𝑦−𝑎𝑥𝑖𝑠

\(A=2 \pi f(y) \int_a^b \sqrt{1+\left (\frac{dx}{dy}\right )^2 } dy \)

#### Exercise

- Let \(𝑓(𝑥)=\sqrt{x}\) over the interval [1,4], Find the surface area of the surface generated by revolving the graph of 𝑓(𝑥) around the 𝑥-axis.
- Let \(𝑓(𝑥)=\sqrt{1-x} \) over the interval [0,1/2], Find the surface area of the surface generated by revolving the graph of 𝑓(𝑥) around the 𝑥-axis.
- Let \(𝑓(y)=\frac{y^3}{3}\) over the interval [0,2], Find the surface area of the surface generated by revolving the graph of 𝑓(y) around the y-axis.
- Let \(𝑓(y)=\sqrt{9-y^2} \) over the interval [0,2], Find the surface area of the surface generated by revolving the graph of 𝑓(y) around the y-axis.

#### Exercise

- Find the volume of the solid generated by revolving the region between the y–axis and curve \(x = 2 \sqrt{y} , 0 \le y \le 4\) about the y–axis.
- Find the volume of solid obtained by rotating about the x–axis the region bounded by curve \(y = \sqrt{x} , x = 0, x = 1\)
- Find the volume of solid generated by revolving the region bounded by y = x
^{2}, y = 0, x = 2 about x–axis.
- Find the volume of solid obtained by rotating the region bounded by \(y = 2 – \frac{x}{2} , y = 0, x = 1, x = 2\) about x–axis.
- Find the volume of solid obtained by rotating the region bounded by \(x = 2 \sqrt{x} , x = 0, y = 9\) about y–axis.
- A pyramid 3 m high has a square base that is 3 m on a side. The cross–section of the pyramid perpendicular to the altitude x m down from the vertex is a square x m on a side. Find the volume of the pyramid.
- A mixing bowl is a hemisphere of radius 5 in. Determine the height of 100 cubic inches of liquid.

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