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The Reimann integral was the first ever accurate definition of the integral of a function on an interval and was discovered by a German Mathematician named Bernhard Reimann in 1854. The Reimann integral was put forward for the analysis in the study of functions for real variables. Now, the Reiman integral basically deals with real values, real variables, and real valued functions. In addition to the study of analytic properties of sequences, it also deals with the study of limits and the convergence of real number series and sequences. The Reimann integral can be calculated by the basic theorem of Calculus or approximated by numerical integration for many other practical applications and functions.
Table of Contents |
Key Takeaways: Reimann integral, Indefinite integral, Calculus, Trigonometric series, Partial fraction, Variable, Derivative, Probability
What is Riemann Integral?
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Reimann integral in simple terms can be stated as the limit of the Reimann sums of a function as the partitions get finer. If the limit ascends then the function is called integrable or, more precisely, Reimann-integrable.
Assume that f(x) is a regular and a non-negative function which has a range of a, ba, b. The area between f and x-axis expresses the integral of ‘f’ with respect to ‘x’. This type of integral is called a definite integral for the given function ‘f’ in the closed interval a, ba, b. And if the function is almost constant in all the sub-intervals, the Riemann Integral works in such cases.
Now, let’s understand it step by step:
- If the rectangle formed is extremely narrow, the approximation error will become nearly negligible.
- The value procured from the measurement of the areas of the narrow rectangles may be the same as the value of the area under the curve ‘f’.
- This process can be divided if we make larger rectangles.
- Hence, to make this method more convincing, the width of the rectangle can be obtained by dividing the interval a, ba, b into smaller portions.
Riemann integral
Riemann Sums
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Before moving forward with Reimann sums, some of the important terms that need to be understood are:
- Partition:
Assume (a, b) belongs to a close interval. The partition of this interval (a, b) can have the sequence of this form:
a = x0 < x1<……………<xn
where every [xi, xi+1] is known as a sub-interval.
- Norm: Norm is also known as mesh. It is expressed as the length of the biggest sub-interval.
- Tagged partition:
Let (a, b) belong to a close interval. Then partition q (x, t) along with the numbers t0, t1……….tn-1 is said as a tagged portion. It needs to fulfil the condition that for every i, ti ∈ [xi, xi+1].
- Riemann Sum
Riemann sum can be expressed as the total sum of real valued function ‘f’ in the given interval a, ba, b with respect to the tagged partition of a, ba, b. The formula for Reimann sum is as given;
\(\sum ^{n-1}_{i=0} f(t_i)(x_{i=1} - x_1)\)
Each term in the formula is the area of the rectangle with the length per height as f(ti) and breadth as xi+1 – xi. So, the final value of the Reimann sum is the sum of areas of all the rectangles which is the area covered under the curve of f(x) within the closed interval a, ba, b.
Riemann Integral Formula
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Let ‘f’ be a real valued function over the interval (a, b) and let us assume ‘L’ be a real number. Then, ‘f’ is called integrable within closed interval (a, b) if and only if there is a Δ>0 for each ∈>0 such that for each partition having a property that ||P|| < Δ , we have here;
|S (f, P) - L| <∈
Where, ‘L’ is called as the integral of ‘f’ over the interval (a, b), which is expressed as below:
L = \(\int_a^b\)f (x) dx
Riemann Integral Properties
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Reimann integral possess three very important properties which are:
- Linearity
- Monotonicity
- Addivity
- Linearity:
If f: a, b →R is integrable
And there exists a c ∈ R, now we can say that cf is also integrable such that,
\(\int_a^b\)cf =c \(\int_a^b\) f and, \(\int_a^b\) f + g= \(\int_a^b\) f + \(\int_a^b\) g
- Monotonicity:
If a condition is satisfied such that f<= g then,
\(\int_a^b\)f≤ abg is also satisfied.
- Addivity:
If a condition is satisfied such that a< c < b then,
\(\int_c^a\)f + \(\int_c^b\)f = \(\int_a^b\)f is also satisfied.
Applications Of Riemann Integral
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The Riemann Integral finds application in many fields, such as:
- It is used in integration and also used in differential calculus,
- It is used in partial differential equations and in representing functions by trigonometric series,
- It is used in measuring distance travelled by some body, as we can easily calculate the average velocity of the journey and total time by the velocity versus time graph. The distance so travelled is expressed by the area under the given curve.
Though having major applications, the Reimann integral is quite challenging to cope with as the definition is a bit sophisticated. So, it is considered a bit inconvenient to make use of the Reimann integral in practical life.
Things To Remember
- The Reimann integrals can be computed only for proper integrals.
- There are two types of integrals: definite integral also known as Reimann integral and indefinite integral also known as antiderivative.
- The definite integral (Riemann Integral) is used in solving many interesting problems in the fields like economics, finance, probability and more.
- There are three prominent methods of integration:
- Integration by substitution
- Integration by Partial fractions
- Integration by Parts
- The value of a definite integral of a function over any specific interval depends on the ‘function’ and ‘the interval’, but not on the variable of integration that one chooses to express the independent variable.
Sample Questions
Ques. State the Applications of Reimann integral? (4 marks)
Ans: There are various applications of Riemann Integral in multiple fields in our routine life. Mostly, they are used in Physics and Mathematical problems:
- Application in integral calculus and differential calculus.
- It is used in partial differential equations
- It is commonly used in Trigonometric series as well.
- Measuring the distance covered by an object as the distance covered can be found by summing up the area under the graph of time and velocity.
Ques. Prove that two indefinite integrals with the equal derivative led to the same family of curves and so they are equivalent. (3 marks)
Ans: Let a nad b be two functions such tha
\(\frac{d}{dx}\)∫ a(x)dx = \(\frac{d}{dx}\)∫ b(x)dx
\(\frac{d}{dx}\) [ ∫a(x)dx = ∫ b(x)dx] = 0
Hence, ∫a(x)dx = ∫ b(x)dx = C ---------- C is any real number.
∫a(x)dx = ∫ b(x)dx + C
So, the families of the curve are given as { ∫a(x)dx + C1, C1 ∈ R} and { ∫b(x)dx + C2, C2 ∈ R} are similar.
Ques. Give the antiderivative for each of the following functions: (3 marks)
a) Cos 2x
b) 3x2 + 4x3
c) 1/x, x≠0
Ans: The antiderivatives for the given functions are:
- Cos 2x
Cos 2x = \(\frac{1}{2} \frac{d}{dx}\)(sin sin 2x) = \(\frac{d}{dx}\) (\(\frac{1}{2} \)sin sin 2x)
Hence, the antiderivative of Cos 2x is \(\frac{1}{2} \) sin sin 2x.
- 3x2 + 4x3
3x2 + 4x3 => \(\frac{d}{dx}\) (x3 + x4) = 3x2 + 4x3
Hence, the antiderivative for 3X2 + 4X3 is x3 + x4
- \(\frac{1}{x}\), x ≠ 0
⇒ \(\frac{d}{dx}\) (log log x) = \(\frac{1}{x}\), x > 0
\(\frac{d}{dx}\) [ log log ( – x) ] = – \(\frac{1}{x}\)( – 1) = \(\frac{1}{x}\), x < 0.
combining both, we get \(\frac{d}{dx}\) (log log |x|) = \(\frac{1}{x}\)
Hence, ∫ \(\frac{1}{x} dx\) = log |x| is one of the antiderivatives of \(\frac{1}{x}\).
Ques. Find the integral of ∫sin3x cos2x dx. (3 marks)
Ans: ∫sin 3x cos2x dx = ∫sin 2x cos2x (sin x) dx
= ∫(1- cos2 x) cos2 x (sin x) dx
Now substitute, t = cos x so that dt = - sin x dx
Hence, ∫sin 2x cos2x (sin x) dx = – ∫(1-t2) t2 dt
= – ∫(t2 – t4) dt = – ( \(\frac{t^3}{3} \)- \(\frac{t^5}{5} \)) + c
= – 13cos3x +15cos5x+c.
Hence, the integral of ∫sin3x cos2 x dx is – \(\frac{1}{3} \)cos3x + \(\frac{1}{5} \)cos5x+c.
Ques. Find the following Integrals: (5 marks)
a) ∫\(\frac{dx}{x^2-16} \)
b) ∫\(\frac{dx}{2x-x^2} \)
Ans: The integrals can be found as:
- \(\int \frac{dx}{x^2 - 16} = \int \frac{dx}{x^2 - 4^2} = \frac{1}{8} log |\frac {x-4}{x+4}| = c.\)
- \(\frac{dx}{\sqrt{2x -x^2}} = \frac{dx}{\sqrt{1}- (x -1)^2}\)
put x – 1 = t, then dx = dt.
Hence, \(\frac{dx}{\sqrt{2x -x^2}} = \frac{dx}{\sqrt{1 - t^2}}\) = sin -1 (t) + c
= sin-1 (x – 1) + c.
Ques. Find: (3 marks)
a) ∫log x dx
b) ∫x ex dx
Ans:
- ∫(log x.1) dx = log x ∫1 dx - ∫ [\(\frac{d}{dx}\)(log logx) ∫1 dx ] dx
= (log x).x - ∫\(\frac{1}{x}\)dx = xlog logx – x + c.
- ∫x ex dx = x ex - ∫1. ex dx = x ex - ex +c.
Ques. Find ∫\(\frac{xSinx^1}{\sqrt{1 - x^2}}dx \) . (3 marks)
Ans: \(\int \frac{xSin^{-1}x}{\sqrt{1 - x^2}}dx \) => integral of the function \(\int \frac{x}{\sqrt{1 - x^2}}dx \)
put t = 1 – x2. Then dt = – 2x dx.
Hence, \(\int \frac{x}{\sqrt{1 - x^2}}dx \) = \(\frac{1}{2} \int \frac{dt}{\sqrt{t}}\) = \(\sqrt{t} = \sqrt{1 - x^2}\)
Hence,
\(\int \frac{xSin^{-1}x}{\sqrt{1 - x^2}}dx \) = (sin-1x) \(( - \sqrt{1-x^2})\) – \(\int \frac{x}{\sqrt{1 - x^2}} (- \sqrt{1-x^2})dx \)
= – \(\sqrt{1 - x^2}\)sin -1 x + x + c = x – \(\sqrt{1 - x^2}\) sin -1 x + c.
Ques. Find: ∫\(\sqrt{3-2x-x^2dx}\). (3 marks)
Ans: ∫\(\sqrt{3-2x-x^2dx}\) = ∫\(\sqrt{4-(x+1)^2dx}\)
put x+1 = y so that dx=dy.
∫\(\sqrt{3-2x-x^2dx}\)= ∫\(\sqrt{4-y^2dy}\)
= \(\frac{1}{2}\)y\(\sqrt{4-y^2}\) + \(\frac{4}{2}\) sin-1\(\frac{y}{2}\) + c
= \(\frac{1}{2}\)(x+1)\(\sqrt{3-2x-x^2dx}\) + 2 sin-1(\(\frac{x+1}{2}\)) + c.
Ques. Evaluate the following integrals: (5 marks)
a) 32∫x2dx
b) ∫cos 6x\(\sqrt{1 + sin6x dx}\)
Ans: a) let I = \(\int _2^3 x^dx, \int _2^3 x^dx = \frac{x^3}{3} =F(x)\)
I = F (3) – F (2) = \(\frac{27}{3} - \frac{8}{3} = \frac{19}{3}.\)
b) ∫ cos 6x \(\sqrt{ 1 + sin6xdx} = \frac {1}{6} \int t ^{\frac{1}{2}} dt\)
= \(\frac{1}{6} \times \frac{2}{3} (t) ^{\frac{3}{2}} + C = \frac{1}{9}\) (1 + sin sin6x) \(^{\frac{3}{2}}\) + C.
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