Riemann Sums and Riemann Integral MCQ Quiz in বাংলা - Objective Question with Answer for Riemann Sums and Riemann Integral - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

mi = \(\rm\displaystyle\inf _{x \in\left[x_{i−1}, x_i\right]}\) f(x)

Mi = \(\rm\displaystyle\sup_{x \in\left[x_{i−1}, x_i\right]}\) f(x)

then L(f, Pn) = \(\rm\displaystyle\sum_{i = 1}^n\) mi Δxi, U(f, Pn) = \(\rm\displaystyle\sum_{i = 1}^n\) Mi Δxi

Here, I is divided into n equal parts.

⇒ Δxi = xi − xi−1 = \(\rm\frac{1}{n}\) and f(x) = x1

Now, mi = \(\rm\displaystyle\inf _{x \in\left[x_{i−1}, x_i\right]}\) f(x) = \(\rm\displaystyle\inf \left(\frac{i−1}{n}, \frac{i}{n}\right) = \frac{i−1}{n}\)

and Mi = \(\rm\displaystyle\sup _{x \in\left[x_{i−1}, x_i\right]}\) f(x) = \(\rm\displaystyle\sup \left(\frac{i−1}{n}, \frac{i}{n}\right) = \frac{i}{n}\)

then L(f, Pn) = \(\rm\displaystyle\sum_{i = 1}^n m_i \Delta x_i = \frac{0}{n} \cdot \frac{1}{n}+\frac{1}{n} \cdot \frac{1}{n}+ \ldots \frac{1}{n} \frac{n−1}{n}\)

⇒ L(f, Pn) = \(\rm\frac{1}{n^2}[1+2+\cdots+(n−1)] = \frac{1}{n^2} \frac{(n−1)(n−1+1)}{2} = \frac{n−1}{2 n}\)

and U(P, Pn) = \(\rm\displaystyle\sum_{i = 1}^n M_i \Delta x_i = \frac{1}{n} \cdot \frac{1}{n}+\cdots+\frac{1}{n} \cdot \frac{n}{n} = \frac{1}{n^2}[1+2+\cdots+n] = \frac{n+1}{2 n}\)

So, An = U(f, Pn) − L(f, Pn) = \(\rm\frac{n+1}{2 n}−\frac{n−1}{2 n} = \frac{2}{2 n} = \frac{1}{n}\) (is strictly monotonically)

then \(\rm\displaystyle\lim_{n \rightarrow \infty} n \cdot A_n = \lim _{n \rightarrow \infty} n \cdot \frac{1}{n}\) = 1 and \(\rm\sum \frac{1}{n}\) is divergent (by p-test)

\(\rm\displaystyle\sum_{n=1}^{\infty}\) An An+1 = \(\rm\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\) and Sm = \(\rm\displaystyle\sum_{k=1}^m \frac{1}{k(k+1)}\)

⇒ {Sm} = \(\rm\left\{1-\frac{1}{m+1}\right\}\) and \(\rm\displaystyle\lim _{n \rightarrow \infty} S_n=\lim _{n \rightarrow \infty}\left(1-\frac{1}{n+1}\right)=1=\sum_{n=1}^{\infty} A_n A_{n+1}\)

opt (3), (4) correct.

Concept:

Partition set of closed interval: Let f be a real-valued function defined on the closed interval [a, b].

Therefore the partition set of [a, b] is given by P = {a = x₀, x₁, x₂, ...., x_{r - 1}, x_r, ...., x_n = b}

where a = x₀ < x₁ < x₂ < .... < x_{r - 1} < x_r < ... x_n = b.

And [x₀, x₁], [x₁, x₂], .....[x_{r - 1}, x_r],.....[x_{n - 1}, x_n] are called segments of partition.

Length of Sub-Interval:

Let [a, b] is a closed interval and P is the partition set of [a, b] given by P = { a = x₀, x₁, x₂,....,x_{r - 1}, x_r,... x_n = b}. Therefore the length of the sub-interval is given by δ_r  = x_r - x_{r - 1}

And the greatest of the lengths of subintervals is called the norm of P which is denoted by || P ||.

Norm of P = || P || = max {δ_r : r = 1, 2, ...n}

The upper and lower Riemann sums are given by

U(f, P) = \( \displaystyle\sum_{r=1}^{n} M_r.δ_r\) and L(f, P) = \( \displaystyle\sum_{r=1}^{n} m_r.δ_r\)

where M_r = Supremum of f in [x_{r - 1}, x_r] and m_r = Infimum of f in [x_{r - 1}, x_r].

Calculation:

Let f(x) is monotonic on [a, b] we want to show that f ∈ R [a, b]

Let P = { a = x0, x1, x2,....,xr - 1, xr,... xn = b} be the partition of [a, b] with a norm of partition μ(P) < δ 

Case I: If f(x) is monotonic increasing in [a, b]

⇒ f(x) is monotonic increasing in r^th sub-interval [x_{r - 1}, x_r] in P of [a, b].

Let M_r and m_r be the l.u.b. and g.l.b. of f(x) in [x_{r - 1}, x_r] of P of [a, b].

δ_r be the length of r^th sub-interval.

Let f(x_r) = M_r and f(x_{r - 1}) = m_r

Now, U(P) - L(P) = \( \displaystyle\sum_{r=1}^{n} M_r.δ_r \ - \ \displaystyle\sum_{r=1}^{n} m_r.δ_r \)

If μ(P) < δ  ⇒ δ_r  < δ 

⇒ U(P) - L(P) < \( \displaystyle\sum_{r=1}^{n} M_r.δ_r \ - \ \displaystyle\sum_{r=1}^{n} m_r.δ_r < \displaystyle\sum_{r=1}^{n} (M_r - m_r).δ \)

⇒ U(P) - L(P) < \( δ \displaystyle\sum_{r=1}^{n} { f(x_r) - f(x _{r - 1})} \)

Putting the value of r

⇒ U(P) - L(P) < \( δ [ f(x_1) - f(x_0) + f(x_2) - f(x_1) + ... + f(x_n) - f(x_{n - 1})]\)

⇒ U(P) - L(P) < δ {f(x_n) - f(x₀)} < δ {f(b) - f(a)}

Taking δ = \(\frac{∈}{ f(b) \ - \ f(a) }\), we get

⇒ U(P) - L(P) < ∈ 

⇒ f ∈ R[a, b]

Case II: If f(x) is monotonically decreasing in [a, b].

f(x) is monotonically decreasing in r^th sub-interval.

Now, Let f(x_{r - 1}) = M_r and f(x_r) = m_r

U(P) - L(P) = \( \displaystyle\sum_{r=1}^{n} M_r.δ_r \ - \ \displaystyle\sum_{r=1}^{n} m_r.δ_r \) = \( \displaystyle\sum_{r=1}^{n} (M_r - m_r).δ_r \)

⇒ μ(P) < δ ⇒ δ_r < δ 

⇒ U(P) - L(P) < \( \displaystyle\sum_{r=1}^{n} (M_r - m_r).δ\) < \( δ \displaystyle\sum_{r=1}^{n} { f(x_{r - 1}) - f(x _r)} \)

⇒ U(P) - L(P) < δ{ f(x_o) - f(x₁) + f(x₁) - f(x₂) + .... + f(x_{n - 1}) - f(x_n)}

⇒ U(P) - L(P) < δ{f(x_o) - f(x_n)} < δ {f(a) - f(b)}

Taking δ = \( \frac{∈}{f(a) \ - \ f(b)}\), we get

⇒ U(P) - L(P) < ∈ 

⇒ f(x) ∈ R[a, b]

We can say that every monotonic function is R- integral.

Concept:

Partition: By partition of [a, b], we mean a finite set P of points x1, x2, ....., xn Where,

a = x0 ≤ x1 ≤ ....≤ xn-1 ≤ xn = b

Darboux sum: Suppose f: [a, b] → R is bounded and P is  a partition of [a, b], then the lower Darboux sum is:

L(f, p) = \(\sum_{i=1}^{n}m_i(x_i\ -\ x_{i-1})\)

where, m_i = inf {f(x) | x ϵ [x_i-1, x_i]}

And upper darboux sum is

U(f, p) = \(\sum_{i=1}^{n}M_i(x_i\ -\ x_{i-1})\)

where, M_i = sup {f(x) | x ϵ [xi-1, xi]}

Note: m(b − a) ≤  L(f, P) ≤  U(f, P) ≤  M(b − a) 

Riemann integral: A bounded function f: [a, b] → R is said to be Riemann integral over [a, b] if 

\(\int_{a}^{\bar{b}}f(x)dx\ =\ \int_{\bar{a}}^{b}f(x)dx\)

and f is denoted by \( \int_{a}^{b}f(x)dx\)

1. The upper integral of f on [a, b]:

\(U\int_{a}^{b}f(x)dx\ =\ inf\ [U(f, p)]\)

2. The upper integral of f on [a, b]:

\(L\int_{a}^{b}f(x)dx\ =\ sup\ [L(f, p)]\)

Where P is a partition of [a, b].

Calculation:

Given that, 

\(\rm g(x) = \left \{ \begin{matrix} \rm cos^2\ x & \rm x \ ϵ\ Q \\ \rm 0 & \rm \ otherwise \end{matrix}\right.\)

and  g ∶ [0, \(\frac{\pi}{2}\)] → R by g(x)

Explanation: 

Given  \(f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right), & {if }\quad x \in (0,1], \\ 0, & {if } \quad x = 0. \end{cases} \)

1. f(x) is continuous on (0,1] , as  \(x^2 \sin\left(\frac{1}{x}\right) \)  is well-defined for \(x \neq 0 \)

2. At x = 0 , f(0) = 0

To check continuity at x = 0 :

\(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 \sin\left(\frac{1}{x}\right) \) 

Since  \( |\sin\left(\frac{1}{x}\right)| \leq 1 \) , we have:

\(|f(x)| \leq x^2 \quad {and } \lim_{x \to 0^+} x^2 = 0 \) 

Thus, f(x) is continuous at x = 0 

Riemann Integrability : 

To show that f(x) is Riemann integrable on [0,1], recall the sufficient condition:

If f(x) is bounded and continuous almost everywhere (or has a finite number of discontinuities),

it is Riemann integrable.

Since f(x) is continuous on [0,1], it is Riemann integrable

Hence Option(1) is the correct answer.

Explanation:

This integral is known as the Gamma function Γ(n):

Γ(n) = \( ∫_{0}^{∞} e^{-x} x^{n-1} dx \)

The Gamma function converges for n > 0

Therefore, The integral is convergent when n ≥ 1

Also, When n > 0, the integral converges

When n ≤ 0, the integral diverges

Hence Option(4) is the correct answer.

Concept:

(i) Every continuous function is Reimann integrable

(ii) If a function has a countable number of discontinuity on [a, b] then f(x) is Riemann integrable on [a, b]

Explanation:

(1): \(\rm f(x)= \begin{cases}1 & \text { if } x \in \mathbb{Q} \cap[0,1] \\ -1 & \text { otherwise }\end{cases} \)

f(x) has infinite number of discontinuity,

so it not not Reimann integrable on [0, 1]

(2): \(\rm f(x)= \begin{cases}x & \text { if } x \in[0,1) \\ 0 & \text { if } x=1\end{cases} \)

\(\lim_{x\to1}f(x)\) = \(\lim_{x\to1}x\) = 1 ≠ f(1)

f(x) is not continuous at x = 1

f has a countable number of discontinuity on [0, 1]

f(x) is Reimann integrable on [0, 1]

(3): \(\rm f(x)=\displaystyle\int_0^x\left|\frac{1}{2}-t\right| d t\)

= \(\displaystyle-\int_0^{\frac12}(\frac{1}{2}-t)d t+\displaystyle\int_{\frac12}^x(\frac{1}{2}-t)d t\) which will give a continuous function

So f(x) is Reimann integrable on [0, 1]

(4): \(\rm f(x)= \begin{cases}x^2 \sin (1 / x) & \text { if } x ≠ 0 \\ 0 & \text { if } x=0\end{cases} \)

\(\lim_{x\to0}f(x)\) = \(\lim_{x\to0}x^2 \sin(1/x)\) = 0 = f(0)

so f(x) is continuous and so Reimann integrable on [0, 1] 

Hence Option(1) is the correct answer.

Explanation:

f is integrable on [a, b] iff for every ϵ > 0 there exists a partition P such that U(P, f) - L(P, f) < ϵ.

(4) is correct

Concept:

Partition: By partition of [a, b], we mean a finite set P of points x₁, x₂, ....., x_n Where,

a = x₀ ≤ x₁ ≤ ....≤ x_n-1 ≤ x_n = b

Riemann integral: A bounded function f: [a, b] → R is said to be Riemann integral over [a, b] if 

\(\int_{a}^{\bar{b}}f(x)dx\ =\ \int_{\bar{a}}^{b}f(x)dx\)

and f is denoted by \( \int_{a}^{b}f(x)dx\)

1. The upper integral of f on [a, b]:

\(U\int_{a}^{b}f(x)dx\ =\ inf\ [U(f, p)]\)

2. The upper integral of f on [a, b]:

\(L\int_{a}^{b}f(x)dx\ =\ sup\ [L(f, p)]\)

Where P is a partition of [a, b].

Conditions of integrability:

• A bounded function is said to be integrable when upper and lower integrals are equal.
• Riemann’s criterion for integrability: f is integrable on [a, b] ⇔ for every ϵ > 0 there exists a partition P such that U(P, f) − L(P, f) < ϵ 
• If f is continuous on [a, b] then f is integrable.
• If f is a monotone function then f is integrable.

Calculation:

From the above discussion, we can say that if upper and lower integrals are the same then integral is said to be Riemann integrable i.e.,

\( \int_{a}^{b}f(x)dx\ =\ U\int_{a}^{b}f(x)dx\ =\ L\int_{a}^{b}f(x)dx\)

A necessary condition for providing the upper and lower integrals of a bounded function exists if and only if

a, b ϵ R, a < b, and f : [a, b] → R to be bounded.

Hence, the correct answer is option 2).

Explanation :

Let us assume function f is increasing function on [a, b]

Now for a given positive number ε; there exists a partition 

P = {a = x₀, x₁, x₂, xn = b} of [a, b] such that

(xᵣ - xᵣ₋₁) < ε / [f(b) - f(a) + 1] for r = 1, 2, , n

Since, the function f being increasing on [a, b] it is bounded

and increasing on each sub interval [xᵣ₋₁, xᵣ]

then, Mᵣ = f(xᵣ) and  mᵣ = f(xᵣ₋₁)

Hence, for this partition P, we have

U(P, f) - L(P, f) = Σ(Mᵣ - mᵣ)(xᵣ - xᵣ₋₁) < ε / [f(b) - f(a) + 1] Σ |f(xᵣ) - f(xᵣ₋₁)|

which implies

U(P, f) - L(P, f)  < ε / [f(b) - f(a) + 1] [f(xₙ) - f(x₀)]

which implies U(P, f) - L(P, f)  < ε / [f(b) - f(a) + 1] [f(b) - f(a)] [Since x₀ = a, xₙ = b]

thus, U(P, f) - L(P, f) < ε

that is, the function f is Riemann integrable on [a, b]. 

⇒ Statement B is correct

Hence Option(3) is the correct answer

Explanation:

1.\(​f(x) = x^2 \sin\left(\frac{1}{x}\right) for x \in (0,1] \) , with f(0) = 0 .

f is bounded and continuous on (0,1], and the discontinuity at x = 0 is removable because:

\(\lim_{x \to 0^+} f(x) = 0 = f(0) \)

Since f is continuous everywhere after removing the discontinuity, f is Riemann integrable on [0,1].  

Statement 1 is correct.

2. f(x) is the Dirichlet function.

For the Dirichlet function,\( f(x)^2 = f(x) \) , because f(x) takes values only in \(\{0,1\} \).  

The function \(f(x)^2 \) is identical to f(x) ,

which is not Riemann integrable because the set of discontinuities (all of [0,1]) is not of measure zero.  

Statement 2 is incorrect.

3. If \(\int_0^1 |f(x)| \, dx = 0 \) , then f(x) = 0 almost everywhere.

The integral of |f(x)| being zero implies that the measure of the set where |f(x)| > 0 is zero.  

Thus, f(x) = 0 almost everywhere on [0,1].  

Statement 3 is correct. 

4. If f has a single removable discontinuity on [0,1], then f is Riemann integrable.

A function with a finite number of discontinuities (including removable ones) is Riemann integrable

because the set of discontinuities has measure zero.  

Statement 4 is correct.

Hence Option(1) , (3) and (4) are correct.