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Integral Calculus is a branch of Calculus that studies the theory, properties, and applications of integrals. Integrals are defined as the values of the function calculated through the process of integration. Integration is a process used to find the antiderivatives, also known as integrals, of a function.
- Integral Calculus forms the basis of mathematical analysis along with Differential Calculus.
- Differential Calculus and Integral Calculus are linked through the Fundamental Theorem of Calculus
- Integral Calculus is concerned with lengths, areas, volumes, and the derivation of the antiderivative formula.
Read More: NCERT Solutions For Class 12 Mathematics Integrals
Key Terms: Integral Calculus, Integration, Differential Calculus, Definite Integral, Indefinite Integral, Fundamental Theorem of Calculus, Calculus, Integral, Differentiation, Integration
What is Integral Calculus?
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Integral Calculus is the study of the properties and applications of the indefinite integral and definite integral. Integration is the process of calculating the value of an integral. Integration is an important concept as it refers to the inverse of differentiation.
- In differential calculus, f’ is referred to as the derivative of the function f.
- In integral calculus, f is referred to as the anti-derivative or primitive of function f’.
- Integration is the process of obtaining f(x) from f'(x).
Example: Given that f(x) = x2
Derivative of f(x) = f'(x) = 2x = g(x)
Now, if g(x) = 2x, then, the anti-derivative of g(x) = ∫ g(x) = x2
Introduction to Integrals Detailed Video Explanation
Read More:
| Relevant Concepts | ||
|---|---|---|
| Integral Calculus Formula | Continuous Integration | Differentiation and Integration Formula |
| Definite Integral Formula | Line Integral | Double Integral |
Fundamental Theorem of Integral Calculus
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Fundamental Theorem of Calculus is a theorem that links the concepts of integration and differentiation. Integrals are defined as the function of the area covered by the curve y = f(x), a ≤ x ≤ b, x-axis, and the ordinates x = a and x = b, where b>a.
Assume x to be a given point in [a,b]. Then, the area of a function is represented as \(\int\limits_a^b f(x) dx\).
The concept of area function, thus, leads to two fundamental theorem of integral calculus:
- First Fundamental Theorem of Integral Calculus
- Second Fundamental Theorem of Integral Calculus
First Fundamental Theorem of Integrals
When the function is continuous on [a,b], then
| \(A(x) =\int\limits_a^b f(x) dx \,for \,all \,x ≥ a\) |
Therefore, A'(x) = f(x) for all x ϵ [a,b].
Second Fundamental Theorem of Integrals
If f is the continuous function of x stated on the closed interval [a,b] and F is 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 defined as the definite integral of f over the range [a,b], with a being the lower limit and b being the upper limit.
Types of Integrals
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Integral Calculus is used to solve the following types of problems:
- Finding a function if its derivative is given.
- Finding the area covered by the graph of a function under given conditions.
Therefore, Integral Calculus is divided into two major types as follows:
- Definite Integral (Value of Integral is Definite)
- Indefinite Integral (Value of Integral is Indefinite with an Arbitrary Constant, C)
Definite Integral
Definite integral is defined as an integral that has a pre-existing value of limits, i.e. it has both upper and lower limits. It is also referred to as Riemann Integral when it is constrained to lie on the real line.
Definite Integrals are represented as follows:
| \(\int_{a}^{b}f(x)dx\) |
Indefinite Integral
Indefinite Integral is a function that takes the antiderivative of another function. They do not have a pre-existing value of limits, i.e. upper and lower limits aren't used to define the indefinite integral. Indefinite integrals represent the family of functions whose derivatives are f, and thus, it returns a function of the independent variable.
If the integration of a function f (x) is F(x), then, it is represented as
| ∫f(x) dx = F(x) + C |
Where
- R.H.S. of the Equation: Integral of (x) with respect to x
- F(x): Antiderivative or Primitive
- f(x): Integrand
- dx: Integrating Agent
- x: Variable of Integration
- C: Constant of Integration
Properties of Integral Calculus
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The important properties of integrals are listed below:
(I) The derivative of an integral is the Integrand itself.
| ∫ f(x) dx = f(x) + C |
(II) Two indefinite integrals with the same derivative are equivalent and lead to the same family of curves.
| ∫ [ f(x) dx – g(x) dx] = 0 |
(III) The integral of the sum or difference of a finite number of functions is equivalent to the sum or difference of the integrals of the individual functions.
| ∫ [ f(x) dx+g(x) dx] = ∫ f(x) dx + ∫ g(x) dx |
(IV) The constant is placed outside the sign of the integral.
| ∫ k f(x) dx = k ∫ f(x) dx; k ∈ R |
The last two properties are combined in order to get the form:
| ∫ [k1f1(x) + k2f2(x) +... knfn(x)] dx = k1∫ f1(x)dx + k2∫ f2(x)dx+ ... kn ∫ fn(x)dx |
Integral Calculus Formulas
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Given below is the list of integral formulas that are considered the standard formulas of integral calculus:
- ∫ xn dx=xn+1 /n+1+C, 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
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Integrals of a function are found using the three main methods of integration listed as follows:
- Integration by Substitution Method
- Integration by Parts
- Integration by Partial Fractions
Integration by Substitution Method
Few integrals can be 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).
Integration by Parts
When two functions are given in a form of a product, integrals are found by the integration by parts method.
| ∫f(x)g(x) dx = f(x)∫ g(x) dx - ∫ (f'(x) ∫g(x) dx) dx |
By Parts Method of integrals Video Explanation
Integration by Partial Fractions
Integration of rational functions whose numerator and denominator comprise positive integral powers of x with constant coefficients is done by resolving them into partial fractions.
To find ∫ f(x)/g(x) dx, the improper rational function is decomposed into a proper rational function and is then integrated.
| ∫f(x)/g(x) dx = ∫ p(x)/q(x) + ∫ r(x)/s(x) |
Where g(x) = a(x) . s(x)
Uses of Integrals
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Integral Calculus is used for two major purposes:
- Integral Calculus is used to calculate f from f’.
- If a function f can be differentiated in the interval of consideration, then f’ is defined.
- The derivatives of a function can be undone with the help of integral calculus in differential calculus.
- Integral Calculus is also used to calculate the area under a curve.
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Application of Integral Calculus
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Given below are some of the most essential applications of Integral Calculus:
- Area between Two Curves
- Centre of Mass
- Surface Area
- Volume
- Kinetic Energy
- Work
- Average Value of a Function
- Distance, Acceleration, and Velocity
Calculus is used in physics, chemistry, medicine, economics, biology, engineering, space exploration, statistics, and pharmacology, among other fields. Integral calculus is used in areas such as the area under curves, arc length, surface area, volume, probability, and so on.
Integral Calculus Examples
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Here are some solved examples on Integral Calculus:
Example 1: Evaluate \(i =\int\limits_2^3 (x+1)\,dx\).
Solution: Using the Second Theorem of the Fundamental Theorem of Integral calculus, we get
\(\int\limits_a^b F(x) dx = f(b) - f(a)\)
\(\int\limits_2^3\) (x+1) dx = f(3) -f(2)
f(x) = x2/2 + x + C
Thus,
- 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 = 7/2
Therefore, the value of the given integral is 7/2.
Example 2: Integrate f(x) = 2x sin(x2+1) with respect to x.
Solution: The derivative of x2+1 is 2x.
Here, we will use Integration by Substitution,
Let x2+1 = z
So, 2x dx = dz
∫ f(x)dx = ∫ 2xsin (x2+1)dx
= ∫ sin zdz
= −cosz + C
= −cos(x2+1) + C
Therefore,
∫ 2x sin (x2+1)dx = −cos(x2+1) + C
Things to Remember
- Integral Calculus is a branch of calculus concerned with the theory and applications of integrals.
- Differential Calculus and Integral Calculus are two important components of Calculus.
- Integrals are the anti-derivatives of a function calculated through Integration.
- Definite Integrals and Indefinite Integrals are the two major types of integrals.
- Integrals can be calculated by substitution method, integration by parts, and integration by partial fractions.
- Fundamental Theorem of Calculus links the concepts of integration and differentiation.
- Integral Calculus is used for calculations involving arc length, pressure, center of mass, volume, area, and work.
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Previous Year Questions
- The area of the region A=[(x,y) : 0≤y≤x∣x∣+1 and −1≤x≤1] in s units is... (JEE Main – 2019)
- \(\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} \)is equal to... (BITSAT - 2011)
- Let [.] denote the greatest integer function then the value of… (AIEEE - 2011)
- \(\displaystyle\int_{1/2}^{2}|\log_{10}\,x|dx=\)...
- The area (in s units) of the region {(x,y) : x≥0, x + y≤3, x2≤4y and... (JEE Main – 2017)
- If \(∫xlog\,(1 + {1 \over x})\)= dx = f(x)log(x+1) + g(x)x2 + Lx+C then... (BITSAT – 2017)
- If a is a positive integer, then the number of values...
- The integral ∫cos (logx) dx is equal to… (JEE Main - 2019)
- For a>0, let the curves C1:y2 = ax and C2:x2 = ay intersect at origin O... (JEE Main – 2020)
- \(\int\limits_{0} ^{1}\frac{dx}{[ax+b(1 x)]^2}\) is equal to…
Sample Questions
Ques. Find the integral of cos2n with respect to n. (3 Marks)
Ans. Let us assume that f(n) = cos2n
We know that, 2 cos2A = cos 2A + 1
Hence, f(n) = (1/2)(cos 2n + 1)
Now, find the integral of f(n).
∫f(n) dn = ∫(1/2)(cos 2n + 1) dn = (1/2) ∫(cos 2n + 1) dn
= (1/2) ∫cos 2n dn + (1/2)∫1 dn = (1/2) (sin 2n/2) + (1/2) n + C
= (sin 2n/4) + (n/4) + C = (1/4)[sin 2n + n] + C
Therefore, the integral of cos2n with respect to n is (1/4)[sin 2n + n] + C.
Ques. Evaluate \(\begin{array}{l}\int_{0}^{\pi}sin x\ dx\end{array}\). (3 Marks)
Ans. We know that, ∫sin x dx = cos x + C. Now,
\(\begin{array}{l}\int_{0}^{\pi}sin x\ dx = [-cosx]_{0}^{\pi}\end{array}\)
= -cos π – (-cos 0)
= -(-1) + 1
= 1 + 1 = 2
Therefore, the value is calculated as 2.
Ques. Integrate 2x cos (x2 – 5) with respect to x. (3 Marks)
Ans. I = ∫2xcos(x2 – 5).dx
Let x2 – 5 = t …..(1)
2x.dx = dt
Substituting these values, we have
I = ∫cos(t).dt
= sint + c …..(2)
Substituting the value of 1 in 2, we have
= sin (x2 – 5) + C
This is the required integration for the given function.
Ques. Determine the integral of f(x) = √x. (3 Marks)
Ans. Given that,
f(x) = √x
∫f(x) dx = ∫√x dx
\(\int \sqrt{x}\ dx = \int x^{\frac{1}{2}}\ dx\)
We know that,
\(\int x^{n}\ dx = \frac{x^{n+1}}{n+1}+C\)
Proceed further to get
\(\int \sqrt{x}\ dx = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\)
\(= \frac{x^{\frac{1+2}{2}}}{\frac{1+2}{2}}+C\)
\(=\frac{2}{3}x^{\frac{3}{2}}+C\)
Therefore, the integral of f(x) = √x is \(\frac{2}{3}x^{\frac{3}{2}}+C\).
Ques. Find the integral of cos 3x. (3 Marks)
Ans. ∫ d/dx(f(x)) =∫ cos 3x
Let us assume that 3x = t
Thus, x = t/3
dx = dt/3
The integral thus becomes ∫1/3(cos t) dt
= 1/3(sin t) + C = 1/3 sin (3x) + C
Therefore, the integral of cos 3x is 1/3 sin (3x) + C.
Ques. Evaluate the given integral: ∫23 y2dy. (3 Marks)
Ans. Assume that I = ∫23 y2dy
We know that,
∫y2dy = y3/3 = F(y)
Using the Second Fundamental Theorem of Calculus,
I = F(3) – F(2) = 27/3 – 8/3 = 19/3
Ques. Evaluate ∫12[ydy/(y+1)(y+2)]. (3 Marks)
Ans. Factoring the term under integral using the partial fractions, we get
y/[(y+1)(y+2)] = [-1/(y+1)]+[2/(y+2)]
∫y/[(y+1)(y+2)] = -log|y+1|+2log|x+2| = F(y)
Thus, using the Second Fundamental Theorem of Calculus, we get;
I = F(2)-F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3] I = – 3 log 3 + log 2 + 2 log 4
I = log(32/27)
Ques. What will be the Integral of e3x? (2 Marks)
Ans. ∫ d/dx(f(x)) = ∫ d/dx( e3x)
As we know, the form of integral is ∫ d/dx( eax) = 1/a eax + C
∫ d/dx( e3x) = 1/3 e3x + C
Therefore, the integral of e3x is 1/3 e3x + C.
Ques. Name the types of Integrals. (2 Marks)
Ans. Integrals can be classified into two types namely Definite Integral and Indefinite Integral.
- Indefinite Integrals: Indefinite integrals are not bound to pre-existing values.
- Definite Integrals: Definite integrals are bound by limits.
Ques. What is the symbol of Integrals? (1 Mark)
Ans. The symbol of integrals is ∫. It means that it is bound to a limit from lower to higher. Integrals represent the area of the curve under the graph of the function.
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