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Definite Integral Definite integral is generally considered to be a tough topic by students. It must be studied after one is thorough with the concepts of indefinite integrals. The topic is flooded with formulae related to change of limits etc. and hence demands consistent practice. It is an important component of the IIT JEE Mathematics syllabus and so students cannot dare to skip this topic. It not only requires learning of all important formulae but also requires a good speed with thorough knowledge of integration. Many students grasp integration but make silly errors when completing the two part calculation. Simple peer assessment could help students to think about setting each step of their work minimizing these small but significant slips. This work could be made more engaging by inviting students to mix and match integration problems with graphical representations. Broadly speaking, this section also forms the root of all scientific organizations and studies. The topics covered under it are listed below: Geometrical Interpretation of Definite Integral Fundamental Theorems of Calculus Properties of Definite Integration Differentiation Under Integral Sign Definite Integral as Limit of Sum Area as Definite Integral Working Rule for Finding Area Objective Problems of Definite Problems Solved Examples of Area Definite Integral Area Bounded by Parabola Straight Line Circle Solved Examples of Definite Integral These heads have been discussed in detail in the coming sections. Here we shall discuss them in brief: What do you mean by a definite integral? A quantity of the form is called the definite integral of f(x) between the limits ‘a’ and ‘b’, where d/dx (F(x)) = f(x). How is indefinite integral different from definite integral? Indefinite integral lays down the foundation for definite integral. Both the concepts are synonymous and interrelated. As the name suggests, while indefinite integral refers to the evaluation of indefinite area, in definite integration the area is to be calculated within specific limits. The figure given below illustrates clearly the difference between definite and indefinite integration: Some of the important properties of definite integrals are listed below: The value of the integral is unaffected by the variable of integration as long as the limits of integration remain the same. If the limits of integration are interchanged, the value of integration remains same but the sign changes. If the value of the integral is zero, then the equation f(x) = 0 has at least one root in (a, b) provided f is continuous in (a, b). The converse of the above result is not true. where c is a point lying inside or outside the interval [a, b], provided the function ‘f’ is piecewise continuous in (a, b). where ‘a’ is the period of the function i.e. f(a + x) = f(x). where f(x) is periodic with period ‘a’. where K = π/2, if both m and n are even = 1 otherwise If g is continuous on [a, b] and f_{1} (x) and f_{2} (x) are differentiable functions whose values lie in [a, b], then d/dx ∫ f_{2}(x) f_{1}(x) g(t) dt = g (f_{2}(x)) f_{2}'(x) – g (f_{1}(x)) f_{1}'(x) Express the given series in the form ∑ 1/n f (r/n) Then the limit is its sum when n→∞, i.e. lim n→∞ h ∑1/n f(r/n) Replace r/n by x and 1/n by dx and lim n→∞ ∑ by the sign of ∫. The lower and the upper limit of integration are the limiting values of r/n for the first and the last term of r respectively. The definite integral can be interpreted to represent the area under the graph. This follows from the definition itself that the definite integral is a sum of the product of the lengths of intervals and the "height" of the function being integrated in that interval including the formula for the area of the rectangle. The figure given below clearly illustrates that as we find the value of the definite integral a to b, we are actually finding the area of the shaded portion. Some Important Expansions
Definite integral is generally considered to be a tough topic by students. It must be studied after one is thorough with the concepts of indefinite integrals. The topic is flooded with formulae related to change of limits etc. and hence demands consistent practice. It is an important component of the IIT JEE Mathematics syllabus and so students cannot dare to skip this topic. It not only requires learning of all important formulae but also requires a good speed with thorough knowledge of integration.
Many students grasp integration but make silly errors when completing the two part calculation. Simple peer assessment could help students to think about setting each step of their work minimizing these small but significant slips. This work could be made more engaging by inviting students to mix and match integration problems with graphical representations. Broadly speaking, this section also forms the root of all scientific organizations and studies.
Geometrical Interpretation of Definite Integral
Fundamental Theorems of Calculus
Properties of Definite Integration
Differentiation Under Integral Sign
Definite Integral as Limit of Sum
Area as Definite Integral
Working Rule for Finding Area
Objective Problems of Definite Problems
Solved Examples of Area Definite Integral
Area Bounded by Parabola Straight Line Circle
Solved Examples of Definite Integral
These heads have been discussed in detail in the coming sections. Here we shall discuss them in brief:
A quantity of the form
is called the definite integral of f(x) between the limits ‘a’ and ‘b’, where d/dx (F(x)) = f(x).
Indefinite integral lays down the foundation for definite integral. Both the concepts are synonymous and interrelated. As the name suggests, while indefinite integral refers to the evaluation of indefinite area, in definite integration the area is to be calculated within specific limits. The figure given below illustrates clearly the difference between definite and indefinite integration:
The value of the integral is unaffected by the variable of integration as long as the limits of integration remain the same.
If the limits of integration are interchanged, the value of integration remains same but the sign changes.
If the value of the integral is zero, then the equation f(x) = 0 has at least one root in (a, b) provided f is continuous in (a, b).
The converse of the above result is not true.
where c is a point lying inside or outside the interval [a, b], provided the function ‘f’ is piecewise continuous in (a, b).
where ‘a’ is the period of the function i.e. f(a + x) = f(x).
where f(x) is periodic with period ‘a’.
where K = π/2, if both m and n are even
= 1 otherwise
If g is continuous on [a, b] and f_{1} (x) and f_{2} (x) are differentiable functions whose values lie in [a, b], then d/dx ∫ f_{2}(x) f_{1}(x) g(t) dt = g (f_{2}(x)) f_{2}'(x) – g (f_{1}(x)) f_{1}'(x)
Express the given series in the form ∑ 1/n f (r/n)
Then the limit is its sum when n→∞, i.e. lim n→∞ h ∑1/n f(r/n)
Replace r/n by x and 1/n by dx and lim n→∞ ∑ by the sign of ∫.
The lower and the upper limit of integration are the limiting values of r/n for the first and the last term of r respectively.
The definite integral can be interpreted to represent the area under the graph. This follows from the definition itself that the definite integral is a sum of the product of the lengths of intervals and the "height" of the function being integrated in that interval including the formula for the area of the rectangle. The figure given below clearly illustrates that as we find the value of the definite integral a to b, we are actually finding the area of the shaded portion.
1-1/2 + 1/3 – 1/4 + 1/5 – ….. ∞ = ln 2
1/1^{2 }+ 1/2^{2} + 1/3^{2} + 1/4^{2} + …. ∞ = π^{2}/6
1/1^{2} - 1/2^{2} + 1/3^{2} - 1/4^{2} + …. ∞ = π^{2}/12
1/1^{2} + 1/3^{2} + 1/5^{2} + 1/7^{2} + …. ∞ = π^{2}/8
1/2^{2} + 1/4^{2} + 1/6^{2} + 1/8^{2} + …. ∞ = π^{2}/24
Given that F(x) = ∫ f(t) dt, where integral runs from 0 to x.
By Leibnitz rule, we have
F’(x) = f(x)
But F(x^{2}) = x^{2} (1+x) = x^{2} + x^{3}
Hence, this gives F(x) = x + x^{3/2}
So, F’(x) = 1+ 3/2 x^{1/2}
Hence, f(x) = F’(x) = 1+ 3/2 x^{1/2}
Hence, f(4) = 1+ 3/2 . (4)^{1/2}
This gives f(4) = 1+ 3/2 . 2 = 4
Hence, option (3) is correct.
________________________________________________________________________________
F(x) = ∫√2-t^{2 }dt, where the integral runs from 1 to x
Hence, f’(x) = √2-x^{2}
Another condition that is given in the question is that x^{2} – f’(x) = 0
Therefore, x^{2 }= √2-x^{2}
Hence, x^{4 }= 2-x^{2}
So, x^{4 }+ x^{2} – 2 = 0
This gives the value of x as ±1.
Q1: In Leibnitz rule,
(a) the function f is continuous on [a, b] and the functions g_{1} and g_{2} are also continuous.
(b) the function f is discontinuous on [a, b] and the functions g_{1} and g_{2} are differentiable in [a, b].
(c) the function f is continuous on [a, b] and the functions g_{1} and g_{2} are discontinuous.
(d) the function f is continuous on [a, b] and the functions g_{1} and g_{2} are differentiable in [a, b].
Q2. If the upper and the lower limits are interchanged, the value of the integral
(a) remains same
(b) may or may not change
(c) value is same but the sign changes
(d) None of these
Q3.
(a) always holds good
(b) true provided f is piecewise continuous
(c) true if f is differentiable
(d) never true
Q4.
(a) f(2a-x) = -f(x)
(b) f(2a-x) = f(x)
(c) f(2a-x) = 0
(d) f(2a-x) = f(a+x)
Q5. Once definite integral is computed, the final result
(a) is a definite value without a constant (may be finite or infinite).
(b) may contain a constant
(c) always contains a constant
(d) contains a finite value along with a constant
Q1(d), Q2( c), Q3(b), Q4(a), Q5(a).
Click here to refer the most Useful Books of Mathematics.
For getting an idea of the type of questions asked, refer the previous year papers.
You might like to refer Indefinite Integral.
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