Riemann Sums and Riemann Integral MCQ Quiz - Objective Question with Answer for Riemann Sums and Riemann Integral - Download Free PDF

Concept:

• Monotonic Function: A function \( f:[0,1] \to \mathbb{R} \) is monotonic if it is either non-increasing or non-decreasing throughout the interval.
• Riemann Integrability: A function is Riemann integrable on a closed interval if the set of its discontinuities has measure zero.
• Discontinuities of Monotonic Functions: A monotonic function can only have jump discontinuities, and the set of such points is at most countable. Therefore, it cannot contain any non-empty open interval.
• Lebesgue Measurability: A function is Lebesgue measurable if the preimage of any Borel set is a Lebesgue measurable set. Monotonic functions are always Lebesgue measurable.
• Borel Measurability: A function is Borel measurable if the preimage of any Borel set lies in the Borel sigma-algebra. All monotonic functions are Borel measurable.

Calculation:

Given,

\( f:[0,1] \to \mathbb{R} \) is a monotonic function

Step 1: Check Riemann integrability

⇒ Monotonic functions have at most countably many discontinuities

⇒ Such sets have measure zero

⇒ Riemann integrable

Step 2: Discontinuities and open sets

⇒ Discontinuity set of monotonic function is at most countable

⇒ Countable set cannot contain non-empty open set

⇒ Statement 2 is TRUE

Step 3: Lebesgue measurability

⇒ Every monotonic function is Lebesgue measurable

⇒ Statement 3 is TRUE

Step 4: Borel measurability

⇒ Monotonic functions are Borel measurable

⇒Statement 4 is TRUE

∴ All four statements are TRUE.

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:

1. \(f(x)=\left\{\begin{array}{ll} 1 & \text { if } x=0, \\ \frac{1}{n} & \text { if } x=\frac{m}{n} \text { for some } m, n ∈ \mathbb{N} \text { with } m \leq n \text { and } \operatorname{gcd}(m, n)=1, \\ 0 & \text { if } x ∈[0,1] \text { is irrational. } \end{array}\right.\)

Since f is discontinuous at all rational

⇒ f has countable discontinuity 

⇒ f is Riemann Integrable 

⇒ Option(1) is correct 

2. \(g(x)=\left\{\begin{array}{ll} 0 & \text { if } x=0 \\ 1 & \text { if } x ∈(0,1] \end{array}\right.\)

g is discontinuous only at x = 0 

⇒ g is also Riemann  integrable 

⇒Option(2) is correct 

3. Since f(g(x)) = 1

f(g(x)) = f ∘ g (x) is also Riemann integrable 

⇒ Option(3) is correct

4. g ∘ f(x) = \(\left\{\begin{array}{ll} 0 & \text { if } f(x) =0 \\ 1 & \text { if } f(x) ∈(0,1] \end{array}\right. \) = \(\left\{\begin{array}{ll} 0 & \text { if } x ∈ \mathbb{Q}^c∩ [0, 1] \\ 1 & \text { if } x ∈ \mathbb{Q}∩ [0, 1] \end{array}\right. \)

⇒ g ∘ f has uncountable discontinuity 

⇒  g ∘ f  is not Riemann integrable 

⇒ Option(4) is not correct 

Hence Option(1) , Option(2) and Option(3)  are correct.

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.

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): d(f, f) = \(\int_0^1|f(x)-f(x)| d x \) = 0

(2): d(f, g) ≥ 0 satisfied 

(4): d(f, g) = d(g, f) & d(f, g) ≤ d(f, h) + d(h, g) satisfied as d satisfy symmetric property and triangular property.

(3): let \(f(x)=1 \)  if \(x \in [0,1]\)

      and  \(g(x)=\begin{cases}1, x\in [0, 1)\\2, x = 1\end{cases}\) 

then  \(f,g \in R[0,1] \) with  \(f \neq g\) 

 but    \(d(f,g)=\) \(\int_0^1|f(x)-g(x)| d x \)

                    \(=\int_0^1|1-1| d x\) = 0 

Here d(f, g) = 0 for f ≠ g

Therefore correct answer is option :3

Concept:

A bounded function on a compact interval [a, b] is Riemann integrable if and only if it is continuous almost everywhere.

Explanation:

 f: [a, b] → R, is continuous then f must be integrable on [a, b]

(3) is correct

Explanation:

In Reiman Integral, if f(x) is a function and P is a partition on a closed interval then U(P, f) is called upper Reimann sum, and L(P, f) is called lower Reimann sum.

(4) is correct

Explanation:

\(\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{k^2+n^2}\)

= \(\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2(1+\frac{k^2}{n^2})}\)

= \(\lim_{n\to\infty}\frac1n\sum_{k=0}^n\frac{1}{(1+\frac{k^2}{n^2})}\)

= \(\int_0^1\frac1{1+x^2}dx\)

= \([\tan^{-1}x]_0^1\)

= π/4

(4) is correct

Concept:

A function f is Riemann integral if and only if f(x) has countable points of discontinuity.

Explanation: 

f(x) = \(\begin{cases}x^2, x \text{ is rational}\\x^3, x \text{ is irrational}\end{cases}\) , x ∈ [0, 1]

So f is continuous only on 0, 1

Therefore f has an uncountable number of points of discontinuity in [0, 1]

Hence f is not Riemann integrable on [0, 1]

(3) is correct

Concept -

If the function \(f(x) = \begin{cases} h(x) & x \in [0,1] \cap Q \\ g(x) & x \in [0,1] \cap Q^c \\ \end{cases}\) then f(x) is continuous at those points which satisfied the equation h(x) = g(x). 

If function f(x) is not continuous at uncountable points then f(x) is not Riemann Integrable.

Explanation -

We have \(f(x) = \begin{cases} x & x \in [0,1] \cap Q \\ 1-x & x \in [0,1] \cap Q^c \\ \end{cases}\)

So f(x) is continuous only at x = 1 - x ⇒ 2x = 1 ⇒ x = 1/2

Hence function is continuous only at one point and discontinuous at uncountable( infinite points).

So function f(x) is not Riemann Integrable on [0,1].

Hence option(2) is correct.

Concept -

(1) Every monotone function has at most countable discontinuous points.

(2) If the function f(x) has countable discontinuous points then f(x) is Riemann Integrable. 

Explanation -

we have the function  f(x) = [x] and we all know the graph of the function given below

From the above graph -

Function f(x) is clearly discontinuous at integer point because the graph is broken at all the integer points. and it is also increasing function.

Hence f(x) is Riemann integrable 

So option(3) is correct.

As, \((x+x^2)-(x^2+x^3)=x-x^3\)

\(=x(2-x^2)\)

So, (x + x²)-(x²+x³)\(\left\{\begin{matrix} >0, \ \text{if} \ 0 < x < 1 \\ < 0, \ \text{if} \ 1 < x < 2 \end{matrix}\right.\)

if M, and m, be the upper and lower bounds of the given 

funtion f(x) in i_r respectively wherer I_r has its usual meaning,

then for all value of r, we have

Mr ₌ \(\left\{\begin{matrix} {x + x^2}, \ \text{if} \ 0 < x < 1 \\ {x^2+x^3}, \ \text{if} \ 1 < x < 2 \end{matrix}\right.\)

and let \(ϕ(x)=\frac{1}{x^{1-m}}\)

Hence \(\frac{fx}{ϕ(x)\rightarrow1} \rm as \)\(x \rightarrow 0\)

as \(\int^{-1/2}_0ϕ (x)dx=\int^{1/2}_0\frac{1}{x^1-m}dx\)

is convergent at 0, if and only if,

1 - m < 1

⇔ 0 < m

⇒ \(\int^{1/2}_0x^{m-1}(1-x)^{(n-1)}dx\)

is converget at 0, if and only if, m > 0

Convergence at 1 We have,

f'(x) = x^m-1(1-x)^n-1

\(=\frac{x^{m-1}}{(1-x)^{1-n}}\)

and let ϕ (x) = \(\frac{1}{1-x^{1-n}}\)

Hence \(\frac{f(x)}{ϕ(x)}\rightarrow1\) as \(x \rightarrow 1\)

as \(\int^1_{1/2} \phi (x) dx =\int^{1}_{1/2}\frac{1}{(1-x)^{1-n}}dx\)

is convergent, if and only if

1 - n < 1 ⇔ 0 < n

⇒ \(\int^1_{1/2}x^{m-1}(1-x)^{n-1}dx\)

converges if and only if

n > 0

Thus, \(\int^1_{x}x^{m-1}(1-x)^{n-1}dx\) exists for opsitive values of m, n only.

and m_r= \(\left\{\begin{matrix} x^2+x^3, \ \text{if} \ 0 < x < 1 \\ x+x^2, \ \text{if} \ 1 < x < 2 \end{matrix}\right.\)

Hence, the upper Riemann itegral is

\(= \int^1_0(x+x^2)dx+\int^2_1(x^2+x^3)dx\)

\(=\left(\frac{x^2}{2}+\frac{x^3}{3}\right)^1_0+\left(\frac{x^3}{3}+\frac{x^4}{4}\right)^2_1\)

\(=\left(\frac{1}{2}+\frac{1}{3}\right)+\left[\left\{\left(\frac{8}{3}\right)+\left(\frac{16}{4}\right)\right\}- \left(\frac{1}{3}+\frac{1}{4}\right)\right]\)

=\(\left(\frac{83}{12}\right)\)

and the lower Riemann intergral is

\(=\int^1_0(x^2+x^3)dx + \int^2_1(x+x^2)dx\)

\(=\left(\frac{x^3}{3}+\frac{x^4}{4}\right)^1_0+\left(\frac{x^2}{2}+\frac{x^3}{3}\right)^2_1\)

\(=\left[\left(\frac{1}{3}+\frac{1}{4}\right)+[\{\frac{4}{2}+\frac{8}{3}\}-\left(\frac{1}{2}+\frac{1}{3}\right)\right]\)

\(=\frac{53}{12}\)

Concept:

A function f(x) is Reimann integrable if and only if it has finite numbes of points of discontinuities

Explanation:

f(x) = \(\sum_{n=1}^{200}\frac{1}{n^6}\chi_{[0, \frac n{200}]}(x)\),  \(\chi_A(x)\) denotes the function which is 1 if x ∈ A and 0 otherwise

So f(x) = \(\begin{cases}\frac{1}{1^6}+\frac{1}{2^6}+...+\frac{1}{200^6}, 0\leq x\leq \frac{1}{200}\\ \frac{1}{2^6}+...+\frac{1}{200^6}, \frac{1}{200}\leq x\leq \frac{2}{200}\\...........................\\ \frac{1}{200^6}, \frac{199}{200}\leq x\leq \frac{200}{200}\end{cases}\)................((i)

So here we can see that f(x) is not continuous at all points \(\frac{n}{200}\), n = 1, 2, ..., 200

(3) is NOT correct

So here we can see that f(x) has a finite number of points of discontinuity.

Hence f(x) is Riemann integral on [0, 1]

(1) is correct

Every Riemann integral is Lebesgue integral

So (2) is correct

Also from (i) we can see that f(x) is monotonic

(4) is correct

Solution-

Let [a,b] be the given integral A Partition P of [a,b] is a finite set of points 

\(a = x_o \leq x_1\leq x_2... 0= b\)

\(L(P,f) = \sum_{i=1}^{n} m_i \Delta x_i\)

Here infimum in x and 0 is o so \(\frac{1}{q}\) at infimum is 0 

then, \(L(P,f) = 0\) 

Therefore, Correct Option is Option 3).