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Integral calculus helps in finding the anti-derivatives of a function. These anti-derivatives are also called the integrals of the function. The process of finding the anti-derivative of a function is called integration. The inverse process of finding derivatives is finding the integrals. The integral of a function represents a family of curves. Finding both derivatives and integrals form the fundamental calculus. In this topic, we will cover the basics of integrals and evaluating integrals.
|1.||What is Integral Calculus?|
|2.||Fundamental Theorems of Integrals|
|3.||Types of Integrals|
|4.||Properties of Integrals|
|6.||Methods of Integrals|
|7.||Applications of Integrals|
|8.||FAQs on Integrals|
What is Integral Calculus?
Integrals are the values of the function found by the process of integration. The process of getting f(x) from f'(x) is called integration. Integrals assign numbers to functions in a way that describe displacement and motion problems, area and volume problems, and so on that arise by combining all the small data. Given the derivative f’ of the function f, we can determine the function f. Here, the function f is called antiderivative or integral of f’.
Example: Given: f(x) = x2 .
Derivative of f(x) = f'(x) = 2x = g(x)
if g(x) = 2x, then anti-derivative of g(x) = ∫ g(x) = x2
Definition of Integral
F(x) is called an antiderivative or Newton-Leibnitz integral or primitive of a function f(x) on an interval I. F'(x) = f(x), for every value of x in I.
Integral is the representation of the area of a region under a curve. We approximate the actual value of an integral by drawing rectangles. A definite integral of a function can be represented as the area of the region bounded by its graph of the given function between two points in the line. The area of a region is found by breaking it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summed up. We specify an integral of a function over an interval on which the integral is defined.
Fundamental Theorems of Integral Calculus
We define integrals as the function of the area bounded by the curve y = f(x), a ≤ x ≤ b, the x-axis, and the ordinates x = a and x =b, where b>a. Let x be a given point in [a,b]. Then \(\int\limits_a^b f(x) dx\) represents the area function. This concept of area function leads to the fundamental theorems of integral calculus.
- First Fundamental Theorem of Integral Calculus
- Second Fundamental Theorem of Integral Calculus
First Fundamental Theorem of Integrals
A(x) = \(\int\limits_a^b f(x) dx\) for all x ≥ a, where the function is continuous on [a,b]. Then A'(x) = f(x) for all x ϵ [a,b]
Second Fundamental Theorem of Integrals
If f is continuous function of x defined on the closed interval [a,b] and F be another function such that d/dx F(x) = f(x) for all x in the domain of f, then \(\int\limits_a^b f(x) dx\) = f(b) -f(a). This is known as the definite integral of f over the range [a,b], a being the lower limit and b the upper limit.
Types of Integrals
Integral calculus is used for solving the problems of the following types.
a) the problem of finding a function if its derivative is given.
b) the problem of finding the area bounded by the graph of a function under given conditions. Thus the Integral calculus is divided into two types.
- Definite Integrals (the value of the integrals are definite)
- Indefinite Integrals (the value of the integral is indefinite with an arbitrary constant, C)
These are the integrals that do not have a pre-existing value of limits; thus making the final value of integral indefinite. ∫g'(x)dx = g(x) + c. Indefinite integrals belong to the family of parallel curves.
The definite integrals have a pre-existing value of limits, thus making the final value of an integral, definite. if f(x) is a function of the curve, then \(\int\limits_a^b f(x) dx = f(b) - f(a)\)
Properties of Integral Calculus
Let us study the properties of indefinite integrals to work on them.
- The derivative of an integral is the integrand itself. ∫ f(x) dx = f(x) +C
- Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. ∫ [ f(x) dx -g(x) dx] =0
- The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the individual functions. ∫ [ f(x) dx+g(x) dx] = ∫ f(x) dx + ∫ g(x) dx
- The constant is taken outside the integral sign. ∫ k f(x) dx = k ∫ f(x) dx, where k ∈ R.
- The previous two properties are combined to get the form: ∫ [k\(_1\)f\(_1\)(x) + k\(_2\)f\(_2\)(x) +... k\(_n\)f\(_n\)(x)] dx = k\(_1\)∫ f\(_1\)(x)dx + k\(_2\)∫ f\(_2\)(x)dx+ ... k\(_n\) ∫ f\(_n\)(x)dx
We can remember the formulas of derivatives of some important functions. Here are the corresponding integrals of these functions that are remembered as standard formulas of integrals.
- ∫ xn dx=xn+1 /n+1+C, where n ≠ -1
- ∫ dx =x+C
- ∫ cosxdx = sinx+C
- ∫ sinx dx = -cosx+C
- ∫ sec2x dx = tanx+C
- ∫ cosec2x dx = -cotx+C
- ∫ sec2x dx = tanx+C
- ∫ secx tanxdx = secx+C
- ∫ cscx cotx dx = -cscx+C
- ∫1/(√(1-x2))= sin-1 x + C
- ∫-1/(√(1-x2))= cos-1 x + C
- ∫1/(1+x2)= tan-1 x + C
- ∫-1/(1+x2)= cot-1 x + C
- ∫1/(x√(x2 -1))= sec-1 x + C
- ∫-1/(x√(x2 -1))= cosec-1 x + C
- ∫ exdx=ex + C
- ∫dx/x=ln|x| + C
- ∫ ax dx=ax/ln a + C
Methods to Find Integrals
There are several methods adopted for finding the indefinite integrals. The prominent methods are:
- Finding integrals by integration by substitution method
- Finding integrals by integration by parts
- Finding integrals by integration by partial fractions.
Finding Integrals by Substitution Method
A few integrals are found by the substitution method. If u is a function of x, then u' = du/dx.
∫ f(u)u' dx = ∫ f(u)du, where u = g(x).
Finding Integrals by Integration by Parts
If two functions are of the product form, integrals are found by the method of integration by parts.
∫f(x)g(x) dx = f(x)∫ g(x) dx - ∫ (f'(x) ∫g(x) dx) dx.
Finding Integrals by Integration by Partial Fractions
Integration of rational algebraic functions whose numerator and denominator contain positive integral powers of x with constant coefficients is done by resolving them into partial fractions.
To find ∫ f(x)/g(x) dx, decompose this improper rational function to a proper rational function and then integrate.
∫f(x)/g(x) dx = ∫ p(x)/q(x) + ∫ r(x)/s(x), where g(x) = a(x) . s(x)
Applications of Integral Calculus
Using integration, we can find the distance given the velocity. Definite integrals form the powerful tool to find the area under simple curves, the area bounded by a curve and a line, the area between two curves, the volume of the solids. The displacement and motion problems also find their applications of integrals. The area of the region enclosed between two curves y = f(x) and y = g(x) and the lines x =a, x =b is given by
Area = \(\int\limits_a^b (f(x) -g(x))dx\)
Let us find the area bounded by the curve y = x and y = x2 that intersect at (0,0)and (1,1).
The given curves are that of a line and a parabola. The area bounded by the curves = \(\int\limits_0^1 (y_2 -y_1)dx\)
Area = \(\int\limits_0^1 (x-x^2)dx\)
= x2 /2- x 3/3
= 1/6 sq units.
- The primitive value of the function found by the process of integration is called an integral.
- An integral is a mathematical object that can be interpreted as an area or a generalization of area.
- When a polynomial function is integrated the degree of the integral increases by 1.
☛ Also Check:
Integral Calculus Examples
Example 1. Find the integral of e3x
∫ d/dx(f(x)) = ∫ d/dx( e3x)
We know this is of the form of integral, ∫ d/dx( eax) = 1/a eax + C
∫ d/dx( e3x) = 1/3 e3x + C
Answer: The integral of e3x = 1/3 e3x + C
Example 2. Find the integral of cos 3x.
∫ d/dx(f(x)) =∫ cos 3x
Let 3x = t
thus x = t/3
dx = dt/3
The given integral becomes ∫1/3(cos t) dt
= 1/3(sin t) + C
= 1/3 sin (3x) + C
Answer: The integral of cos 3x = 1/3 sin (3x) + C
Example 3. Evaluate the integral i = \(\int\limits_2^3\) (x+1) dx
By the 2nd theorem of fundamentals of integrals we know that \(\int\limits_a^b F(x) dx = f(b) - f(a)\)
\(\int\limits_2^3\) (x+1) dx = f(3) -f(2)
f(x) = x2/2 + x + C
f(3) = 32/2 +3 = 9/2 + 3 = 15/2
f(2)= 22/2 + 2 = 4/2 + 2 = 4
f(3) -f(2) = 15/2 - 4
Answer: The value of the given integral I = 7/2
FAQs on Integral Calculus
What Are Integrals?
Integrals are the values of the function found by the process of integration. An integral is defined as the area of the region under the curve that is represented as a function y = f(x).
What is The Integrals Symbol Called?
The ntegrals symbol is ∫. This means that it is bound to a limit from the lower to higher and that the integrals represent the area of the curve under the graph of the function.
What Are The Types of Integrals?
The two types of integrals are definite integral and indefinite integral. The definite integrals are bound by the limits. The indefinite integrals are not bound to pre-existing values.
Can an Integral Have Two Answers?
Yes, an indefinite integral can have infinite answers depending upon the value of the constant term; while a definite integral will be a constant value.
What is a Double Integral Used For?
A double integral is used in order to calculate the areas of regions, find the volumes of a given surface, or also the mean value of any given function in a plane region.
How Do you Find The Integrals?
Finding integrals is the inverse operation of finding the derivatives. A few integrals are remembered as formulas. For example, ∫ xn = xn+1 / (n+1) + C. Thus x6 = x6+1 / 6+1 = x7 / 7 + C. A few integrals use the techniques of integration by parts, integration by partial fractions, substitution method, and so on.
How Do You Use Integrals using Trigonometry?
Use the trigonometric identities and simplify the function into integrable function and then apply the formulas and adopt the integration procedures to find the integrals using trigonometry.
What is The Integral of sin x?
The integral of sine x is -cos x + C. ∫ sin x dX = -cos x + C.
What is Integral Calculus Used For?
We use definite integrals to find the area under the curve or between the curves that are defined by the functions, we find their indefinite integrals using the formulas and the techniques and then find their difference of the integrals applying the limits. We use definite integrals for computing the volumes of 3-d solids. Given the velocity, we can find the distance as the distance is the integral of velocity.