Definite Integral Calculator

The subject of mathematics is incredibly diverse. There are near-infinite concepts and topics, and one of them is integral. We can understand an integral as either a numerical value equal to the area under the graph of a function for some interval, known as the definite integral. We can integrate a definite integral of any function by using the indefinite integral and a corollary to the fundamental theorem of calculus. Before getting into the nuances of the definite integral calculator, let us analyze what integration is. 

The definition of integration is related to the application of definite integrals as well indefinite integrals. In simple words, integration is the calculation of an integral. As we have already discussed, an integral is used to find many valuable quantities such as areas, volumes, displacement, and so forth. It is noteworthy to illustrate that when we say  integra, we usually mean definite integrals. As such, the definite integral solver becomes crucial in Calculus and mathematics. 

Integration represents the sum of discrete data. It is calculated to extract the functions that will describe aspects such as area, displacement, volume, and so forth that occur due to the amalgamation of small data that we cannot measure singularly. Now that we have touched upon the fundamentals of integration, let us try and explore the nitty-gritty of integral calculus. 

A Brief Discussion on Integral Calculus

As per pioneering mathematician Bernhard Riemann – an integral is premised on a limiting or restricting procedure that assumes the area of a curvilinear region by breaking down the part into thin slabs of vertical length. For example, we can investigate the slope by using a slope formula, but if we want to find an area of a curve, we have to be equipped with differential calculus. You can use the Wolfram integral calculator to grasp the topic of integral and integration further. 

Definite Integral Calculator – Standard Format and Formula

The definite integral is also deemed as the Riemann integral. The definite integral of a function is equal to the area of the region bounded by the curve. A definite integral relates to numerical values and is used in pure mathematics, applied mathematics, statistics, science, and so forth. To get into the crux of the definite integral solver, we have to know the formula of definite integrals which are –

\[\int_{b}^{a}f(x)dx\] = \[-\int_{a}^{b}f(x)dx\]

\[\int_{b}^{a}f(x)dx\] = 0

\[\int_{a}^{b}cf(x)dx\] = \[c\int_{a}^{b}f(x)dx\]

\[\int_{a}^{b}f(x)\pm g(x)dx\] = \[\int_{a}^{b}f(x)dx\] $\pm$ \[\int_{a}^{b}g(x)dx\]

\[\int_{a}^{b}f(x)dx\] = \[\int_{a}^{c}f(x)dx\] + \[\int_{c}^{b}f(x)dx\]

Wolfram Definite Calculator – Trigonometric Functions 

Wolfram Alpha is a computational knowledge engine and answer engine developed by Wolfram research. It answers questions related to definite integers, trigonometry, and so forth. It is an important tool for people who want answers to complex questions. As such, we can understand the definite integral calculator with steps. 

Applications of Definite Integral

The use of definite integrals is manifold and diverse. They are used in probability theory to determine the probability of some random variable falling within a specific range. Integrals can also compute the area of a two-dimensional region with a curved boundary and calculate the volume of three-dimensional objects with a curved perimeter. Integrals are also utilized in physics in kinematics, displacement, time, velocity, and so forth. They also have a crucial role in thermodynamics. You can opt for a wolfram integral calculator to know more about the topic.

Definite Integral Solver to Find the Area of a Circle 

It is noteworthy to mention that definite integrals can be easily used to calculate the area of a circle. 

(Image will be uploaded soon)

The area of the circle is extracted by calculating the area of the part of the circle in the first quadrant, and the equation used is x²+y²=a². We will find the area of the one-quarter circle and multiply it by 4 to determine the circle's total area since circles are symmetric around the x and y axes.

Find the value of y in the given equation above.

After that, the equation is changed to the equation of a curve \[Y=\sqrt{a^2-x^2}\]. 

Then, we utilize the concept of definite integral to extract the equation of the curve and give  the limits from '0' to 'a'. The area of a circle is four times the area of a quadrant. The area of the quadrant is calculated by integrating the equation of the curve across the limits in the first quadrant.

∴ Area of circle =4 × Area of first quadrant

\[A=\int_{0}^{a}y.dx\]

Here, putting the value of y from above, we get,

     = \[4\int_{0}^{a}\sqrt{a^2-x^2}.dx\]

= \[4[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}]_0^a\]

     = \[4[0+\frac{a^2}{2}sin^{-1}\frac{a}{a}-(0+\frac{a^2}{2}sin^{-1}\frac{0}{a})]\]

     = \[4[\frac{a^2}{2}sin^{-1}1-0]\]

     = \[4(\frac{a^2}{2})(\frac{\pi }{2})\]

     = \[\pi a^2\]

Thus, the area of the circle is \[\pi a^2\] square units.

 Area of a Parabola by Using a Definite Integral

We know that a parabola comprises an axis that divides it into two equal or symmetric parts. 

(Image will be uploaded soon)

A parabola that is equal along the x-axis and has an equation of y²=4ax. Then we transform the equation into y=√4ax. We begin by finding the area of the parabola in the first quadrant by utilizing the definite integral formulas vis-à-vis the x-axis and the limits from '0' to 'a'. After that, the definite integral can be extracted within the boundary, and we have to double it to extract the area of the whole parabola. 

\[A=2\int_{0}^{a}\sqrt{4ax}.dx\]

= \[4\sqrt{a}\int_{0}^{a}\sqrt{x}.dx\]

= \[4\sqrt{a}[\frac{2}{3}.x^\frac{3}{2}]_0^a\]

= \[4\sqrt{a} (\frac{2}{3}a^\frac{3}{2})-0\]

= \[\frac{8a^2}{3}\]

Hence, the area under the curve covered by the parabola is \[\frac{8a^2}{3}\] square units.

Area of An Ellipse By Using a Definite Integral

We know that the ellipse equation with the major axis of length 2a and a minor axis of 2b is  \[\frac{x^2}{a^2} + \frac{y^2}{b^2}=1\] . Then we have to transform the equation to \[y=\frac{b}{a}\sqrt{(a^2-x^2)}\] . 

(Image will be uploaded soon)

After that, the definite integral concept has to be appropriated to calculate the area bounded by the ellipse in the first coordinate and the x-axis. Then, we have to multiply the equation by 4 to obtain the area of the ellipse. 

\[A=4\int_{0}^{a}y.dx\]

= \[4\int_{0}^{a}\frac{b}{a}(a^2-x^2).dx\]

= \[4\frac{b}{a}[\frac{x}{2}.\sqrt{a^2-x^2} + \frac{a^2}{2}sin^{-1}\frac{x}{a}]_0^a\]

= \[4\frac{b}{a}(\frac{a}{2}\times0)+\frac{a^2}{2}sin^{-1}-0]\]

= \[4\frac{b}{a}\times\frac{a^2}{2}\times\frac{\pi }{2}\]

= \[\pi ab\]

Thus, the area of the ellipse is \[\pi ab\] sq units.

In conclusion, definite integral is a crucial part of mathematics and forms the basis of various branches of the discipline. It is imperative that we know the basics of the topic.